NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

 

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability: In CBSE class 11 you have already learnt the differentiation of certain functions like polynomial functions and trigonometric functions. In this article, you will get NCERT solutions for class 12 maths chapter 5 continuity and differentiability. Questions based on the topics like continuity, differentiability, and relations between them are covered in the solutions of NCERT for class 12 maths chapter 5 continuity and differentiability. In the NCERT textbook, there are 48 solved examples to understand the concepts. First, try to solve examples and understand the concept, then it will be easy to solve NCERT exercise questions. If you are finding difficulties in solving them, you can take help from CBSE NCERT solutions for class 12 maths chapter 5 continuity and differentiability article. Here, you will also learn the differentiation of inverse trigonometric functions, exponential functions, and logarithmic functions. Check all NCERT solutions at a single place which will help you to learn CBSE science and maths.

This chapter "continuity and differentiability" is a continuation of the differentiation of functions that you have already learnt in NCERT class XI. This chapter alone has 9% weightage in the 12th board final examination and the next chapters of calculus(44 % weightage in the final exam) also depend on the concepts of this chapter. This chapter is very important for the exam point of view as well as the application point of view. To get command on this chapter you should solve the miscellaneous exercise also. In this solutions of NCERT for class 12 maths chapter 5 continuity and differentiability article, you will get solutions for miscellaneous exercise too. This chapter contains 8 exercises with 121 questions. You will find all these CBSE NCERT solutions for class 12 maths chapter 5 continuity and differentiability explained in a detailed manner.

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

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

Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability 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

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


Hence, function is continuous for all values of x < 0

case (ii)  x = 0

L.H.L at x= 0

R.H.L. at x = 0

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

case (iii)  k > 0


Hence , function is continuous for all values of x > 0

case (iv) k < 1


 Hence , function is continuous for all values of x < 1

case (v)  k > 1


 Hence , function is continuous for all values of x > 1

case (vi)  x = 1



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 byf (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 byf (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

CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 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
f(t) = 2\sqrt\cot t
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}}                      

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 solutions for class 12 maths chapter 5 Continuity and Differentiability 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}}

Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability 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:6Differentiate 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

CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 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

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 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)}