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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

Edited By Ramraj Saini | Updated on Sep 14, 2023 07:52 PM IST | #CBSE Class 12th

NCERT Continuity And Differentiability Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 5 are provided here. These NCERT solutions are created bu expert team at careers360 considering the latest syllabus of CBSE 2023-24. Questions based on the topics like continuity, differentiability, and relations between them are covered in the NCERT solutions for class 12 maths chapter 5. In NCERT Class 12 maths book, there are 48 solved examples to understand the concepts of continuity and differentiability class 12. If you are finding difficulties in solving them, you can take help from NCERT maths chapter 5 class 12 solutions.

In NCERT class 11 Maths solutions, you have already learned the differentiation of certain functions like polynomial functions and trigonometric functions. In this chapter, you will get NCERT solutions for class 12 maths chapter 5 continuity and differentiability. If you are interested in the chapter 5 class 12 maths NCERT solutions then you can check NCERT solutions for class 12 other subjects.

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Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae

>> Continuity: A function f(x) is continuous at a point x = a if:

  • f(a) exists (finite, definite, and real).

  • lim(x → a) f(x) exists.

  • lim(x → a) f(x) = f(a).

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
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>> Discontinuity: f(x) is discontinuous in an interval if it is discontinuous at any point in that interval.

Algebra of Continuous Functions:

Sum, difference, product, and quotient of continuous functions are continuous.

Differentiation:

The derivative of f(x) at x = a, denoted as f'(a), represents the slope of the tangent line to the graph.

Chain Rule:

If f = v o u, where t = u(x), and if both dt/dx and dv/dx exist, then: df/dx = dv/dt * dt/dx.

Derivatives of Some Standard Functions:

  • d/dx(xn) = nxn-1

  • d/dx(sin x) = cos x

  • d/dx(cos x) = -sin x

  • d/dx(tan x) = sec2 x

  • d/dx(cot x) = -csc2 x

  • d/dx(sec x) = sec x * tan x

  • d/dx(csc x) = -csc x * cot x

  • d/dx(ax) = ax * ln(a)

  • d/dx(ex) = ex

  • d/dx(ln x) = 1/x

JEE Main Important Physics formulas

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

Mean Value Theorem:

Mean Value Theorem states that if f(x) is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that: f'(c) = (f(b) - f(a)) / (b - a).

Rolle's Theorem:

Rolle's Theorem states that if f(x) is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists some c in (a, b) such that f'(c) = 0.

Lagrange's Mean Value Theorem:

Lagrange's Mean Value Theorem states that if f(x) is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that:

f'(c) = (f(b) - f(a)) / (b - a).

Free download Continuity And Differentiability Class 12 NCERT Solutions for CBSE Exam.

NCERT Continuity And Differentiability Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT Continuity And Differentiability Class 12 Solutions : Excercise: 5.1

Question:1 . Prove that the function f ( x) = 5 x -3 is continuous at x = 0, at\: \: x = - 3 and at x = 5

Answer:

Given function is
f ( x) = 5 x -3
f(0) = 5(0)-3 = -3
\lim_{x\rightarrow 0} f(x) = 5(0)-3 = -3
\lim_{x\rightarrow 0} f(x) =f(0)
Hence, function is continous at x = 0

f(-3)= 5(-3)-3=-15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = f(-3)
Hence, function is continous at x = -3

f(5)= 5(5)-3=25-3=22\\\Rightarrow \lim_{x\rightarrow 5} f(x) = 5(5)-3 = 25-3=-22\\ \Rightarrow \lim_{x\rightarrow 5} f(x) = f(5)
Hence, function is continuous at x = 5

Question:2 . Examine the continuity of the function f (x) = 2x ^2 - 1 \: \: at\: \: x = 3.

Answer:

Given function is
f(x) = 2x^2-1
at x = 3
f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\ \lim_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17
\lim_{x\rightarrow 3}f(x) = f(3)
Hence, function is continous at x = 3

Question:3 Examine the following functions for continuity.
(a) f (x) = x - 5

Answer:

Given function is
f(x) = x-5
Our function is defined for every real number say k
and value at x = k , f(k) = k-5
and also,
\lim_{x\rightarrow k} f(x) = k -5\\ \lim_{x\rightarrow k} f(x) = f(k)
Hence, the function f(x) = x-5 is continuous at every real number

Question:3 b) Examine the following functions for continuity.

f (x) = \frac{1}{x-5} , x \neq 5

Answer:

Given function is
f(x ) = \frac{1}{x-5}
For every real number k , k \neq 5
We get,
f(k) = \frac{1}{k-5}\\ \lim_{x\rightarrow k}f(x ) = \frac{1}{k-5}\\ \lim_{x\rightarrow k}f(x ) = f(k)
Hence, function f(x ) = \frac{1}{x-5} continuous for every real value of x, x \neq 5

Question:3 c) Examine the following functions for continuity.

f (x) = \frac{x ^2-25}{x+5}, x \neq -5

Answer:

Given function is
f(x ) = \frac{x^2-25}{x+5}
For every real number k , k \neq -5
We gwt,
f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = f(k)
Hence, function f(x ) = \frac{x^2-25}{x+5} continuous for every real value of x , x \neq -5

Question:3 d) Examine the following functions for continuity. f (x) = | x - 5|

Answer:

Given function is
f (x) = | x - 5|
for x > 5 , f(x) = x - 5
for x < 5 , f(x) = 5 - x
SO, different cases are their
case(i) x > 5
for every real number k > 5 , f(x) = x - 5 is defined
f(k) = k - 5\\ \lim_{x\rightarrow k }f(x) = k -5\\ \lim_{x\rightarrow k }f(x) = f(k)
Hence, function f(x) = x - 5 is continous for x > 5

case (ii) x < 5
for every real number k < 5 , f(x) = 5 - x is defined
f(k) = 5-k\\ \lim_{x\rightarrow k }f(x) = 5 -k\\ \lim_{x\rightarrow k }f(x) = f(k)
Hence, function f(x) = 5 - x is continous for x < 5

case(iii) x = 5
for x = 5 , f(x) = x - 5 is defined
f(5) = 5 - 5=0\\ \lim_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim_{x\rightarrow 5 }f(x) = f(5)
Hence, function f(x) = x - 5 is continous for x = 5

Hence, the function f (x) = | x - 5| is continuous for each and every real number

Question:4 . Prove that the function f (x) = x^n is continuous at x = n, where n is a positive integer

Answer:

GIven function is
f (x) = x^n
the function f (x) = x^n is defined for all positive integer, n
f(n) = n^n\\ \lim_{x\rightarrow n}f(x) = n^n\\ \lim_{x\rightarrow n}f(x) = f(n)
Hence, the function f (x) = x^n is continuous at x = n, where n is a positive integer

Question:5. Is the function f defined by
f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.
continuous at x = 0? At x = 1? At x = 2?

Answer:

Given function is
f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.
function is defined at x = 0 and its value is 0
f(0) = 0\\ \lim_{x\rightarrow 0}f(x) = f(x) = 0\\ \lim_{x\rightarrow 0}f(x) = f(0)
Hence , given function is continous at x = 0

given function is defined for x = 1
Now, for x = 1 Right-hand limit and left-hand limit are not equal
f(1) = 1\\ \lim_{x\rightarrow 1^-}f(x) = f(x) = 1\\ \lim_{x\rightarrow 1^+}f(x) =f(5) = 5
R.H.L \neq L.H.L.
Therefore, given function is not continous at x =1
Given function is defined for x = 2 and its value at x = 2 is 5
f(2) = 2\\ \lim_{x\rightarrow 2}f(x) = f(5) = 5\\\lim_{x\rightarrow 2}f(x) = f(2)
Hence, given function is continous at x = 2

Question:6. Find all points of discontinuity of f, where f is defined by

f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.

Answer:

Given function is
f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
f(k) = 2k-3\\ \lim_{x\rightarrow k}f(x) = 2k-3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
f(k) = 2k +3\\ \lim_{x\rightarrow k}f(x) = 2k+3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

\lim_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ \lim_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1
Right hand limit at x= 2 \neq Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

Question:7. Find all points of discontinuity of f, where f is defined by

f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.

Answer:

Given function is
f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.
GIven function is defined for every real number k
Different cases are their
case (i) k < -3
f(k) = -k + 3\\ \lim_{x\rightarrow k}f(x) = -k + 3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for every value of k < -3

case(ii) k = -3
f(-3) = -(-3) + 3 = 6\\ \lim_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\ \lim_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\ R.H.L. = L.H.L. = f(-3)
Hence, given function is continous for x = -3

case(iii) -3 < k < 3
f(k) = -2k \\ \lim_{x\rightarrow k}f(x) = -2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in -3 < k < 3 given function is continous

case(iv) k = 3
f(3) = 6x+2 = 6\times3+2 =18+2=20\\ \lim_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\ \lim_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\ R.H.L. = f(3) \neq L.H.L.
Hence . x = 3 is the point of discontinuity

case(v) k > 3
f(k) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each and every value of k > 3

Question:8. Find all points of discontinuity of f, where f is defined by

f (x )= \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.

Answer:

Given function is
f (x ) \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.
if x > 0 , f(x)=\frac{x}{x} = 1
if x < 0 , f(x)=\frac{-(x)}{x} = -1
given function is defined for every real number k
Now,
case(i) k < 0
f(k) = -1\\ \lim_{x\rightarrow k }f(x) = -1\\ \lim_{x\rightarrow k }f(x) = f(k)
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
f(k) = 1\\ \lim_{x\rightarrow k }f(x) = 1\\ \lim_{x\rightarrow k }f(x) = f(k)
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
f(0) = 0\\ \lim_{x\rightarrow 0^- }f(x) = -1\\ \lim_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.
Hence, 0 is the only point of discontinuity

Question:9. Find all points of discontinuity of f, where f is defined by

f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.

Answer:

Given function is
f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.
if x < 0 , f (x) =\frac{x }{|x|} = \frac{x}{-(x)} = -1
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for each and every value of x
Hence, no point of discontinuity

Question:10. Find all points of discontinuity of f, where f is defined by

f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.

Answer:

Given function is
f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
f(k) = k+1\\ \lim_{x\rightarrow k}f(x) = k+1\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
f(k) = k^2 ++1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

\lim_{x\rightarrow 1^-}f(x) = x^2+1 = 1^2 + 1 = 1 + 1 = 2\\ \lim_{x\rightarrow 1^+}f(x) = x+1 = 1+1 = 2\\ f(1) = 1^2+1 = 2 \\ R.H.L. = L.H.L. = f(1)

Hence, at x = 2 given function is continuous
Therefore, no point of discontinuity

Question:11. Find all points of discontinuity of f, where f is defined by

f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.

Answer:

Given function is
f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
f(k) = k^2+1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
f(k) = k^3 -3\\ \lim_{x\rightarrow k}f(x) = k^3-3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

\lim_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\ \lim_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.
Hence, given function is continuous at x = 2
There, no point of discontinuity

Question:12. Find all points of discontinuity of f, where f is defined by

f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.

Answer:

Given function is
f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
f(k) = k^2\\ \lim_{x\rightarrow k}f(x) = k^2\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
f(k) = k^{10} -1\\ \lim_{x\rightarrow k}f(x) = k^{10}-1\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

\lim_{x\rightarrow 1^-}f(x) = x^{10}-1 = 1^{10} - 1 = 1 - 1 = 0\\ \lim_{x\rightarrow 1^+}f(x) = x^2 = 1^2 = 1\\ f(1) = x^{10}-1 = 0\ f(1) = L.H.L. \neq R.H.L.

Hence, x = 1 is the point of discontinuity

Question:13. Is the function defined by

f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.

a continuous function?

Answer:

Given function is
f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
f(k) = k-5\\ \lim_{x\rightarrow k}f(x) = k-5\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
f(k) = k+5\\ \lim_{x\rightarrow k}f(x) = k+5\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

\lim_{x\rightarrow 1^-}f(x) = x+5 = 1 + 5 = 1 + 5 = 6\\ \lim_{x\rightarrow 1^+}f(x) = x-5 = 1-5 = -4\\ f(1) = x+5 =1+5= 6 \\ L.H.L. = f(1) \neq R.H.S.

Hence, x = 1 is the point of discontinuity

Question:14. Discuss the continuity of the function f, where f is defined by

f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.

Answer:

Given function is
f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.
GIven function is defined for every real number k
Different cases are their
case (i) k < 1
f(k) = 3\\ \lim_{x\rightarrow k}f(x) = 3\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for every value of k < 1

case(ii) k = 1
f(1) = 3 \\ \lim_{x\rightarrow 1^-}f(x) = 3\\ \lim_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)
Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity

case(iii) 1 < k < 3
f(k) = 4 \\ \lim_{x\rightarrow k}f(x) = 4\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in 1 < k < 3 given function is continous

case(iv) k = 3
f(3) =5\\ \lim_{x\rightarrow 3^-}f(x) = 4\\ \lim_{x\rightarrow 3^+}f(x) =5\\ R.H.L. = f(3) \neq L.H.L.
Hence. x = 3 is the point of discontinuity

case(v) k > 3
f(k) = 5 \\ \lim_{x\rightarrow k}f(x) = 5 \\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for each and every value of k > 3
case(vi) when k < 3

f(k) = 4 \\ \lim_{x\rightarrow k}f(x) = 4\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in k < 3 given function is continous

Question:15 Discuss the continuity of the function f, where f is defined by f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.

Answer:

Given function is
gif
Given function is satisfies for the all real values of x
case (i) k < 0
gif
gif
Hence, function is continuous for all values of x < 0

case (ii) x = 0
gif
L.H.L at x= 0
gif
R.H.L. at x = 0
gif
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii) k > 0
gif
gif
Hence , function is continuous for all values of x > 0

case (iv) k < 1
gif
gif
Hence , function is continuous for all values of x < 1

case (v) k > 1
gif
gif
Hence , function is continuous for all values of x > 1

case (vi) x = 1
gif
gif
gif
Hence, function is not continuous at x = 1

Question:16. Discuss the continuity of the function f, where f is defined by

f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.

Answer:

Given function is
f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.
GIven function is defined for every real number k
Different cases are their
case (i) k < -1
f(k) = -2\\ \lim_{x\rightarrow k}f(x) = -2\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continuous for every value of k < -1

case(ii) k = -1
f(-1) = -2 \\ \lim_{x\rightarrow -1^-}f(x) = -2\\ \lim_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)
Hence, given function is continous at x = -1

case(iii) k > -1
f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for all values of x > -1

case(vi) -1 < k < 1
f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in -1 < k < 1 given function is continous

case(v) k = 1
f(1) =2x = 2(1)=2\\ \lim_{x\rightarrow 1^-}f(x) = 2x=2(1)=2\\ \lim_{x\rightarrow 1^+}f(x) =2\\ R.H.L. = f(1) = L.H.L.
Hence.at x =1 function is continous

case(vi) k > 1
f(k) = 2 \\ \lim_{x\rightarrow k}f(x) = 2 \\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, given function is continous for each and every value of k > 1
case(vii) when k < 1

f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)
Hence, for every value of k in k < 1 given function is continuous

Therefore, continuous at all points

Question:17. Find the relationship between a and b so that the function f defined by
f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.
is continuous at x = 3.

Answer:

Given function is
f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.
For the function to be continuous at x = 3 , R.H.L. must be equal to L.H.L.
\lim_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim_{x\rightarrow 3^+}f(x) = bx+3=3b+3
For the function to be continuous
\lim_{x\rightarrow 3^-}f(x) = \lim_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}

Question:18. For what value of l is the function defined by
f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.
continuous at x = 0? What about continuity at x = 1?

Answer:

Given function is
f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.
For the function to be continuous at x = 0 , R.H.L. must be equal to L.H.L.
\lim_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim_{x\rightarrow 0^+}f(x) = 4x+1=1
For the function to be continuous
\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^+}f(x) \\ 0\neq 1
Hence, for no value of function is continuous at x = 0

For x = 1
f(1)=4x+1=4(1)+1=5\\ \lim_{x\rightarrow 1}f(x) =4+1=5 \\\ \lim_{x\rightarrow 1}f(x) = f(x)
Hence, given function is continuous at x =1

Question:19. Show that the function defined by g (x) = x- [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Answer:

Given function is
g (x) = x- [x]
Given is defined for all real numbers k
\lim_{x\rightarrow k^-}f(x) = k - (k-1) = k-k+1 =1\\ \lim_{x\rightarrow k^+}f(x) = k - k = 0\\ \lim_{x\rightarrow k^-}f(x) \neq \lim_{x\rightarrow k^+}f(x)
Hence, by this, we can say that the function defined by g (x) = x- [x] is discontinuous at all integral points

Question:20. Is the function defined by f (x) = x^2 - sin x + 5 continuous at x = \pi ?

Answer:

Given function is
f (x) = x^2 - sin x + 5
Clearly, Given function is defined at x = \pi
f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim_{x\rightarrow \pi}f(x) = f(\pi)
Hence, the function defined by f (x) = x^2 - sin x + 5 continuous at x = \pi

Question:21. Discuss the continuity of the following functions:
a) f (x) = \sin x + \cos x

Answer:

Given function is
f (x) = \sin x + \cos x
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h
=\sin c\cos 0 + \cos c \sin 0 = \sin c
\lim_{x\rightarrow c}g(x) = g(c)
Hence, function g(x) = \sin x is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
=\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \cos x is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) + h(x) = sin x + cos x is also a continuous function

Question:21. b) Discuss the continuity of the following functions:
f (x) = \sin x - \cos x

Answer:

Given function is
f (x) = \sin x - \cos x
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h
=\sin c\cos 0 + \cos c \sin 0 = \sin c
\lim_{x\rightarrow c}g(x) = g(c)
Hence, function g(x) = \sin x is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
=\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \cos x is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

Question:21 c) Discuss the continuity of the following functions:
f (x) = \sin x \cdot \cos x

Answer:

Given function is
f (x) = \sin x . \cos x
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h
=\sin c\cos 0 + \cos c \sin 0 = \sin c
\lim_{x\rightarrow c}g(x) = g(c)
Hence, function g(x) = \sin x is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
=\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \cos x is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x).h(x) = sin x .cos x is also a continuous function

Question:22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Answer:

We, know that if two function g(x) and h(x) are continuous then
\frac{g(x)}{h(x)} , h(x) \neq0\ is \ continuous\\ \frac{1}{h(x)} , h(x) \neq 0\ is \ continuous\\ \frac{1}{g(x)} , g(x) \neq0\ is \ continuous\\
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h
=\sin c\cos 0 + \cos c \sin 0 = \sin c
\lim_{x\rightarrow c}g(x) = g(c)
Hence, function g(x) = \sin x is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
=\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, the function h(x) = \cos x is a continuous function
We proved independently that sin x and cos x is a continous function
So, we can say that
cosec x = \frac{1}{\sin x} = \frac{1}{g(x)} is also continuous except at x=n\pi
sec x = \frac{1}{\cos x} = \frac{1}{h(x)} is also continuous except at x=\frac{(2n+1) \pi}{2}
cot x = \frac{\cos x}{\sin x} = \frac{h(x)}{g(x)} is also continuous except at x=n\pi

Question:23. Find all points of discontinuity of f, where

f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.

Answer:

Given function is
f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.
\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\\ \lim_{x\rightarrow 0^+}f(x) = x + 1 = 1\\ \lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^+}f(x)
Hence, the function is continuous
Therefore, no point of discontinuity

Question:24. Determine if f defined by
f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.
is a continuous function?

Answer:

Given function is
f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.
Given function is defined for all real numbers k
when x = 0
f(0) = 0\\ \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1)
\lim_{x\rightarrow 0}f(x) = f(0)
Hence, function is continuous at x = 0
when x \neq 0
f(k) = k^2\sin \frac{1}{k}\\ \lim_{x\rightarrow k}f(x)=\lim_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim_{x\rightarrow k} = f(k)
Hence, the given function is continuous for all points

Question:25 . Examine the continuity of f, where f is defined by

f (x) = \left\{\begin{matrix} \sin x - \cos x & if x \neq 0 \\ -1 & if x = 0 \end{matrix}\right.

Answer:

Given function is
f (x) = \sin x - \cos x
Given function is defined for all real number
We, know that if two function g(x) and h(x) are continuous then g(x)+h(x) , g(x)-h(x) , g(x).h(x) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h
=\sin c\cos 0 + \cos c \sin 0 = \sin c
\lim_{x\rightarrow c}g(x) = g(c)
Hence, function g(x) = \sin x is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
=\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \cos x is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

When x = 0
f (0) = -1\\ \lim_{x\rightarrow 0^-}f(x) = \sin 0 - \cos 0 = -1\\ \lim_{x\rightarrow 0^+}f(x) = \sin 0 - \cos 0 = -1 \\ R.H.L. = L.H.L. = f(0)
Hence, function is also continuous at x = 0

Question:26. Find the values of k so that the function f is continuous at the indicated point in Exercises

f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right. \: \: \: at \: \: x = \pi /2

Answer:

Given function is
f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right.
When x = \frac{\pi}{2}
f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim_{x\rightarrow \frac{\pi}{2}}f(x)= \lim_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\
For the function to be continuous
\lim_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6
Therefore, the values of k so that the function f is continuous is 6

Question:27 . Find the values of k so that the function f is continuous at the indicated point in Exercises

f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right. \: \: at \: \: x = 2

Answer:

Given function is
f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right.
When x = 2
For the function to be continuous
f(2) = R.H.L. = LH.L.
f(2) = 4k\\ \lim_{x\rightarrow 2^-}f(x)= 4k\\ \lim_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim_{x\rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}
Hence, the values of k so that the function f is continuous at x= 2 is \frac{3}{4}

Question:28 . Find the values of k so that the function f is continuous at the indicated point in Exercises

f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right. \: \: at \: \: x = \pi

Answer:

Given function is
f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right.
When x = \pi
For the function to be continuous
f( \pi ) = R.H.L. = LH.L.
f(\pi) = k\pi+1\\ \lim_{x\rightarrow \pi^-}f(x)= k\pi+1\\ \lim_{x\rightarrow \pi^+}f(x) = \cos \pi = -1\\ f(\pi) = \lim_{x\rightarrow \pi^-}f(x) = \lim_{x\rightarrow \pi^+}f(x)\\ k\pi+1 = -1\\ k = \frac{-2}{\pi}
Hence, the values of k so that the function f is continuous at x= \pi is \frac{-2}{\pi}

Question:29 Find the values of k so that the function f is continuous at the indicated point in Exercises

f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right. \: \: at x = 5

Answer:

Given function is
f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right.
When x = 5
For the function to be continuous
f(5) = R.H.L. = LH.L.
f(5) = 5k+1\\ \lim_{x\rightarrow 5^-}f(x)= 5k+1\\ \lim_{x\rightarrow 5^+}f(x) = 3(5)-5 = 15-5=10\\ f(5) = \lim_{x\rightarrow 5^-}f(x) = \lim_{x\rightarrow 5^+}f(x)\\ 5k+1 = 10\\ k = \frac{9}{5}
Hence, the values of k so that the function f is continuous at x= 5 is \frac{9}{5}

Question:30 Find the values of a and b such that the function defined by
f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.
is a continuous function.

Answer:

Given continuous function is
f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.
The function is continuous so
\lim_{x\rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)\\ and\\ \lim_{x\rightarrow 10^-}f(x)=\lim_{x\rightarrow 10^+}f(x)
\lim_{x\rightarrow 2^-}f(x) = 5\\ \lim_{x\rightarrow 2^+}f(x)=ax+b=2a+b\\ 2a+b = 5 \ \ \ \ \ \ \ \ \ \ \ -(i)\\ and\\ \lim_{x\rightarrow 10^-}f(x)=ax+b=10a+b\\ \lim_{x\rightarrow 10^+}f(x)=21\\ 10a+b=21 \ \ \ \ \ \ \ \ -(ii)
By solving equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right. is a continuous function is 2 and 1 respectively

Question:31. Show that the function defined by f (x) = \cos (x^2 ) is a continuous function.

Answer:

Given function is
f (x) = \cos (x^2 )
given function is defined for all real values of x
Let x = k + h
if x\rightarrow k , \ then \ h \rightarrow 0
f(k) = \cos k^2\\ \lim_{x \rightarrow k}f(x) = \lim_{x \rightarrow k}\cos x^2 = \lim_{h \rightarrow 0}\cos (k+h)^2 = \cos k^2\\ \lim_{x \rightarrow k}f(x) = f(k)
Hence, the function f (x) = \cos (x^2 ) is a continuous function

Question:32. Show that the function defined by f (x) = |\cos x | is a continuous function.

Answer:

Given function is
f (x) = |\cos x |
given function is defined for all values of x
f = g o h , g(x) = |x| and h(x) = cos x
Now,
g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}
g(x) is defined for all real numbers k
case(i) k < 0
g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k < 0

case (ii) k > 0
g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k > 0

case (iii) k = 0
g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) = cos x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h
=\cos c\cos 0 + \sin c \sin 0 = \cos c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \cos x is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = g o h is also continuous

Question:33 . Examine that sin | x| is a continuous function.

Answer:

Given function is
f(x) = sin |x|
f(x) = h o g , h(x) = sin x and g(x) = |x|
Now,

g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}
g(x) is defined for all real numbers k
case(i) k < 0
g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k < 0

case (ii) k > 0
g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k > 0

case (iii) k = 0
g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) = sin x
Let suppose x = c + h
if x \rightarrow c , \ then \ h \rightarrow 0
h(c) = \sin c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h
=\sin c\cos 0 + \cos c \sin 0 = \sin c
\lim_{x\rightarrow c}h(x) = h(c)
Hence, function h(x) = \sin x is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x) = h o g is also continuous

Question:34. Find all the points of discontinuity of f defined by f (x) = | x| - | x + 1|.

Answer:

Given function is
f (x) = | x| - | x + 1|
Let g(x) = |x| and h(x) = |x+1|
Now,
g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}
g(x) is defined for all real numbers k
case(i) k < 0
g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k < 0

case (ii) k > 0
g(k) = k\\ \lim_{x\rightarrow k}g(x) = k\\ \lim_{x\rightarrow k}g(x) = g(k)
Hence, g(x) is continuous when k > 0

case (iii) k = 0
g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x

Now,
h(x)\begin{cases} -(x+1) & \text{ if } x<-1 \\ 0 & \text{ if } x= -1\\ (x+1)& \text{ if } x>-1 \end{cases}
g(x) is defined for all real numbers k
case(i) k < -1
h(k) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = h(k)
Hence, h(x) is continuous when k < -1

case (ii) k > -1
h(k) = k+1\\ \lim_{x\rightarrow k}h(x) = k+1\\ \lim_{x\rightarrow k}h(x) = h(k)
Hence, h(x) is continuous when k > -1

case (iii) k = -1
h(-1) = 0\\ \lim_{x\rightarrow -1^-}h(x) = -(x-1) = 0\\ \lim_{x\rightarrow -1^+}h(x ) = x+1 = 0\\ \lim_{x\rightarrow -1^-}h(x) = h(0) = \lim_{x\rightarrow -1^+}h(x )
Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x
g(x) is continuous and h(x) is continuous
Therefore, f(x) = g(x) - h(x) = |x| - |x+1| is also continuous


NCERT class 12 maths chapter 5 question answer: Excercise: 5.2

Question:1. Differentiate the functions with respect to x in

\sin (x^2 +5 )

Answer:

Given function is
f(x)=\sin (x^2 +5 )
when we differentiate it w.r.t. x.
Lets take t = x^2+5 . then,
f(t) = \sin t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} (By chain rule)
\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)
\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x
Therefore, the answer is 2x \cos (x^2+5)

Question:2. Differentiate the functions with respect to x in

\cos ( \sin x )

Answer:

Given function is
f(x)= \cos ( \sin x )
Lets take t = \sin x then,
f(t) = \cos t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} ( By chain rule)
\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)
\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x
Therefore, the answer is -\sin(\sin x).\cos x

Question:3. Differentiate the functions with respect to x in

\sin (ax +b )

Answer:

Given function is
f(x) = \sin (ax +b )
when we differentiate it w.r.t. x.
Lets take t = ax+b . then,
f(t) = \sin t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} (By chain rule)
\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)
\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a
Therefore, the answer is a \cos (ax+b)

Question:4 . Differentiate the functions with respect to x in

\sec (\tan (\sqrt x) )

Answer:

Given function is
f(x)=\sec (\tan (\sqrt x) )
when we differentiate it w.r.t. x.
Lets take t = \sqrt x . then,
f(t) = \sec (\tan t)
take \tan t = k . then,
f(k) = \sec k
\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} (By chain rule)
\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)
(\because k = \tan t \ and \ t = \sqrt x)
\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)
\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}
Now,
\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} =\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}
Therefore, the answer is \frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}

Question:5. Differentiate the functions with respect to x in

\frac{\sin (ax +b )}{\cos (cx + d)}

Answer:

Given function is
f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}
We know that,
f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}
g(x) = \sin(ax+b) and h(x) = \cos(cx+d)
Lets take u = (ax+b) \ and \ v = (cx+d)
Then,
\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c
g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx} (By chain rule)
\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)
\frac{du}{dx} = \frac{d(ax+b)}{dx} = a
g^{'}(x)=a\cos (ax+b) -(i)
Similarly,
h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}
\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))
\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c
h^{'}(x)=-c\sin(cx+d) -(ii)
Now, put (i) and (ii) in
f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}
= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}
= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)
Therefore, the answer is a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)

Question:6. Differentiate the functions with respect to x in

\cos x^3 . \sin ^ 2 ( x ^5 )

Answer:

Given function is
f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )
Differentitation w.r.t. x is
f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)
g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)
Lets take u = x^3 \ and \ v = x^5
Our functions become,
\cos x^3 = \cos u and \sin^2(x^5) = \sin^2v
Now,
g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx} ( By chain rule)
\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)
\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2
g^{'}(x) = -\sin x^3.3x^2 -(i)
Similarly,
h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}
\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5 \ \ \ (\because v = x^5)

\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4
h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5 -(ii)
Put (i) and (ii) in
f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) = -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5\cos x^5
Therefore, the answer is 10x^4\sin x^5\cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5

Question:7. Differentiate the functions with respect to x in

2 \sqrt { \cot ( x^2 )}

Answer:

Give function is
f(x)=2 \sqrt { \cot ( x^2 )}
Let's take t = x^2
Now, take \cot t = k^2
f(k) = 2k
Differentiation w.r.t. x
\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx} -(By chain rule)
\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2
\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}} \ \ \ (\because t = x^2)
\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x
So,
\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} } ( Multiply and divide by \sqrt 2 and multiply and divide \sqrt {\cot x^2} by \sqrt{\sin x^2 )
(\because \cot x = \frac{\cos x}{\sin x} \ and \ cosec x = \frac{1}{\sin x } )
=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}} \ \ \ \ (\because 2\sin x\cos x=\sin2x)
There, the answer is \frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}

Question:8 Differentiate the functions with respect to x in

\cos ( \sqrt x )

Answer:

Let us assume : y\ =\ \cos ( \sqrt x )

Differentiating y with respect to x, we get :

\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}

or =\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}

or =\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}



Question:9 . Prove that the function f given by f (x) = |x-1 | , x \epsilon R is not differentiable at x = 1.

Answer:

Given function is
f (x) = |x-1 | , x \epsilon R
We know that any function is differentiable when both
\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h} and \lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h} are finite and equal
Required condition for function to be differential at x = 1 is

\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}
Now, Left-hand limit of a function at x = 1 is
\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^-}\frac{|h|-0}{h}
= \lim_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)
Right-hand limit of a function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^+}\frac{|h|-0}{h}
=\lim_{h\rightarrow 0^-}\frac{h}{h} = 1
Now, it is clear that
R.H.L. at x= 1 \neq L.H.L. at x= 1
Therefore, function f (x) = |x-1 | is not differentiable at x = 1

Question:10. Prove that the greatest integer function defined by f (x) = [x] , 0 < x < 3 is not differentiable at

x = 1 and x = 2.

Answer:

Given function is
f (x) = [x] , 0 < x < 3
We know that any function is differentiable when both
\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h} and \lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h} are finite and equal
Required condition for function to be differential at x = 1 is

\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}
Now, Left-hand limit of the function at x = 1 is
\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^-}\frac{0-1}{h}
=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \Rightarrow 1+h<1, \therefore [1+h] =0)
Right-hand limit of the function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^+}\frac{1-1}{h}
=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 1+h>1, \therefore [1+h] =1)
Now, it is clear that
R.H.L. at x= 1 \neq L.H.L. at x= 1 and L.H.L. is not finite as well
Therefore, function f(x) = [x] is not differentiable at x = 1
Similary, for x = 2
Required condition for function to be differential at x = 2 is

\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}
Now, Left-hand limit of the function at x = 2 is
\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^-}\frac{1-2}{h}
=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \Rightarrow 2+h<2, \therefore [2+h] =1)
Right-hand limit of the function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^+}\frac{2-2}{h}
=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 2+h>2, \therefore [2+h] =2)
Now, it is clear that
R.H.L. at x= 2 \neq L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function f(x) = [x] is not differentiable at x = 2


NCERT class 12 maths chapter 5 question answer: Exercise: 5.3

Question:1. Find dy/dx in the following:

2 x + 3 y = \sin x

Answer:

Given function is
2 x + 3 y = \sin x
We can rewrite it as
3y = \sin x - 2x
Now, differentiation w.r.t. x is
3\frac{dy}{dx} = \frac{d(\sin x - 2x)}{dx} = \cos x - 2
\frac{dy}{dx} = \frac{\cos x-2}{3}
Therefore, the answer is \frac{\cos x-2}{3}

Question:2. Find dy/dx in the following: 2 x + 3y = \sin y

Answer:

Given function is
2 x + 3 y = \sin y
We can rewrite it as
\sin y - 3y = 2x
Now, differentiation w.r.t. x is
\frac{dy}{dx}(\sin y - 3y) = \frac{d( 2x)}{dx}

(\cos y\frac{dy}{dx} - 3\frac{dy}{dx}) = 2
\frac{dy}{dx} = \frac{2}{\cos y -3}
Therefore, the answer is \frac{2}{\cos y -3}

Question:3. Find dy/dx in the following: ax + by ^2 = \cos y

Answer:

Given function is
ax + by ^2 = \cos y
We can rewrite it as
by^2-\cos y = -ax
Now, differentiation w.r.t. x is
\frac{dy}{dx}(2by - (-\sin y)) = \frac{d( -ax)}{dx} = -a
\frac{dy}{dx} = \frac{-a}{2b y +\sin y}
Therefore, the answer is \frac{-a}{2b y +\sin y}

Question:4. Find dy/dx in the following:

xy + y^2 = \tan x + y

Answer:

Given function is
xy + y^2 = \tan x + y
We can rewrite it as
xy+y^2-y= \tan x
Now, differentiation w.r.t. x is
y+\frac{dy}{dx}(x+2y-1) = \frac{d( \tan x)}{dx} = \sec^2 x
\frac{dy}{dx} = \frac{\sec^2 x- y}{x+2y-1}
Therefore, the answer is \frac{\sec^2 x- y}{x+2y-1}

Question:5. Find dy/dx in the following: x^2 + xy + y^2 = 100

Answer:

Given function is
x^2 + xy + y^2 = 100
We can rewrite it as
xy + y^2 = 100 - x^2
Now, differentiation w.r.t. x is
y+\frac{dy}{dx}(x+2y) = \frac{d( 100-x^2)}{dx} = -2x
\frac{dy}{dx} = \frac{-2 x- y}{x+2y}
Therefore, the answer is \frac{-2 x- y}{x+2y}

Question:6 Find dy/dx in the following:

x ^3 + x^2 y + xy^2 + y^3 = 81

Answer:

Given function is
x ^3 + x^2 y + xy^2 + y^3 = 81
We can rewrite it as
x^2 y + xy^2 + y^3 = 81 - x^3
Now, differentiation w.r.t. x is
\frac{d(x^2 y + xy^2 + y^3)}{dx} = \frac{d(81 - x^3)}{dx}
2xy+y^2+\frac{dy}{dx}(x^2+2xy+3y^2) = -3x^2\\ \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}
Therefore, the answer is \frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}

Question:7 . Find dy/dx in the following: \sin ^ 2 y + \cos xy = k

Answer:

Given function is
\sin ^ 2 y + \cos xy = k
Now, differentiation w.r.t. x is
\frac{d(\sin^2y+\cos xy)}{dx} = \frac{d(k)}{dx}
2\sin y \cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0\\ \frac{dy}{dx}(2\sin y \cos y-x\sin xy)= y\sin xy\\ \frac{dy}{dx} = \frac{y\sin xy}{2\sin y \cos y-x\sin xy} = \frac{y\sin xy}{\sin 2y -x\sin xy} \ \ \ \ \ \ (\because 2\sin x\cos y = \sin 2x)
Therefore, the answer is \frac{y\sin xy}{\sin 2y -x\sin xy}

Question:8. Find dy/dx in the following:

\sin ^2 x + \cos ^ 2 y = 1

Answer:

Given function is
\sin ^2 x + \cos ^ 2 y = 1
We can rewrite it as
\cos ^ 2 y = 1-\sin^2x
Now, differentiation w.r.t. x is
\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}
2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)
Therefore, the answer is \frac{\sin 2x}{\sin 2y }

Question:9 Find dy/dx in the following:

y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )

Answer:

Given function is
y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )
Lets consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2 x})
Our equation reduces to
y = \sin^{-1}(\sin 2t)
y = 2t
Now, differentiation w.r.t. x is
\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}
Therefore, the answer is \frac{2}{1+x^2}

Question:10. Find dy/dx in the following:
y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right ) , - \frac{1}{\sqrt 3 } < x < \frac{1}{\sqrt 3 }

Answer:

Given function is
y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right )
Lets consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{3x-x^3}{1-3x^2} = \frac{3\tan t-\tan^3t }{1-3\tan^2t} = \tan3t \ \ \ \ \ \ (\because \tan 3x = \frac{3\tan x-\tan^3x }{1-3\tan^2x} )
Our equation reduces to
y = \tan^{-1}(\tan 3t)
y = 3t
Now, differentiation w.r.t. x is
\frac{d(y)}{dx} = \frac{d(3t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 3.\frac{1}{1+x^2} = \frac{3}{1+x^2}
Therefore, the answer is \frac{3}{1+x^2}

Question:11. Find dy/dx in the following:

y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right ) , 0 < x < 1

Answer:

Given function is
y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )
Let's consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )
Our equation reduces to
y = \cos^{-1}(\cos 2t)
y = 2t
Now, differentiation w.r.t. x is
\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}
Therefore, the answer is \frac{2}{1+x^2}

Question:12 . Find dy/dx in the following: y = \sin ^{-1 } \left ( \frac{1- x ^2 }{1+ x^2} \right ) , 0< x < 1

Answer:

Given function is
y = \sin ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )
We can rewrite it as
\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )
Let's consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )
Our equation reduces to
\sin y = \cos 2t
Now, differentiation w.r.t. x is
\frac{d(\sin y)}{dx} = \frac{d(\cos2t)}{dt}.\frac{dt}{dx}
\cos y\frac{dy}{dx} = 2(-\sin 2t).\frac{1}{1+x^2} = \frac{-2\sin2t}{1+x^2} = \frac{-2.\frac{2\tan t}{1+\tan^2t}}{1+x^2} =\frac{-2.\frac{2x}{1+x^2}}{1+x^2} =\frac{-4x}{(1+x^2)^2}
(\because \sin 2x = \frac{2\tan x}{1+\tan^2x} \ and \ x = \tan t)
\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )\Rightarrow \cos y = \frac{2x}{1+x^2}
\frac{2x}{1+x^2}\frac{dy}{dx} = \frac{-4x}{(1+x^2)^2}
\frac{dy}{dx} = \frac{-2}{(1+x^2)}
Therefore, the answer is \frac{-2}{1+x^2}

Question:13. Find dy/dx in the following:

y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right ) , -1 < x < 1

Answer:

Given function is
y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right )
We can rewrite it as
\cos y = \left ( \frac{2x}{1+ x^2 } \right )
Let's consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2x} )
Our equation reduces to
\cos y = \sin 2t
Now, differentiation w.r.t. x is
\frac{d(\cos y)}{dx} = \frac{d(\sin2t)}{dt}.\frac{dt}{dx}
(-\sin y)\frac{dy}{dx} = 2(\cos 2t).\frac{1}{1+x^2} = \frac{2\cos2t}{1+x^2} = \frac{2.\frac{1-\tan^2 t}{1+\tan^2t}}{1+x^2} =\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2} =\frac{2(1-x^2)}{(1+x^2)^2}
(\because \cos 2x = \frac{1-\tan^2 x}{1+\tan^2x} \ and \ x = \tan t)
\cos y = \ \left ( \frac{2 x }{1+ x^2 } \right )\Rightarrow \sin y = \frac{1-x^2}{1+x^2}
-\frac{1-x^2}{1+x^2}\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}
\frac{dy}{dx} = \frac{-2}{(1+x^2)}
Therefore, the answer is \frac{-2}{1+x^2}

Question:14 . Find dy/dx in the following:

y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} ) , -\frac{1}{\sqrt2} < x \frac{1}{\sqrt 2 }

Answer:

Given function is
y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} )
Lets take x = \sin t
Then,
\frac{d(x)}{dx} = \frac{(\sin t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)
1 =\cos t.\frac{dt}{dx}
\frac{dt}{dx} = \frac{1}{\cos t } = \frac{1}{\sqrt{1-\sin ^2t}} = \frac{1}{\sqrt{1-x^2}}
(\because \cos x = \sqrt{1-\sin^2x} \ and \ x = \sin t )
And
2x\sqrt{1-x^2} = 2\sin t \sqrt{1-\sin^2t} = 2\sin t \sqrt{\cos^2 t} = 2\sin t\cos t =\sin 2t
(\because \cos x = \sqrt{1-\sin^2x} \ and \ 2\sin x\cos x = \sin2x )
Now, our equation reduces to
y = \sin ^ { -1 } ( \sin 2t )
y = 2t
Now, differentiation w.r.t. x
\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 2.\frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}
Therefore, the answer is \frac{2}{\sqrt{1-x^2}}

Question:15 . Find dy/dx in the following:

y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right ) , 0 < x < 1/ \sqrt 2

Answer:

Given function is
y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right )
Let's take x = \cos t
Then,
\frac{d(x)}{dx} = \frac{(\cos t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)
1 =-\sin t.\frac{dt}{dx}
\frac{dt}{dx} = \frac{-1}{\sin t } = \frac{-1}{\sqrt{1-\cos ^2t}} = \frac{-1}{\sqrt{1-x^2}}
(\because \sin x = \sqrt{1-\cos^2x} \ and \ x = \cos t )
And
\frac{1}{2x^2-1} =\frac{1}{2\cos^2 t - 1} = \frac{1}{\cos2t} = \sec2t
(\because \cos 2x = \sqrt{2\cos^2x-1} \ and \ \cos x = \frac{1}{\sec x} )

Now, our equation reduces to
y = \sec ^{-1} \left ( \sec 2t \right )
y = 2t
Now, differentiation w.r.t. x
\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 2.\frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}}
Therefore, the answer is \frac{-2}{\sqrt{1-x^2}}


NCERT class 12 maths chapter 5 question answer: Exercise 5.4

Question:1. Differentiate the following w.r.t. x:

\frac{e ^x }{\sin x }

Answer:

Given function is
f(x)=\frac{e ^x }{\sin x }
We differentiate with the help of Quotient rule
f^{'}(x)=\frac{\frac{d(e^x)}{dx}.\sin x-e^x.\frac{(\sin x)}{dx} }{\sin^2 x }
=\frac{e^x.\sin x-e^x.\cos }{\sin^2 x } = \frac{e^x(\sin x-\cos x)}{\sin^2x}
Therefore, the answer is \frac{e^x(\sin x-\cos x)}{\sin^2x}

Question:2 . Differentiate the following w.r.t. x:

e ^{\sin ^{-1}x}

Answer:

Given function is
f(x)=e ^{\sin ^{-1}x}
Let g(x)={\sin ^{-1}x}
Then,
f(x)=e^{g(x)}
Now, differentiation w.r.t. x
f^{'}(x)=g^{'}(x).e^{g(x)} -(i)
g(x) = \sin^{-1}x \Rightarrow g^{'}(x ) = \frac{1}{\sqrt{1-x^2}}
Put this value in our equation (i)
f^{'}(x) = \frac{1}{\sqrt{1-x^2}}.e^{\sin^{-1}x} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}

Question:3 . Differentiate the following w.r.t. x:

e ^x^{ ^3}

Answer:

Given function is
f(x)=e ^{x^3}
Let g(x)=x^3
Then,
f(x)=e^{g(x)}
Now, differentiation w.r.t. x
f^{'}(x)=g^{'}(x).e^{g(x)} -(i)
g(x) = x^3 \Rightarrow g^{'}(x ) =3x^2
Put this value in our equation (i)
f^{'}(x) =3x^2.e^{x^3}
Therefore, the answer is 3x^2.e^{x^3}

Question:4. Differentiate the following w.r.t. x:

\sin ( \tan ^ { -1} e ^{-x })

Answer:

Given function is
f(x)=\sin ( \tan ^ { -1} e ^{-x })
Let's take g(x ) = \tan^{-1}e^{-x}
Now, our function reduces to
f(x) = \sin(g(x))
Now,
f^{'}(x) = g^{'}(x)\cos(g(x)) -(i)
And
g(x)=\tan^{-1}e^{-x}\\\Rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}}
Put this value in our equation (i)
f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})
Therefore, the answer is \frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})

Question:5 . Differentiate the following w.r.t. x:

\log (\cos e ^x )

Answer:

Given function is
f(x)=\log (\cos e ^x )
Let's take g(x ) = \cos e^{x}
Now, our function reduces to
f(x) = \log(g(x))
Now,
f^{'}(x) = g^{'}(x).\frac{1}{g(x)} -(i)
And
g(x)=\cos e^{x}\\\Rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x
Put this value in our equation (i)
f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)
Therefore, the answer is -e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N

Question:6 . Differentiate the following w.r.t. x:

e ^x + e ^{x^2} + .....e ^{x^5}

Answer:

Given function is
f(x)= e ^x + e ^{x^2} + .....e ^{x^5}
Now, differentiation w.r.t. x is
f^{'}(x)= \frac{d(e^x)}{dx}.\frac{d(x)}{dx}+\frac{d(e^{x^2})}{dx}.\frac{d(x^2)}{dx}+\frac{d(e^{x^3})}{dx}.\frac{d(x^3)}{dx}+\frac{d(e^{x^4})}{dx}.\frac{d(x^4)}{dx}+\frac{d(e^{x^5})}{dx}.\frac{d(x^5)}{dx}
=e^x.1+e^{x^2}.2x+e^{x^3}.3x^2+e^{x^4}.4x^3+e^{x^5}.5x^4
=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}
Therefore, answer is e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}

Question:7 . Differentiate the following w.r.t. x:

\sqrt { e ^{ \sqrt x }} , x > 0

Answer:

Given function is
f(x)=\sqrt { e ^{ \sqrt x }}
Lets take g(x ) = \sqrt x
Now, our function reduces to
f(x) = \sqrt {e^{g(x)}}
Now,
f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}} -(i)
And
g(x)=\sqrt x\\\Rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}
Put this value in our equation (i)
f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}
Therefore, the answer is \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0

Question:8 Differentiate the following w.r.t. x: \log ( \log x ) , x > 1

Answer:

Given function is
f(x)=\log ( \log x )
Lets take g(x ) = \log x
Now, our function reduces to
f(x) = \log(g(x))
Now,
f^{'}(x) = g^{'}(x).\frac{1}{g(x)} -(i)
And
g(x)=\log x\\\Rightarrow g^{'}(x) = \frac{1}{x}
Put this value in our equation (i)
f^{'}(x) =\frac{1}{x}.\frac{1}{\log x} = \frac{1}{x\log x}
Therefore, the answer is \frac{1}{x\log x}, \ \ x>1

Question:9. Differentiate the following w.r.t. x:

\frac{\cos x }{\log x} , x > 0

Answer:

Given function is
f(x)=\frac{\cos x }{\log x}
We differentiate with the help of Quotient rule
f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }
=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}
Therefore, the answer is \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}

Question:10. Differentiate the following w.r.t. x:

\cos ( log x + e ^x ) , x > 0

Answer:

Given function is
f(x)=\cos ( log x + e ^x )
Lets take g(x) = ( log x + e ^x )
Then , our function reduces to
f(x) = \cos (g(x))
Now, differentiation w.r.t. x is
f^{'}(x) = g^{'}(x)\(-\sin) (g(x)) -(i)
And
g(x) = ( log x + e ^x )
g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x
Put this value in our equation (i)
f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)
Therefore, the answer is -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0


Class 12 Maths Chapter 5 NCERT solutions: Exercise: 5.5

Question:1 Differentiate the functions w.r.t. x. \cos x . \cos 2x .\cos 3x

Answer:

Given function is
y=\cos x . \cos 2x .\cos 3x
Now, take log on both sides
\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \log y = \log \cos x + \log \cos 2x + \log \cos 3x
Now, differentiation w.r.t. x
\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}\\ \frac{1}{y}\frac{dy}{dx} = -(\tan x+\tan 2x+\tan 3x) \ \ \ \ \ \ (\because \frac{\sin x }{\cos x} =\tan x)\\ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)\\ \frac{dy}{dx}= -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)
There, the answer is -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)

Question:2. Differentiate the functions w.r.t. x.

\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}

Answer:

Given function is
y=\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}
Take log on both the sides
\log y=\frac{1}{2}\log\left ( \frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)} \right )\\ \log y = \frac{1}{2} (\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5))\\
Now, differentiation w.r.t. x is
\frac{d(\log y)}{dx} = \frac{1}{2} (\frac{d(\log(x-1))}{dx}+\frac{d(\log(x-2))}{dx}-\frac{d(\log(x-3))}{dx}-\frac{d(\log(x-4))}{dx}-\\ \frac{d(\log(x-5))}{dx})
\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx}=y\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx} = \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})
Therefore, the answer is \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})

Question:3 Differentiate the functions w.r.t. x. (\log x ) ^{\cos x}

Answer:

Given function is
y=(\log x ) ^{\cos x}
take log on both the sides
\log y=\cos x\log (\log x )
Now, differentiation w.r.t x is
\frac{d(\log y)}{dx} = \frac{d(\cos x\log(\log x))}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)(\log(\log x)) + \cos x.\frac{1}{\log x}.\frac{1}{x}\\ \frac{dy}{dx}= y( \cos x.\frac{1}{\log x}.\frac{1}{x}-\sin x\log(\log x) )\\ \frac{dy}{dx} = (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )
Therefore, the answer is (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )

Question:4 Differentiate the functions w.r.t. x. x ^x - 2 ^{ \sin x }

Answer:

Given function is
y = x ^x - 2 ^{ \sin x }
Let's take t = x^x
take log on both the sides
\log t=x\log x\\
Now, differentiation w.r.t x is
\log t=x\log x\\ \frac{d(\log t)}{dt}.\frac{dt}{dx} = \frac{d(x\log x)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)\\ \frac{1}{t}.\frac{dt}{dx} = \log x +1\\ \frac{dt}{dx} = t(\log x+1)\\ \frac{dt}{dx}= x^x(\log x+1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x )
Similarly, take k = 2^{\sin x}
Now, take log on both sides and differentiate w.r.t. x
\log k=\sin x\log 2\\ \frac{d(\log k)}{dk}.\frac{dk}{dx} = \frac{d(\sin x\log 2)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)\\ \frac{1}{k}.\frac{dk}{dx} = \cos x \log 2\\ \frac{dk}{dx} = k(\cos x\log 2)\\ \frac{dk}{dx}= 2^{\sin x}(\cos x\log 2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because k = 2^{\sin x} )
Now,
\frac{dy}{dx} = \frac{dt}{dx}-\frac{dk}{dx}\\ \frac{dy}{dx} = x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)

Therefore, the answer is x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)

Question:5 Differentiate the functions w.r.t. x. ( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4

Answer:

Given function is
y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4
Take log on both sides
\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]\\ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)
Now, differentiate w.r.t. x we get,
\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}\\ \frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )\\ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)
Therefore, the answer is (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)

Question:6 Differentiate the functions w.r.t. x. ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }

Answer:

Given function is
y = ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }
Let's take t = ( x+ \frac{1}{x} ) ^ x
Now, take log on both sides
\log t =x \log ( x+ \frac{1}{x} )
Now, differentiate w.r.t. x
we get,
\frac{1}{t}.\frac{dt}{dx}=\log \left ( x+\frac{1}{x} \right )+x(1-\frac{1}{x^2}).\frac{1}{\left ( x+\frac{1}{x} \right )} = \frac{x^2-1}{x^2+1}+\log \left ( x+\frac{1}{x} \right )\\ \frac{dt}{dx} = t(\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))\\ \frac{dt}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))
Similarly, take k = x^{1+\frac{1}{x}}
Now, take log on both sides
\log k = ({1+\frac{1}{x}})\log x
Now, differentiate w.r.t. x
We get,
\frac{1}{k}.\frac{dk}{dx}=\frac{1}{x} \left ( 1+\frac{1}{x} \right )+(-\frac{1}{x^2}).\log x = \frac{x^2+1}{x^2}+\frac{-1}{x^2}.\log x\\ \frac{dk}{dx} = t(\frac{x^2+1}{x^2}+\left (\frac{-1}{x^2} \right )\log x)\\ \frac{dk}{dx} = x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )

Therefore, the answer is \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )

Question:7 Differentiate the functions w.r.t. x. (\log x )^x + x ^{\log x }

Answer:

Given function is
y = (\log x )^x + x ^{\log x }
Let's take t = (\log x)^x
Now, take log on both the sides
\log t = x \log(\log x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}= \log (\log x)+\frac{1}{\log x}\\ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}=(\log x)^x(\log (\log x))+ (\log x )^{x-1}
Similarly, take k = x^{\log x}
Now, take log on both sides
\log k = \log x \log x = (\log x)^2
Now, differentiate w.r.t. x
We get,
\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x
Therefore, the answer is (\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x

Question:8 Differentiate the functions w.r.t. x. (\sin x )^x + \sin ^{-1} \sqrt x

Answer:

Given function is
(\sin x )^x + \sin ^{-1} \sqrt x
Lets take t = (\sin x)^x
Now, take log on both the sides
\log t = x \log(\sin x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \log (\sin x) + x.\cos x.\frac{1}{\sin x}= \log (\sin x)+x.\cot x \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)\\ \frac{dt}{dx}= t.(\log (\sin x)+x.\cot x)\\ \frac{dt}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)
Similarly, take k = \sin^{-1}\sqrt x
Now, differentiate w.r.t. x
We get,
\frac{dk}{dt} = \frac{1}{\sqrt{1-(\sqrt x)^2}}.\frac{1}{2\sqrt x}= \frac{1}{2\sqrt{x-x^2}}\\ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}}\\
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}
Therefore, the answer is (\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}

Question:9 Differentiate the functions w.r.t. x x ^ { \sin x } + ( \sin x )^ \cos x

Answer:

Given function is
y = x ^ { \sin x } + ( \sin x )^ \cos x

Now, take t = x^{\sin x}
Now, take log on both sides
\log t = \sin x \log x
Now, differentiate it w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\ \frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\ \frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )
Similarly, take k = (\sin x)^{\cos x}
Now, take log on both the sides
\log k = \cos x \log (\sin x)
Now, differentiate it w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ \frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )
Now,
\frac{dy}{dx} = x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )
Therefore, the answer is x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )

Question:10 Differentiate the functions w.r.t. x. x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }

Answer:

Given function is
x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }
Take t = x^{x\cos x}
Take log on both the sides
\log t =x\cos x \log x
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \cos x\log x-x\sin x\log x + \frac{1}{x}.x.\cos x\\ \frac{dt}{dx}= t.\left (\log x(\cos x-x\sin x)+ \cos x \right ) = x^{x\cos x}\left ( \log x(\cos x-x\sin x)+ \cos x \right )
Similarly,
take k = \frac{x^2+1}{x^2-1}
Now. differentiate it w.r.t. x
we get,
\frac{dk}{dx} = \frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1)^2} = \frac{2x^3-2x-2x^3-2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} = x^{x\cos x}\left ( \log x(\cos x-x\sin x)+ \cos x \right )-\frac{4x}{(x^2-1)^2}
Therefore, the answer is x^{x\cos x}\left ( \cos x(\log x+1)-x\sin x\log x\right )-\frac{4x}{(x^2-1)^2}

Question:11 Differentiate the functions w.r.t. x. ( x \cos x )^ x + ( x \sin x )^{1/ x}

Answer:

Given function is
f(x)=( x \cos x )^ x + ( x \sin x )^{1/ x}
Let's take t = (x\cos x)^x
Now, take log on both sides
\log t =x\log (x\cos x) = x(\log x+\log \cos x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\ \frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) \ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\ \frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\ \frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))
Similarly, take k = (x\sin x)^{\frac{1}{x}}
Now, take log on both the sides
\log k = \frac{1}{x}(\log x+\log \sin x)
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\ \frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x}) \ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\ \frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\ \frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
Now,
\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
Therefore, the answer is (x\cos x)^x(1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}

Question:12 Find dy/dx of the functions given in Exercises 12 to 15

x ^ y + y ^ x = 1 .

Answer:

Given function is
f(x)=x ^ y + y ^ x = 1
Now, take t = x^y
take log on both sides
\log t = y\log x
Now, differentiate w.r.t x
we get,
\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})
Similarly, take k = y^x
Now, take log on both sides
\log k = x\log y
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})
Now,
f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0

( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0\\ \frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ \frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}

Therefore, the answer is \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}

Question:13 Find dy/dx of the functions given in Exercises 12 to 15.

y^x = x ^y

Answer:

Given function is
f(x)\Rightarrow x ^ y = y ^ x
Now, take t = x^y
take log on both sides
\log t = y\log x
Now, differentiate w.r.t x
we get,
\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})
Similarly, take k = y^x
Now, take log on both sides
\log k = x\log y
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})
Now,
f^{'}(x)\Rightarrow \frac{dt}{dx}= \frac{dk}{dx}

( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) = (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dy}{dx}(x^y(\log x)-xy^{x-1}) = (y^x(\log y)-yx^{y-1})\\ \frac{dy}{dx}= \frac{ y^x(\log y)-yx^{y-1}}{(x^y(\log x)-xy^{x-1})} = \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )

Therefore, the answer is \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )

Question:14 Find dy/dx of the functions given in Exercises 12 to 15. ( \cos x )^y = ( \cos y )^x

Answer:

Given function is
f(x)\Rightarrow (\cos x) ^ y = (\cos y) ^ x
Now, take log on both the sides
y\log \cos x = x \log \cos y
Now, differentiate w.r.t x
\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}
By taking similar terms on the same side
We get,
(\frac{dy}{dx}(\log \cos x)-y\tan x) = (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )} = \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}

Therefore, the answer is \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}

Question:15 Find dy/dx of the functions given in Exercises 12 to 15. xy = e ^{x-y}

Answer:

Given function is
f(x)\Rightarrow xy = e ^{x-y}
Now, take take log on both the sides
\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)
Now, differentiate w.r.t x
\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}
By taking similar terms on same side
We get,
(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}
Therefore, the answer is \frac{y}{x}.\frac{x-1}{y+1}

Question:16 Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) and hence find

f ' (1)

Answer:

Given function is
y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)
Take log on both sides
\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)
NOW, differentiate w.r.t. x
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ \frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )
Therefore, f^{'}(x)= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )
Now, the vale of f^{'}(1) is
f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+ \frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\ f^{'}(1)=16.\frac{15}{2} = 120

Question:17 (1) Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
(i) by using product rule

Answer:

Given function is
f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)
Now, we need to differentiate using the product rule
f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\
= (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\ = 5x^4 -20x^3+45x^2-52x+11
Therefore, the answer is 5x^4 -20x^3+45x^2-52x+11

Question:17 (2) Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.

Answer:

Given function is
f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)
Multiply both to obtain a single higher degree polynomial
f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)
= x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72
= x^5-5x^4+15x^3-26x^2+11x+72
Now, differentiate w.r.t. x
we get,
f^{'}(x)=5x^4-20x^3+45x^2-52x+11
Therefore, the answer is 5x^4-20x^3+45x^2-52x+11

Question:17 (3) Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?

Answer:

Given function is
y=(x^2 - 5x + 8) (x^3 + 7x + 9)
Now, take log on both the sides
\log y = \log (x^2-5x+8)+\log (x^3+7x+9)
Now, differentiate w.r.t. x
we get,
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11
Therefore, the answer is 5x^4-20x^3+45x^2-56x+11
And yes they all give the same answer

Question:18 If u, v and w are functions of x, then show that \frac{d}{dx} ( u,v,w) = \frac{du}{dx} v. w +u . \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx} in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Answer:

It is given that u, v and w are the functions of x
Let y = u.v.w
Now, we differentiate using product rule w.r.t x
First, take y = u.(vw)
Now,
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u -(i)
Now, again by the product rule
\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v
Put this in equation (i)
we get,
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)
Hence, by product rule we proved it

Now, by taking the log
Again take y = u.v.w
Now, take log on both sides
\log y = \log u + \log v + \log w
Now, differentiate w.r.t. x
we get,
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} = (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)
Hence, we proved it by taking the log


Class 12 Maths Chapter 5 NCERT solutions: Exercise:5.6

Question:1 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

x = 2at^2, y = at^4

Answer:

Given equations are
x = 2at^2, y = at^4
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(2at^2)}{dt}= 4at
Similarly,
\frac{dy}{dt}=\frac{d(at^4)}{dt}= 4at^3
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{4at^3}{4at} = t^2
Therefore, the answer is \frac{dy}{dx}= t^2

Question:2 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

x= a \cos \theta , y = b \cos \theta

Answer:

Given equations are
x= a \cos \theta , y = b \cos \theta
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}
Therefore, answer is \frac{dy}{dx}= \frac{b}{a}

Question:3 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx . x = \sin t , y = \cos 2 t

Answer:

Given equations are
x = \sin t , y = \cos 2 t
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t
Similarly,
\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t \ \ \ \ \ (\because \sin 2x = \sin x\cos x)
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t
Therefore, the answer is \frac{dy}{dx} = -4\sin t

Question:4 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx

x = 4t , y = 4/t

Answer:

Given equations are
x = 4t , y = 4/t
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4
Similarly,
\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}
Therefore, the answer is \frac{dy}{dx} = \frac{-1}{t^2}

Question:5 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta

Answer:

Given equations are
x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}
Therefore, answer is \frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}


Question:6 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )

Answer:

Given equations are
x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)
Similarly,
\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})
Therefore, the answer is \frac{dy}{dx}=-\cot \frac{\theta}{2}

Question:7 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}

Answer:

Given equations are
x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}
=\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}
=\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)
Similarly,
\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}
=\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}
=\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}} = \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}
= \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} = \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}
=\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}
(\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)
=\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}

\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t \left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )

Therefore, the answer is \frac{dy}{dx} = -\cot 3t

Question:8 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a ( \cos t + \log \tan t/2 ),y = a \sin t

Answer:

Given equations are
x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})
= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})
=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})
Similarly,
\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t
Therefore, the answer is \frac{dy}{dx} = \tan t

Question:9 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a \sec \theta , y = b \ tan \theta

Answer:

Given equations are
x = a \sec \theta , y = b \ tan \theta
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}
Therefore, the answer is \frac{dy}{dx} = \frac{b cosec \theta}{a}

Question:10 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )

Answer:

Given equations are
x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta
Therefore, the answer is \frac{dy}{dx}= \tan \theta

Question:11 If x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}} , show that dy/dx = - y /x

Answer:

Given equations are
x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}

xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\ sin^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c

differentiating with respect to x

x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}


Class 12 Maths Chapter 5 NCERT solutions: Exercise: 5.7

Question:1 Find the second order derivatives of the functions given in Exercises 1 to 10.

x^2 + 3x+ 2

Answer:

Given function is
y=x^2 + 3x+ 2
Now, differentiation w.r.t. x
\frac{dy}{dx}= 2x+3
Now, second order derivative
\frac{d^2y}{dx^2}= 2
Therefore, the second order derivative is \frac{d^2y}{dx^2}= 2

Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.

x ^{20}

Answer:

Given function is
y=x ^{20}
Now, differentiation w.r.t. x
\frac{dy}{dx}= 20x^{19}
Now, the second-order derivative is
\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}
Therefore, second-order derivative is \frac{d^2y}{dx^2}= 380x^{18}

Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.

x \cos x

Answer:

Given function is
y = x \cos x
Now, differentiation w.r.t. x
\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x
Now, the second-order derivative is
\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x
Therefore, the second-order derivative is \frac{d^2y}{dx^2}= -2\sin x - x\sin x

Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.

\log x

Answer:

Given function is
y=\log x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{1}{x}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{x^2}
Therefore, second order derivative is \frac{d^2y}{dx^2}= \frac{-1}{x^2}

Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.

x ^3 \log x

Answer:

Given function is
y=x^3\log x
Now, differentiation w.r.t. x
\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2
Now, the second-order derivative is
\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)
Therefore, the second-order derivative is \frac{d^2y}{dx^2} = x(6.\log x+5)

Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.

e ^x \sin5 x

Answer:

Given function is
y= e^x\sin 5x
Now, differentiation w.r.t. x
\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)
Now, second order derivative is
\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))
= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)
=2e^x(5\cos5x-12\sin5x)
Therefore, second order derivative is \frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)

Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.

e ^{6x}\cos 3x

Answer:

Given function is
y= e^{6x}\cos 3x
Now, differentiation w.r.t. x
\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)
Now, second order derivative is
\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)
= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)
e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)
Therefore, second order derivative is \frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)

Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.

\tan ^{-1} x

Answer:

Given function is
y = \tan^{-1}x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}

Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.

\log (\log x )

Answer:

Given function is
y = \log(\log x)
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}

Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.

\sin (\log x )

Answer:

Given function is
y = \sin(\log x)
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}
Now, second order derivative is
Using Quotient rule
\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}

Question:11 If y = 5 \cos x - 3 \sin x prove that \frac{d^2y}{dx^2}+y = 0

Answer:

Given function is
y = 5 \cos x - 3 \sin x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x
Now, the second-order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}=-5\cos x+3\sin x
Now,
\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0
Hence proved

Question:12 If y = \cos ^{-1} x Find \frac{d ^2 y }{dx^2 } in terms of y alone.

Answer:

Given function is
y = \cos ^{-1} x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2} -(i)
Now, we want \frac{d^2y}{dx^2} in terms of y
y = \cos ^{-1} x
x = \cos y
Now, put the value of x in (i)
\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y
(\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)
Therefore, answer is \frac{d^2y}{dx^2} = -2\cot y cosec y

Question:13 If y = 3 \cos (\log x) + 4 \sin (\log x) , show that x^2 y_2 + xy_1 + y = 0

Answer:

Given function is
y = 3 \cos (\log x) + 4 \sin (\log x)
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}
=\frac{4\cos (\log x)-3\sin(\log x)}{x} -(i)
Now, second order derivative is
By using the Quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}
=\frac{-\sin(\log x)+7\cos (\log x)}{x^2} -(ii)
Now, from equation (i) and (ii) we will get y_1 \ and \ y_2
Now, we need to show
x^2 y_2 + xy_1 + y = 0
Put the value of y_1 \ and \ y_2 from equation (i) and (ii)
x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x) +4\sin(\log x)
-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\ cos (\log x) +4\sin(\log x)
=0
Hence proved

Question:14 If y = A e ^{mx} + Be ^{nx} , show that \frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0

Answer:

Given function is
y = A e ^{mx} + Be ^{nx}
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx} -(ii)
Now, we need to show
\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0
Put the value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx} from equation (i) and (ii)
m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})
m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx} +mnBe^{nx}
=0
Hence proved

Question:15 If y = 500 e ^{7x} + 600 e ^{- 7x } , show that \frac{d^2 y}{dx ^2} = 49 y
Answer:

Given function is
y = 500 e ^{7x} + 600 e ^{- 7x }
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}
= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x} -(ii)
Now, we need to show
\frac{d^2 y}{dx ^2} = 49 y
Put the value of \frac{d^2y}{dx^2} from equation (ii)
24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})
= 24500e^{7x}+29400e^{-7x}
Hence, L.H.S. = R.H.S.
Hence proved

Question:16 If e ^y (x+1) = 1 show that \frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2

Answer:

Given function is
e ^y (x+1) = 1
We can rewrite it as
e^y = \frac{1}{x+1}
Now, differentiation w.r.t. x
\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2} -(ii)
Now, we need to show
\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2
Put value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx} from equation (i) and (ii)
\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2
=\frac{1}{(x+1)^2}
Hence, L.H.S. = R.H.S.
Hence proved

Question:17 If y = (\tan^{-1} x)^2 show that (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2

Answer:

Given function is
y = (\tan^{-1} x)^2
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2} -(i)
Now, the second-order derivative is
By using the quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2} -(ii)
Now, we need to show
(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2
Put the value from equation (i) and (ii)
(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \Rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2
Hence, L.H.S. = R.H.S.
Hence proved

Class 12 Maths Chapter 5 NCERT solutions: Excercise: 5.8

Question:1 Verify Rolle’s theorem for the function f (x) = x^2 + 2x - 8, x \epsilon [- 4, 2].

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
f (x) = x^2 + 2x - 8
Now, being a polynomial function, f (x) = x^2 + 2x - 8 is both continuous in [-4,2] and differentiable in (-4,2)
Now,
f (-4) = (-4)^2 + 2(-4) - 8= 16-8-8=16-16=0
Similalrly,
f (2) = (2)^2 + 2(2) - 8= 4+4-8=8-8=0
Therefore, value of f (-4) = f(2)=0 and value of f(x) at -4 and 2 are equal
Now,
According to roll's theorem their is point c , c \ \epsilon (-4,2) such that f^{'}(c)=0
Now,
f^{'}(x)=2x+2\\ f^{'}(c)=2c+2\\ f^{'}(c)=0\\ 2c+2=0\\ c = -1
And c = -1 \ \epsilon \ (-4,2)
Hence, Rolle's theorem is verified for the given function f (x) = x^2 + 2x - 8

Question:2 (1) Examine if Rolle’s theorem is applicable to any of the following functions. Can
you say some thing about the converse of Rolle’s theorem from these example?
f (x) = [x] \: \: for \: \: x \epsilon [ 5,9]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, the function is not differential in (5,9)
Hence, Rolle's theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [5,9]

Question:2 (2) Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?

f (x) = [x] \: \:for \: \: x \epsilon [ -2,2]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Rolle's theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [-2,2]

Question:2 (3) Examine if Rolle’s theorem is applicable to any of the following functions. Can
you say some thing about the converse of Rolle’s theorem from these example?
f (x) = x^2 - 1 \: \:for \: \: x \epsilon [ 1,2]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then there exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
f (x) = x^2-1
Now, being a polynomial , function f (x) = x^2-1 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=1^2-1 = 1-1 = 0
And
f(2)=2^2-1 = 4-1 = 3
Therefore, f(1)\neq f(2)
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [-2,2]

Question:3 If f ; [ -5 ,5] \rightarrow R is a differentiable function and if f ' (x) does not vanish
anywhere, then prove that f (-5) \neq f(5)

Answer:

It is given that
f ; [ -5 ,5] \rightarrow R is a differentiable function
Now, f is a differential function. So, f is also a continuous function
We obtain the following results
a ) f is continuous in [-5,5]
b ) f is differentiable in (-5,5)
Then, by Mean value theorem we can say that there exist a c in (-5,5) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(5)-f(-5)}{5-(-5)}\\ f^{'}(c)= \frac{f(5)-f(-5)}{10}\\ 10f^{'}(c)= f(5)-f(-5)
Now, it is given that f ' (x) does not vanish anywhere
Therefore,
10f^{'}(c)\neq 0\\ f(5)-f(-5) \neq 0\\ f(5)\neq f(-5)
Hence proved

Question:4 Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3 in the interval [a, b], where
a = 1 and b = 4.

Answer:

Condition for M.V.T.
If f ; [ a ,b] \rightarrow R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c in (a,b) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
It is given that
f (x) = x^2 - 4x - 3 and interval is [1,4]
Now, f is a polynomial function , f (x) = x^2 - 4x - 3 is continuous in[1,4] and differentiable in (1,4)
And
f(1)= 1^2-4(1)-3= 1-7= -6
and
f(4)= 4^2-4(4)-3= 16-16-3= 16-19=-3
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(4)-f(1)}{4-1}\\ f^{'}(c)= \frac{-3-(-6)}{3}\\ f^{'}(c)= \frac{3}{3}\\ f^{'}(c)= 1
Now,
f^{'}(x) =2x-4\\ f^{'}(c)-2c-4\\ 1=2c-4\\ 2c=5\\ c=\frac{5}{2}
And c=\frac{5}{2} \ \epsilon \ (1,4)
Hence, mean value theorem is verified for the function f (x) = x^2 - 4x - 3

Question:5 Verify Mean Value Theorem, if f (x) = x^3 - 5x^2- 3x in the interval [a, b], where
a = 1 and b = 3. Find all c \epsilon (1,3) for which f '(c) = 0.

Answer:

Condition for M.V.T.
If f ; [ a ,b] \rightarrow R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, their exist a c in (a,b) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
It is given that
f (x) = x^3 - 5x^2- 3x and interval is [1,3]
Now, f being a polynomial function , f (x) = x^3 - 5x^2- 3x is continuous in[1,3] and differentiable in (1,3)
And
f(1)= 1^3-5(1)^2-3(1)= 1-5-3=1-8=-7
and
f(3)= 3^3-5(3)^2-3(3)= 27-5.9-9= 18-45=-27
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(3)-f(1)}{3-1}\\ f^{'}(c)= \frac{-27-(-7)}{2}\\ f^{'}(c)= \frac{-20}{2}\\ f^{'}(c)= -10
Now,
f^{'}(x) =3x^2-10x-3\\ f^{'}(c)=3c^2-10c-3\\ -10=3c^2-10c-3\\ 3c^2-10c+7=0\\ 3c^2-3c-7c+7=0\\ (c-1)(3c-7)=0\\ c = 1 \ \ \ and \ \ \ c = \frac{7}{3}
And c=1,\frac{7}{3} \ and \ \frac{7}{3}\ \epsilon \ (1,3)
Hence, mean value theorem is varified for following function f (x) = x^3 - 5x^2- 3x and c=\frac{7}{3} is the only point where f '(c) = 0

Question:6 Examine the applicability of Mean Value Theorem for all three functions given in
the above exercise 2.

Answer:

According to Mean value theorem function
f:[a,b]\rightarrow R must be
a ) continuous in given closed interval say [a,b]
b ) differentiable in given open interval say (a,b)
Then their exist a c \ \epsilon \ (x,y) such that
f^{'}(c)= \frac{f(b)-f(a)}{b-a}
If all these conditions are satisfies then we can verify mean value theorem
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (5,9)
Hence, Mean value theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [5,9]

Similaly,
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Mean value theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [-2,2]

Similarly,
Given function is
f (x) = x^2-1
Now, being a polynomial , function f (x) = x^2-1 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=1^2-1 = 1-1 = 0
And
f(2)=2^2-1 = 4-1 = 3
Now,
f^{'}(c)= \frac{f(b)-f(a)}{b-a}
f^{'}(c)= \frac{f(2)-f(1)}{2-1}\\ f^{'}(c)=\frac{3-0}{1}\\ f^{'}(c)= 3
Now,
f^{'}(x)= 2x\\ f^{'}(c)=2c\\ 3=2c\\ c=\frac{3}{2}
And c=\frac{3}{2} \ \epsilon \ (1,2)
Therefore, mean value theorem is applicable for the function f (x) = x^2-1


NCERT class 12 continuity and differentiability ncert solutions Miscellaneous Excercise

Question:1 Differentiate w.r.t. x the function in Exercises 1 to 11.

( 3x^2 - 9x + 5 )^9

Answer:

Given function is
f(x)=( 3x^2 - 9x + 5 )^9
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d((3x^2-9x+5)^9)}{dx}= 9(3x^2-9x+5)^8.(6x-9)
= 27(2x-3)(3x^2-9x+5)^8
Therefore, differentiation w.r.t. x is 27(3x^2-9x+5)^8(2x-3)

Question:2 Differentiate w.r.t. x the function in Exercises 1 to 11.

\sin ^3 x + \cos ^6 x

Answer:

Given function is
f(x)= \sin ^3 x + \cos ^6 x
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^3x +\cos^6x)}{dx}=3\sin^2x.\frac{d(\sin x)}{dx}+6\cos^5x.\frac{d(\cos x)}{dx}
=3\sin^2x.\cos x+6\cos^5x.(-\sin x)
=3\sin^2x\cos x- 6\cos^5x\sin x = 3\sin x\cos x(\sin x- 2\cos ^4x)

Therefore, differentiation w.r.t. x is 3\sin x\cos x(\sin x- 2\cos ^4x)

Question:3 Differentiate w.r.t. x the function in Exercises 1 to 11.

( 5 x) ^{ 3 \cos 2x }

Answer:

Given function is
y=( 5 x) ^{ 3 \cos 2x }
Take, log on both the sides
\log y = 3\cos 2x\log 5x
Now, differentiation w.r.t. x is
By using product rule
\frac{1}{y}.\frac{dy}{dx} = 3.(-2\sin 2x)\log 5x + 3\cos 2x.\frac{1}{5x}.5= -6\sin2x\log 5x +\frac{3\cos 2x}{x}\\ \frac{dy}{dx} = y.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )\\ \frac{dy}{dx} = (5x)^{3\cos 2x}.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )

Therefore, differentiation w.r.t. x is (5x)^{3\cos 2x}.\left ( \frac{3\cos 2x}{x}-6\sin2x\log 5x \right )

Question:4 Differentiate w.r.t. x the function in Exercises 1 to 11.

\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1

Answer:

Given function is
f(x)=\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^{-1}x\sqrt x)}{dx}=\frac{1}{\sqrt{1-(x\sqrt x)^2}}.\frac{d(x\sqrt x)}{dx}
=\frac{1}{\sqrt{1-x^3}}.\left ( 1.\sqrt x+x\frac{1}{2\sqrt x} \right )
=\frac{1}{\sqrt{1-x^3}}.\left ( \frac{3\sqrt x}{2} \right )
=\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}

Therefore, differentiation w.r.t. x is \frac{3}{2}.\sqrt{\frac{x}{1-x^3}}

Question:5 Differentiate w.r.t. x the function in Exercises 1 to 11.

\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2

Answer:

Given function is
f(x)=\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2
Now, differentiation w.r.t. x is
By using the Quotient rule
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}})}{dx}=\frac{\frac{d(\cos^{-1}\frac{x}{2})}{dx}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{d(\sqrt{2x+7})}{dx}}{(\sqrt{2x+7})^2}\\ f^{'}(x) = \frac{\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{1}{2}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{1}{2.\sqrt{2x+7}}.2}{2x+7}\\ f^{'}(x)= -\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]

Therefore, differentiation w.r.t. x is -\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]

Question:6 Differentiate w.r.t. x the function in Exercises 1 to 11.

\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2

Answer:

Given function is
f(x)=\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2
Now, rationalize the [] part
\left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ]= \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} .\frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}\right ]

=\frac{(\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x })^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2} \ \ \ \ \ \ (Using \ (a-b)(a+b)=a^2-b^2)

=\frac{((\sqrt { 1+ \sin x })^2+ (\sqrt { 1- \sin x })^2+2(\sqrt { 1+ \sin x })(\sqrt { 1- \sin x }))}{1+\sin x-1+\sin x}
(Using \ (a+b)^2=a^2+b^2+2ab)
=\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x} }{2\sin x}

=\frac{2(1+\cos x)}{2\sin x} = \frac{1+\cos x}{\sin x}

=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \ \ \ \ \ (\because 2\cos^2= 1+\cos2x \ and \ \sin2x = 2\sin x\cos x)

=\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot \frac{x}{2}
Given function reduces to
f(x) = \cot^{-1}(\cot \frac{x}{2})\\ f(x) = \frac{x}{2}
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{x}{2})}{dx} = \frac{1}{2}
Therefore, differentiation w.r.t. x is \frac{1}{2}

Question:7 Differentiate w.r.t. x the function in Exercises 1 to 11. ( \log x )^{ \log x } , x > 1

Answer:

Given function is
y=( \log x )^{ \log x } , x > 1
Take log on both sides
\log y=\log x\log( \log x )
Now, differentiate w.r.t.
\frac{1}{y}.\frac{dy}{dx}= \frac{1}{x}.\log (\log x)+\log x.\frac{1}{\log x}.\frac{1}{x} = \frac{\log x+1}{x}
\frac{dy}{dx} = y.\left ( \frac{\log x+1}{x} \right )\\
\frac{dy}{dx} = (\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\
Therefore, differentiation w.r.t x is (\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\

Question:8 \cos ( a \cos x + b \sin x ) , for some constant a and b.

Answer:

Given function is
f(x)=\cos ( a \cos x + b \sin x )
Now, differentiation w.r.t x
f^{'}(x)= \frac{d(f(x))}{dx}= \frac{d(\cos(a\cos x+ b \sin x))}{dx}
= -\sin(a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}
= -\sin(a\cos x+b\sin x).(-a\sin x+b\cos x)
= (a\sin x-b\cos x)\sin(a\cos x+b\sin x).
Therefore, differentiation w.r.t x (a\sin x-b\cos x)\sin(a\cos x+b\sin x)

Question: 9 (\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}

Answer:

Given function is
y=(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}
Take log on both the sides
\log y=(\sin x - \cos x)\log (\sin x - \cos x)
Now, differentiate w.r.t. x
\frac{1}{y}.\frac{dy}{dx} = \frac{d(\sin x-\cos x)}{dx}.\log(\sin x- \cos x)+(\sin x- \cos x).\frac{d(\log(\sin x- \cos x))}{dx}
\frac{1}{y}.\frac{dy}{dx} =(\cos x -(-\sin x)).\log(\sin x-\cos x)+(\sin x- \cos x).\frac{(\cos x -(-\sin x))}{(\sin x- \cos x)}
\frac{dy}{dx} =y.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )
\frac{dy}{dx} =(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )
Therefore, differentiation w.r.t x is (\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right ), sinx>cosx

Question:10 x ^x + x ^a + a ^x + a ^a , for some fixed a > 0 and x > 0

Answer:

Given function is
f(x)=x ^x + x ^a + a ^x + a ^a
Lets take
u = x^x
Now, take log on both sides
\log u = x \log x
Now, differentiate w.r.t x
\frac{1}{u}.\frac{du}{dx}= \frac{dx}{dx}.\log x+x.\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx}= 1.\log x+x.\frac{1}{x}\\ \\ \frac{du}{dx}= y.(\log x+1)\\ \\ \frac{du}{dx}= x^x.(\log x+1) -(i)
Similarly, take v = x^a
take log on both the sides
\log v = a\log x
Now, differentiate w.r.t x
\frac{1}{v}.\frac{dv}{dx}= a.\frac{d(\log x)}{dx}=a.\frac{1}{x}= \frac{a}{x}\\ \\ \frac{dv}{dx}= v.\frac{a}{x}\\ \\ \frac{dv}{dx}= x^a.\frac{a}{x} -(ii)

Similarly, take z = a^x
take log on both the sides
\log z = x\log a
Now, differentiate w.r.t x
\frac{1}{z}.\frac{dz}{dx}=\log a.\frac{d(x)}{dx}=\log a.1= \log a\\ \\ \frac{dz}{dx}= z.\log a\\ \\ \frac{dz}{dx}= a^x.\log a -(iii)

Similarly, take w = a^a
take log on both the sides
\log w = a\log a= \ constant
Now, differentiate w.r.t x
\frac{1}{w}.\frac{dw}{dx}= a.\frac{d(a\log a)}{dx}= 0\\ \\ \frac{dw}{dx} = 0 -(iv)
Now,
f(x)=u+v+z+w
f^{'}(x) = \frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}
Put values from equation (i) , (ii) ,(iii) and (iv)
f^{'}(x)= x^x(\log x+1)+ax^{a-1}+a^x\log a
Therefore, differentiation w.r.t. x is x^x(\log x+1)+ax^{a-1}+a^x\log a

Question: 11 x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3

Answer:

Given function is
f(x)=x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3
take u=x ^{x^2 -3}
Now, take log on both the sides
\log u=(x^2-3)\log x
Now, differentiate w.r.t x
\frac{1}{u}.\frac{du}{dx}= \frac{d(x^2-3)}{dx}.\log x+(x^2-3).\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx} = 2x.\log x+(x^2-3).\frac{1}{x}\\ \\ \frac{1}{u}.\frac{du}{dx} = \frac{2x^2\log x+x^2-3}{x}\\ \\ \frac{du}{dx}= u.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\ \\ \frac{du}{dx}= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\ \\ -(i)
Similarly,
take v=(x-3)^x^2\\
Now, take log on both the sides
\log v=x^2\log (x-3)
Now, differentiate w.r.t x
\frac{1}{v}.\frac{dv}{dx}= \frac{d(x^2)}{dx}.\log (x-3)+x^2.\frac{d(\log (x-3))}{dx}\\ \\ \frac{1}{v}.\frac{dv}{dx} = 2x.\log (x-3)+x^2.\frac{1}{(x-3)}\\ \\ \frac{1}{v}.\frac{dv}{dx} = 2x\log(x-3)+\frac{x^2}{x-3}\\ \\ \frac{dv}{dx}= v.\left ( 2x\log(x-3)+\frac{x^2}{x-3} \right )\\ \\ \frac{dv}{dx}= (x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )\\ \\ -(ii)
Now
f(x)= u + v
f^{'}(x)= \frac{du}{dx}+\frac{dv}{dx}
Put the value from equation (i) and (ii)
f^{'}(x)= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )
Therefore, differentiation w.r.t x is x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )

Question:12 Find dy/dx if y = 12 (1 - \cos t), x = 10 (t - \sin t), -\frac{\pi }{2} <t< \frac{\pi }{2}

Answer:

Given equations are
y = 12 (1 - \cos t), x = 10 (t - \sin t),
Now, differentiate both y and x w.r.t t independently
\frac{dy}{dt}=\frac{d(12(1-\cos t))}{dt}= -12(-\sin t)=12\sin t
And
\frac{dx}{dt}=\frac{d(10(t-\sin t))}{dt}= 10-10\cos t
Now
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5}.\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{6}{5}.\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}\\ \\
(\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)
\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}
Therefore, differentiation w.r.t x is \frac{6}{5}.\cot \frac{t}{2}

Question:13 Find dy/dx if y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1

Answer:

Given function is
y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1
Now, differentiatiate w.r.t. x
\frac{dy}{dx}= \frac{d(sin ^{-1} x + sin^{-1} \sqrt{1- x^2})}{dx} = \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}}.\frac{d(\sqrt{1-x^2})}{dx}\\ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-1+x^2}}.\frac{1}{2\sqrt{1-x^2}}.(-2x)\\ \\ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\\ \frac{dy}{dx}= 0
Therefore, differentiatiate w.r.t. x is 0

Question:14 If x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}

Answer:

Given function is
x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0
x\sqrt{1+y} = - y\sqrt{1+x}
Now, squaring both sides
(x\sqrt{1+y})^2 = (- y\sqrt{1+x})^2\\ x^2(1+y)=y^2(1+x)\\ x^2+x^2y=y^2x+y^2\\ x^2-y^2=y^2x-x^2y\\ (x-y)(x+y) = -xy(x-y) \\ x+y =-xy\\ y = \frac{-x}{1+x}
Now, differentiate w.r.t. x is
\frac{dy}{dx} = \frac{d(\frac{-x}{1+x})}{dx}= \frac{-1.(1+x)-(-x).(1)}{(1+x)^2}= \frac{-1}{(1+x)^2}
Hence proved

Question:15 If (x - a)^2 + (y - b)^2 = c^2 , for some c > 0, prove that \frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}}\: is a constant independent of a and b.

Answer:

Given function is
(x - a)^2 + (y - b)^2 = c^2
(y - b)^2 = c^2-(x - a)^2 - (i)
Now, differentiate w.r.t. x
\frac{d((x-a)^2)}{dx}+\frac{((y-b)^2)}{dx}=\frac{d(c^2)}{dx}\\ \\ 2(x-a)+2(y-b).\frac{dy}{dx}=0\\ \\ \frac{dy}{dx} = \frac{a-x}{y-b} -(ii)
Now, the second derivative
\frac{d^2y}{dx^2} = \frac{\frac{d(a-x)}{dx}.(y-b) -(a-x).\frac{d(y-b)}{dx}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} =\frac{ (-1).(y-b)-(a-x).\frac{dy}{dx}}{(y-b)^2}\\ \\
Now, put values from equation (i) and (ii)
\frac{d^2y}{dx^2} =\frac{-(y-b)-(a-x).\frac{a-x}{y-b}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} = \frac{-((y-b)^2+(a-x)^2)}{(y-b)^\frac{3}{2}} = \frac{-c^2}{(y-b)^\frac{3}{2}} (\because (x - a)^2 + (y - b)^2 = c^2)
Now,
\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}} = \frac{\left ( 1+\left ( \frac{x-a}{y-b} \right )^2 \right )^\frac{3}{2}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{\frac{\left ( (y-b)^2 +(x-a)^2\right )^\frac{3}{2}}{(y-b)^\frac{3}{2}}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{(c^2)^\frac{3}{2}}{-c^2}= \frac{c^3}{-c^2}= c (\because (x - a)^2 + (y - b)^2 = c^2)
Which is independent of a and b
Hence proved

Question:16 If \cos y = x \cos (a + y) , with \cos a \neq \pm 1 , prove that \frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }

Answer:

Given function is
\cos y = x \cos (a + y)
Now, Differentiate w.r.t x
\frac{d(\cos y)}{dx} = \frac{dx}{dx}.\cos(a+y)+x.\frac{d(\cos (a+y))}{dx}\\ \\ -\sin y \frac{dy}{dx} = 1.\cos (a+y)+x.(-\sin(a+y)).\frac{dy}{dx}\\ \\ \frac{dy}{dx}.(x\sin(a+y)-\sin y)= \cos(a+y)\\ \\ \frac{dy}{dx}.(\frac{\cos y}{\cos (a+b)}.\sin(a+y)-\sin y)= \cos(a+b) \ \ \ \ \ (\because x = \frac{\cos y}{\cos (a+b)})\\ \\ \frac{dy}{dx}.(\cos y\sin(a+y)-\sin y\cos(a+y))=\cos^2(a+b)\\ \\ \frac{dy}{dx}.(\sin(a+y-y))=\cos^2(a+b) \ \ \ \ \ \ \ (\because \cos A\sin B-\sin A\cos B = \sin(A-B))\\ \\ \frac{dy}{dx}= \frac{\cos^2(a+y)}{\sin a}
Hence proved

Question:17 If x = a (\cos t + t \sin t) and y = a (\sin t - t \cos t), find \frac{d^2 y }{dx^2 }

Answer:

Given functions are
x = a (\cos t + t \sin t) and y = a (\sin t - t \cos t)
Now, differentiate both the functions w.r.t. t independently
We get
\frac{dx}{dt} = \frac{d(a(\cos t +t\sin t))}{dt}= a(-\sin t)+a(\sin t+t\cos t)
=-a\sin t+a\sin t+at\cos t = at\cos t
Similarly,
\frac{dy}{dt} = \frac{d(a(\sin t - t\cos t))}{dt}= a\cos t -a(\cos t+t(-\sin t))
= a\cos t -a\cos t+at\sin t =at\sin t
Now,
\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at \cos t} = \tan t
Now, the second derivative
\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \sec^2 t.\frac{dt}{dx}=\frac{\sec^2t.\sec t }{at}=\frac{\sec^3t}{at}
(\because \frac{dx}{dt} = at\cos t \Rightarrow \frac{dt}{dx}= \frac{1}{at\cos t}=\frac{\sec t}{at})
Therefore, \frac{d^2y}{dx^2}=\frac{\sec^3t}{at}

Question:18 If f (x) = |x|^3 , show that f ''(x) exists for all real x and find it.

Answer:

Given function is
f (x) = |x|^3
f(x)\left\{\begin{matrix} -x^3 & x<0\\ x^3 & x>0 \end{matrix}\right.
Now, differentiate in both the cases
f(x)= x^3\\ f^{'}(x)=3x^2\\ f^{''}(x)= 6x
And
f(x)= -x^3\\ f^{'}(x)=-3x^2\\ f^{''}(x)= -6x
In both, the cases f ''(x) exist
Hence, we can say that f ''(x) exists for all real x
and values are
f^{''}(x)\left\{\begin{matrix} -6x &x<0 \\ 6x& x>0 \end{matrix}\right.

Question:19 Using mathematical induction prove that \frac{d}{dx} (x^n) = nx ^{n-1} for all positive integers n.

Answer:

Given equation is
\frac{d}{dx} (x^n) = nx ^{n-1}
We need to show that \frac{d}{dx} (x^n) = nx ^{n-1} for all positive integers n
Now,
For ( n = 1) \Rightarrow \frac{d(x^{1})}{dx}= 1.x^{1-1}= 1.x^0=1
Hence, true for n = 1
For (n = k) \Rightarrow \frac{d(x^{k})}{dx}= k.x^{k-1}
Hence, true for n = k
For ( n = k+1) \Rightarrow \frac{d(x^{k+1})}{dx}= \frac{d(x.x^k)}{dx}
= \frac{d(x)}{dx}.x^k+x.\frac{d(x^k)}{dx}
= 1.x^k+x.(k.x^{k-1}) = x^k+k.x^k= (k+1)x^k
Hence, (n = k+1) is true whenever (n = k) is true
Therefore, by the principle of mathematical induction we can say that \frac{d}{dx} (x^n) = nx ^{n-1} is true for all positive integers n

Question:20 Using the fact that \sin (A + B) = \sin A \cos B + \cos A \sin B and the differentiation,
obtain the sum formula for cosines.

Answer:

Given function is
\sin (A + B) = \sin A \cos B + \cos A \sin B
Now, differentiate w.r.t. x
\frac{d(\sin(A+B))}{dx} = \frac{d\sin A}{dx}.\cos B+\sin A.\frac{d\cos B}{dx}+\frac{d\cos A}{dx}.\sin B+\cos A.\frac{d\sin B}{dx}
\cos (A+b)\frac{d(A+B)}{dx} =\frac{dA}{dx}(\cos A\cos B-\sin A\cos B)+\frac{dB}{dx}(\cos A \sin B-\sin A\sin B)
=(\cos A \sin B-\sin A\sin B).\frac{d(A+B)}{dx}
\cos(A+B)= \cos A\sin B-\sin A\cos B
Hence, we get the formula by differentiation of sin(A + B)

Question:21 Does there exist a function which is continuous everywhere but not differentiable
at exactly two points? Justify your answer.

Answer:

Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere and sum of two continuous function is also a continuous function
Therefore, our function f(x) is continuous
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
\lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}= \lim_{h\rightarrow 0^-}\frac{|h|+|h+1|-|1|}{h}
=\lim_{h\rightarrow 0^-}\frac{-h-(h+1)-1}{h}= 0 (|h| = - h \ because\ h < 0)
R.H.L. at x = 0
\lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}= \lim_{h\rightarrow 0^+}\frac{|h|+|h+1|-|1|}{h}
=\lim_{h\rightarrow 0^+}\frac{h+h+1-1}{h}= \lim_{h\rightarrow 0^+}\frac{2h}{h}= 2 (|h| = h \ because \ h > 0)
R.H.L. is not equal to L.H.L.
Hence.at x = 0 is the function is not differentiable
Now, Similarly
R.H.L. at x = -1
\lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^+}\frac{f(-1+h)-f(-1)}{h}= \lim_{h\rightarrow 0^+}\frac{|-1+h|+|h|-|-1|}{h}
=\lim_{h\rightarrow 0^+}\frac{1-h+h-1}{h}= \lim_{h\rightarrow 0^+}\frac{0}{h}= 0 (|h| = h \ because \ h > 0)
L.H.L. at x = -1
\lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\rightarrow 0^-}\frac{|-1+h|+|h|-|1|}{h}
=\lim_{h\rightarrow 1^+}\frac{1-h-h-1}{h}= \lim_{h\rightarrow 0^+}\frac{-2h}{h}= -2 (|h| = - h \ because\ h < 0)
L.H.L. is not equal to R.H.L, so not differentiable at x=-1

Hence, exactly two points where it is not differentiable

Question:22 If y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix} , prove that dy/dx = \begin{vmatrix} f '(x) & g'(x) & h' (x) \\ l& m &n \\ a& b &c \end{vmatrix}

Answer:

Given that
y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}
We can rewrite it as
y = f(x)(mc-bn)-g(x)(lc-an)+h(x)(lb-am)
Now, differentiate w.r.t x
we will get
\frac{dy}{dx} = f^{'}(x)(mc-bn)-g^{'}(x)(lc-an)+h^{'}(x)(lb-am) \Rightarrow \begin{bmatrix} f^{'}(x) &g^{'}(x) &h^{'}(x) \\ l&m &n \\ a& b &c \end{bmatrix}
Hence proved

Question:23 If y = e ^{a \cos ^{-1}x} , -1 \leqx \leq 1 , show that ( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0

Answer:

Given function is
y = e ^{a \cos ^{-1}x} , -1 \leqx \leq 1

Now, differentiate w.r.t x we will get
\frac{dy}{dx}= \frac{d(e^{a\cos^{-1}x})}{dx}.\frac{d(a\cos^{-1}x)}{dx} = e^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}} -(i)
Now, again differentiate w.r.t x
\frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx}= \frac{-ae^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}.\sqrt{1-x^2}+ae^{a\cos^{-1}x}.\frac{1.(-2x)}{2\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}
=\ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2} -(ii)
Now, we need to show that
( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0
Put the values from equation (i) and (ii)
(1-x^2).\left ( \ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2} \right )-x.\left ( \frac{-ae^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x}
a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}+\left ( \frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x} = 0
Hence proved

If you are looking for continuity and differentiability class 12 NCERT solutions of exercises then these are listed below.

Topics of NCERT class 12 maths chapter 5 Continuity and Differentiability

5.1 Introduction

5.2 Continuity

5.2.1 Algebra of continuous functions

5.3. Differentiability

5.3.1 Derivatives of composite functions

5.3.2 Derivatives of implicit functions

5.3.3 Derivatives of inverse trigonometric functions

5.4 Exponential and Logarithmic Functions

5.5. Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Form

The mathematical definition of Continuity and Differentiability -

Let f be a real function and c be a point in the domain of f. Then f is continuous at c if \lim_{x\rightarrow c}f(x)=f(c) . A function f is differentiable at point c in its domain if it is continuous at point c. A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b].

'Continuity and differentiability' is one of the very important and time-consuming chapters of the NCERT Class 12 maths syllabus. It contains 8 exercises with 121 questions and also 23 questions in the miscellaneous exercise. In this article, you will find all NCERT solutions for class 12 maths chapter 5 continuity and differentiability including miscellaneous exercises.

Also read,

NCERT exemplar solutions class 12 maths chapter 5

NCERT class 12 maths ch 5 question answer - Topics

The main topics covered in chapter 5 maths class 12 are:

  • Continuity

A function is continuous at a given point if the left-hand limit, right-hand limit and value of function exist and are equal. In this class 12 NCERT topics elaborate concepts related to continuity, point of discontinuity, algebra of continuous function. Continuity and Differentiability class 12 solutions include a comprehensive module of quality questions.

  • Differentiability

This ch 5 maths class 12 discuss differentiability concepts of different functions including derivatives of composite functions, derivatives of implicit functions, derivatives of inverse trigonometric functions. To get command on these concepts you can refer to NCERT solutions for class 12 maths chapter 5.

  • Exponential and Logarithmic Functions

This ch 5 maths class 12 also includes concepts of exponential and logarithmic functions including natural log and their graphical representation. maths class 12 chapter 5 also contains fundamental properties of the logarithmic function. You can refer to class 12 NCERT solutions for questions about these concepts.

  • Logarithmic Differentiation

this class 12 ncert chapter discusses a special technique of differentiation known as logarithmic differentiation. to get command of these concepts you can go through the NCERT solution for class 12 maths chapter 5.

  • Derivatives of Functions in Parametric Forms

concepts to differentiate a function which is not implicit and explicit but given in the parametric form are explained in this chapter. Continuity and Differentiability class 12 solutions include problems to understand the concepts.

ch 5 maths class 12 also discuss in detail the concepts of second-order derivative, mean value theorem, Rolle's theorem. for questions on these concepts, you can browse NCERT solutions for class 12 chapter 5.

Topics enumerated in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 5

Also read,

NCERT solutions for class 12 maths - Chapter wise

NCERT solutions for class 12 subject wise

NCERT Solutions class wise

NCERT Books and NCERT Syllabus

Tips to Use NCERT Solutions for Class 12 Maths Chapter 5

NCERT solutions for class 12 maths chapter 5 continuity and differentiability are very helpful in the preparation of this chapter. But here are some tips to get command on this chapter.

  • You should make sure that concepts related to 'limit' are clear to you as it forms the base for continuity.
  • First, go for the theorem and solved examples of continuity given in the NCERT textbook then try to solve exercise questions. You may find some difficulties in solving them. Go through the NCERT solutions for class 12 maths chapter 5 continuity and differentiability, it will help you to understand the concepts in a much easy way.
  • This chapter seems very easy but at the same time, the chances of silly mistakes are also high. So, it is advised to understand the theory and concepts properly before practicing questions of NCERT.
  • Once you are good in continuity, then go for the differentiability. Practice more and more questions to get command on it.
  • Differentiation is mostly formula-based, so practice NCERT questions, it won't take much effort to remember the formulas.

Also check,

Happy learning!!!

Frequently Asked Question (FAQs)

1. What are the important topics in chapter Continuity and Differentiability?

Basic concepts of continuity and differentiability, derivatives of composite functions, derivatives of implicit functions, derivatives of inverse trigonometric functions, exponential and logarithmic functions, logarithmic differentiation, derivatives of functions in parametric form are the important topics in this chapter. Practice these class 12 maths ch 5 question answer to command the concepts. 

2. What are the reasons to opt for ncert continuity and differentiability solution?

The maths chapter 5 class 12 NCERT solutions created by the experts at Careers360 offer numerous advantages to students preparing for their board exams. These solutions provide comprehensive explanations of each topic, which help students achieve high scores. Additionally, the solutions are based on the latest CBSE syllabus for the 2022-23 academic year. Furthermore, these solutions also assist students in preparing for other competitive exams such as JEE Main and JEE Advanced. For ease, Students can study continuity and differentiability pdf both online and offline

3. Which is the best book for CBSE class 12 maths ?

NCERT is the best book for CBSE class 12 maths. Most of the questions in CBSE board exam are directly asked from NCERT textbook. All you need to do is rigorous practice of all the problems given in the NCERT textbook.

4. How many exercises are included in ncert solutions class 12 maths chapter 5?

According to the given information, there are 8 exercises in NCERT Solutions for maths chapter 5 class 12 . The following is the number of questions in each exercise:

  1. Exercise 5.1: 34 questions

  2. Exercise 5.2: 10 questions

  3. Exercise 5.3: 15 questions

  4. Exercise 5.4: 10 questions

  5. Exercise 5.5: 18 questions

  6. Exercise 5.6: 11 questions

  7. Exercise 5.7: 17 questions

  8. Exercise 5.8: 6 questions

Additionally, there is a Miscellaneous Exercise with 23 questions.

5. What is the weightage of the chapter Continuity and Differentiability for CBSE board exam ?

Generally, Continuity and differntiability has 9% weightage in the 12th board final examination. if you want to obtain meritious marks or full marks then you should have good command on concepts that can be developed by practice therefore you should practice NCERT solutions and NCERT exercise solutions.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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