# NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

NCERT solutions for class 11 maths chapter 13 Limits and Derivatives: This chapter introduces an important area of mathematics called calculus. This is a branch of mathematics which deal with the study of change in the value of a function as the points in the domain change.  In this article, you will get NCERT solutions for class 11 maths chapter 13 limits and derivatives. This chapter starts with an intuitive idea of derivative (without actually defining it) then covers the definition of a limit, algebra of limits, the definition of derivative, and algebra derivatives. You will also learn derivatives of certain standard functions. In solutions of NCERT for class 11 maths chapter 13 limits and derivatives, you will get questions related to all the above topics. This chapter is very important for class 11 final examination and also in various competitive exams like JEE Main, JEE Advanced, VITEEE, BITSAT, etc. The knowledge of this chapter is required to study any chapter of calculus that you will be studying in the upcoming class 12 also. There are 43 questions in 2 exercises. The first exercise of this chapter deals with problems on limits and the second exercise deals with problems on derivatives. All these questions are solved in the CBSE NCERT solutions for class 11 maths chapter 13 limits and derivatives. Check all NCERT solutions from class 6 to 12 to understand the concepts in a much easy way. There are two exercises and a miscellaneous exercise of this chapter are explained below.

Exercise:13.1

Exercise:13.2

Miscellaneous Exercise

Topics of NCERT Grade 11 Maths Chapter-13 Limits and Derivatives

13.1 Introduction

13.2 Intuitive Idea of Derivatives

13.3 Limits

13.4 Limits of Trigonometric Functions

13.5 Derivatives

## NCERT solutions for class 11 maths chapter 13 limits and derivatives-Exercise: 13.1

Question:1 Evaluate the following limits

$\lim_{x\rightarrow 3} x +3$

$\lim_{x\rightarrow 3} x +3$

$\Rightarrow \lim_{x\rightarrow 3} 3 +3$

$\Rightarrow 6$ Answer

Below you can find the solution:

$\lim_{x \rightarrow \pi } \left ( x - 22/7 \right )=\pi-\frac{22}{7}$

The limit

$\lim_{r \rightarrow 1} \pi r^2=\pi(1)^2=\pi$

Answer is $\pi$

The limit

$\lim_{x \rightarrow {4}} \frac{4x+3 }{x-2}$

$\Rightarrow \frac{4(4)+3 }{(4)-2}$

$\Rightarrow \frac{19 }{2}$   (Answer)

The limit

$\lim_{x \rightarrow {4}} \frac{x^{10}+ x^5 + 1 }{x-1}$

$\Rightarrow \frac{(-1)^{10}+ (-1)^5 + 1 }{(-1)-1}$

$\Rightarrow \frac{1-1+1}{-2}$

$\Rightarrow -\frac{1}{2}$ (Answer)

The limit

$\lim_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }$

Lets put

$x+1=y$

since we have changed the function, its limit will also change,

so

$x\rightarrow 0,y\rightarrow 0+1=1$

So our function have became

$\lim_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }$

Now As we know the property

$\lim_{x \rightarrow 1 }\frac{ x^5 -a^n}{x-a }=na^{n-1}$

$\lim_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }=5(1)^5=5$

Hence,

$\lim_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }=5$

The limit

$\lim_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}$

$\Rightarrow \lim_{x \rightarrow 2} \frac{(x-2)(3x+5) }{(x-2)(x+2)}$

$\Rightarrow \lim_{x \rightarrow 2} \frac{(3x+5) }{(x+2)}$

$\Rightarrow \frac{(3(2)+5) }{((2)+2)}$

$\Rightarrow \frac{11 }{4}$ (Answer)

The limit

$\lim_{x \rightarrow 3} \frac{x ^4 -81}{2x^2 -5x -3}$

At x = 2 both numerator and denominator becomes zero, so lets factorise the function

$\lim_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$

$\lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

Now we can put the limit directly, so

$\lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

$\Rightarrow \frac{((3)+3)((3)^2+9)}{(2(3)+1)}$

$\Rightarrow \frac{6\times18}{7}$

$\Rightarrow \frac{108}{7}$

The limit,

$\lim_{x \rightarrow 0 } \frac{ax +b}{cx+1}$

$\Rightarrow \frac{a(0) +b}{c(0)+1}$

$\Rightarrow b$ (Answer)

The limit

$\lim_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Here on directly putting limit , both numerator and the deniminator becomes zero so we factorize the function and then put the limit.

$\lim_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}=\lim_{z\rightarrow 1} \frac{z^{(1/6)^2}-1^2}{z^{1/6}-1}$

$=\lim_{z\rightarrow 1} \frac{(z^{(1/6)}-1)(z^{(1/6)}+1)}{z^{1/6}-1}$

$=\lim_{z\rightarrow 1} (z^{(1/6)}+1)}$

$= (1^{(1/6)}+1)}$

$=1+1$

$=2$  (Answer)

The limit:

$\lim_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

Since Denominator is not zero on directly putting the limit, we can directly put the limits, so,

$\lim_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

$=\frac{a(1) ^2 +b(1) + c }{c(1)^2 + b(1) + a },$

$=\frac{a+b+c }{a+b+c },$

$=1$  (Answer)

$\lim_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}$

Here, since denominator becomes zero on putting the limit directly, so we first simplify the function and then put the limit,

$\lim_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{x}{2}}{x+2}$

$=\lim_{x\rightarrow -2} \frac{\frac{x+2}{2x}}{x+2}$

$=\lim_{x\rightarrow -2} \frac{1}{2x}$

$= \frac{1}{2(-2)}$

$= -\frac{1}{4}$  (Answer)

The limit

$\lim_{x \rightarrow 0 } \frac{\sin ax }{bx }$

Here on directly putting the limits, the function becomes $\frac{0}{0}$ form. so we try to make the function in the form of $\frac{sinx}{x}$. so,

$\lim_{x \rightarrow 0 } \frac{\sin ax }{bx }$

$=\lim_{x \rightarrow 0 } \frac{\sin ax(ax) }{bx(ax) }$

$=\lim_{x \rightarrow 0 } \frac{\sin ax }{ax }\frac{a}{b}$

As  $\lim_{x\rightarrow 0}\frac{sinx}{x}=1$

$=1\cdot\frac{a}{b}$

$=\frac{a}{b}$   (Answer)

The limit,

$\lim_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0$

On putting the limit directly, the function takes the zero by zero  form  So,we convert it in the form of $\frac{sina}{a}$.and then put the limit,

$\Rightarrow \lim_{x\rightarrow {0}}\frac{\frac{sinax}{ax}}{\frac{sinbx}{bx}}\cdot\frac{ax}{bx}$

$=\frac{\lim_{ax\rightarrow {0}}\frac{sinax}{ax}}{\lim_{bx\rightarrow {0}}\frac{sinbx}{bx}}\cdot\frac{a}{b}$

$=\frac{a}{b}$   (Answer)

The limit

$\lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}$

$\Rightarrow \lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{ ( \pi -x)}\times\frac{1}{\pi}$

$= 1\times\frac{1}{\pi}$

$= \frac{1}{\pi}$  (Answer)

The limit

$\lim_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

the function behaves well on directly putting the limit,so we put the limit directly. So.

$\lim_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

$=\frac{\cos (0) }{\pi -(0) }$

$=\frac{1 }{\pi }$   (Answer)

The limit:

$\lim_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit

$\lim_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

$\lim_{x\rightarrow 0}$ $\frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}$

$=\lim_{x\rightarrow 0}$    $\frac{\frac{(sin^2x)}{x^2}}{\frac{(sin^2(\frac{x}{2}))}{(\frac{x}{2})^2}}\times\frac{x^2}{(\frac{x}{2})^2}$

$=\frac{1^2}{1^2}\times 4$

$= 4$  (Answer)

$\lim_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

The function takes the form zero by zero when we put the limit directly in the function,. since function consist of sin function and cos function, we try to make the function in the form of $\frac{sinx}{x}$ as we know that it tends to 1 when x tends to 0.

So,

$\lim_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

$=\frac{1}{b}\lim_{x \rightarrow 0}\frac{x(a+ \cos x) }{ \sin x }$

$=\frac{1}{b}\lim_{x \rightarrow 0}\frac{x }{ \sin x }\times(a+ \cos x)$

$=\frac{1}{b}\times1\times(a+ \cos (0))$

$=\frac{a+1}{b}$   (Answer)

$\lim_{x \rightarrow 0} x \sec x$

As function doesn't create any abnormality on putting the limit directly,we can put limit directly. So,

$\lim_{x \rightarrow 0} x \sec x$

$=(0) \sec (0)$

$=(0)\times 1$

$=0$(Answer)

$\lim_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

The function takes the zero by zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So

$\lim_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

$=\lim_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot ax+ bx }{ax + \frac{\sin bx}{bx}\cdot bx }$

$=\lim_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot a+ b }{a + \frac{\sin bx}{bx}\cdot b }$

$=\frac{1\cdot a+b}{a+1\cdot b}$

$=\frac{a+b}{a+ b}$

$=1$  (Answer)

$\lim_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

On putting the limit directly the function takes infinity by infinity form, So we simplify the function and then put the limit

$\lim_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

$=\lim_{x \rightarrow 0} \left (\frac{1}{sinx}-\frac{cosx}{sinx}\right )$

$=\lim_{x \rightarrow 0} \left (\frac{1-cosx}{sinx}\right )$

$=\lim_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{sinx}\right )$

$=\lim_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(\frac{x}{2})^2}{sinx} \right )$

$=\lim_{x \rightarrow 0} \frac{2}{4}\left (\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(x)}{sinx} \right )\cdot x$

$=\frac{2}{4}\times (1)^2\times0$

$=0$   (Answer)

$\lim_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }$

The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits,

So

Let's put

$y=x-\frac{\pi}{2}$

Since we are changing the variable, limit will also change.

as

$x\rightarrow \frac{\pi}{2},y=x-\frac{\pi}{2}\rightarrow \frac{\pi}{2}-\frac{\pi}{2}=0$

So function in new variable becomes,

$\lim_{y \rightarrow 0 } \frac{\tan 2(y+\frac{\pi}{2}) }{y+\pi/2- \pi /2 }$

$=\lim_{y \rightarrow 0 } \frac{\tan (2y+\pi) }{y }$

As we know tha property   $tan(\pi+x)=tanx$

$=\lim_{y \rightarrow 0 } \frac{\tan (2y) }{y }$

$=\lim_{y \rightarrow 0 } \frac{sin2y}{2y}\cdot\frac{2}{cos2y}$

$=1\times 2$

$=2$   (Answer)

Given Function

$f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.$

Now,

Limit at x = 0  :

$at\:x=0^-$

:$\lim_{x\rightarrow{0^-}}f(x)=\lim_{x\rightarrow{0^-}}(2x+3)=2(0)+3=3$

$at\:x=0^+$

$\lim_{x\rightarrow{0^+}}f(x)=\lim_{x\rightarrow{0^+}}3(x+1)=3(0+1)=3$

Hence limit at x = 0 is 3.

Limit at x = 1

$at\:x=1^+$

$\lim_{x\rightarrow{1^+}}f(x)=\lim_{x\rightarrow{1^+}}3(x+1)=3(1+1)=6$

$at\:x=1^-$

$\lim_{x\rightarrow{1^-}}f(x)=\lim_{x\rightarrow{1^-}}3(x+1)=3(1+1)=6$

Hence limit at x = 1 is 6.

$\lim_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.$

Limit at $x=1^+$

$\lim_{x \rightarrow 1^+} f (x ) = \lim_{x \rightarrow 1} (-x^2-1)=-(1)^2-1=-2$

Limit at $x=1^-$

$\lim_{x \rightarrow 1^-} f (x ) = \lim_{x \rightarrow 1} (x^2-1)=(1)^2-1=0$

As we can see that Limit at $x=1^+$ is not equal to Limit at $x=1^-$,The limit of this function at x = 1 does not exists.

$\lim_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or  Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} \frac{|x|}{x}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} \frac{|x|}{x}=\frac{-x}{x}=-1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

$\lim_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or  Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} \frac{x}{|x|}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} \frac{x}{|x|}=\frac{x}{-x}=-1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

$\lim_{x \rightarrow 5 } f (x) , where f( x) = |x| -5$

The right-hand Limit or  Limit at $x=5^+$

$\lim_{x \rightarrow 5^+} f (x) = \lim_{x \rightarrow 5^+} |x|-5=5-5=0$

The left-hand limit or Limit at $x=5^-$

$\lim_{x \rightarrow 5^-} f (x) = \lim_{x \rightarrow 5^-}|x|-5=5-5=0$

Since Left-hand limit and right-hand limit are equal, The limit of this function at x = 5 is 0.

Question:28 Suppose

$f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$   f (x) = f (1) what are possible values of a and b?

Given,

$f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$

And

$\lim_{x\rightarrow 1} f(x)=f(1)$

Since the limit exists,

left-hand limit = Right-hand limit = f(1).

Left-hand limit  = f(1)

$\lim_{x\rightarrow 1^-} f(x)= \lim_{x\rightarrow 1}(a+bx)=a+b(1)=a+b=4$

Right-hand limit

$\lim_{x\rightarrow 1^+} f(x)= \lim_{x\rightarrow 1}(b-ax)=b-a(1)=b-a=4$

From both equations, we get that,

$a=0$ and $b=4$

Hence the possible value of a and b are 0 and 4 respectively.

Given,

$f (x) = (x - a_1 ) (x - a_2 )...(x - a_n ) .$

Now,

$\\\lim_{x \rightarrow a _ 1 }f(x)=\lim_{x \rightarrow a _ 1 }[(x - a_1 ) (x - a_2 )...(x - a_n ) ]\\.=[\lim_{x \rightarrow a _ 1 }(x - a_1 )][\lim_{x \rightarrow a _ 1 }(x - a_2 )][\lim_{x \rightarrow a _ 1 }(x - a_n )] \\=0$

Hence

$\lim_{x \rightarrow a _ 1 }f(x)=0$

Now,

$\lim_{ x \rightarrow a } f (x)=\lim_{ x \rightarrow a } (x-a_1)(x-a_2)...(x-a_n)$

$\lim_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$

Hence

$\lim_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$.

$f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.$

Limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,

Case 1: when a = 0

The right-hand Limit or  Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} |x|-1=1-1=0$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} |x|+1=0+1=1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Case 2: When a < 0

The right-hand Limit or  Limit at $x=a^+$

$\lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|-1=a-1$

The left-hand limit or Limit at $x=a^-$

$\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|-1=a-1$

Since LHL = RHL, Limit exists at x = a and is equal to a-1.

Case 3: When a > 0

The right-hand Limit or  Limit at $x=a^+$

$\lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|+1=a+1$

The left-hand limit or Limit at $x=a^-$

$\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|+1=a+1$

Since LHL = RHL, Limit exists at x = a and is equal to a+1

Hence,

The limit exists at all points except at x=0.

Given

$\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$

Now,

$\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \frac{\lim_{x \rightarrow 1}(f (x)-2)}{\lim_{x \rightarrow 1}(x^2-1)}=\pi$

${\lim_{x \rightarrow 1}(f (x)-2)}=\pi{\lim_{x \rightarrow 1}(x^2-1)}$

${\lim_{x \rightarrow 1}(f (x)-2)}=\pi{(1-1)}$

${\lim_{x \rightarrow 1}(f (x)-2)}=0$

${\lim_{x \rightarrow 1}f (x)}=2$

Question:32 If

$f (x) = \left\{\begin{matrix} mx^2 + n & x < 0 \\ nx + m & 0 \leq x \leq 1 \\ nx^3+m & x > 1 \end{matrix}\righ t.$
. For what integers m and n does both $\lim_{x \rightarrow 0}f (x) \: \: and\: \: \lim_{x \rightarrow 1} f (x)$  exist ?

Given,

$f (x) = \left\{\begin{matrix} mx^2 + n & x < 0 \\ nx + m & 0 \leq x \leq 1 \\ nx^3+m & x > 1 \end{matrix}\righ t.$

Case 1: Limit at x = 0

The right-hand Limit or  Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} nx+m=n(0)+m=m$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} mx^2+n=m(0)^2+n=n$

Hence Limit will exist at x = 0 when m = n .

Case 2: Limit at x = 1

The right-hand Limit or  Limit at $x=1^+$

$\lim_{x \rightarrow 1^+} f (x) = \lim_{x \rightarrow 1^+} nx^3+m=n(1)^3+m=n+m$

The left-hand limit or Limit at $x=1^-$

$\lim_{x \rightarrow 1^-} f (x) = \lim_{x \rightarrow 1^-} nx+m=n(1)+m=n+m$

Hence Limit at 1 exists at all integers.

## NCERT solutions for class 11 maths chapter 13 limits and derivatives-Exercise: 13.2

F(x)=$x ^ 2 -2 \: \:$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 10:

$f'(10)=\lim_{h\rightarrow 0}\frac{f(10+h)-f(10)}{h}$

$f'(10)=\lim_{h\rightarrow 0}\frac{(10+h)^2-2-((10)^2-2)}{h}$

$f'(10)=\lim_{h\rightarrow 0}\frac{100+20h+h^2-2-100 +2}{h}$

$f'(10)=\lim_{h\rightarrow 0}\frac{20h+h^2}{h}$

$f'(10)=\lim_{h\rightarrow 0}20+h$

$f'(10)=20+0$

$f'(10)=20$

Question:2 Find the derivative of x at x = 1.

Given

f(x)= x

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 1:

$f'(1)=\lim_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}$

$f'(1)=\lim_{h\rightarrow 0}\frac{(1+h)-(1)}{h}$

$f'(1)=\lim_{h\rightarrow 0}\frac{(h)}{h}$

$f'(1)=1$ (Answer)

Question:3 Find the derivative of 99x at x = l00.

f(x)= 99x

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of f(x) at x = 100:

$f'(100)=\lim_{h\rightarrow 0}\frac{f(100+h)-f(100)}{h}$

$f'(100)=\lim_{h\rightarrow 0}\frac{99(100+h)-99(100)}{h}$

$f'(100)=\lim_{h\rightarrow 0}\frac{99h}{h}$

$f'(100)=99$

Given

f(x)=$x ^3 -27$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{(x+h)^3-27-((x)^3-27)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+h^3+3x^2h+3h^2x-27+x^3+27}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h^3+3x^2h+3h^2x}{h}$

$f'(x)=\lim_{h\rightarrow 0}{h^2+3x^2+3hx}$

$f'(x)=3x^2$

f(x)=$( x-1)(x-2)$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^2+xh-2x+hx+h^2-2h-x-h+2-x^2+2x+x-2}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2hx+h^2-3h}{h}$

$f'(x)=\lim_{h\rightarrow 0}{2x+h-3}$

$f'(x)=2x-3$  (Answer)

f(x)= $1 / x ^2$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1/(x+h)^2-1/(x^2)}{h}$

$f'(x)=\lim_{h\rightarrow 0} \frac{\frac{x^2-(x+h)^2}{(x+h)^2x^2}}{h}$

$f'(x)=\lim_{h\rightarrow 0} \frac{x^2-x^2-2xh-h^2}{h(x+h)^2x^2}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-2xh-h^2}{h(x+h)^2x^2}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-2x-h}{(x+h)^2x^2}$

$f'(x)= \frac{-2x-0}{(x+0)^2x^2}$

$f'(x)= \frac{-2}{x^3}$  (Answer)

Given:

$f(x)=\frac{x +1}{x-1}$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^2-x+hx-h+x-1-x^2-xh+x-x-h+1}{(x-1)(x+h-1)h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-2h}{(x-1)(x+h-1)h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-2}{(x-1)(x+h-1)}$

$f'(x)=\frac{-2}{(x-1)(x+0-1)}$

$f'(x)=\frac{-2}{(x-1)^2}$

$f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1$

As we know, the property,

$f'(x^n)=nx^{n-1}$

applying that property we get

$f '(x) = 100\frac{x^{99}}{100}+ 99\frac{x ^ {98}}{99} + ....+ 2\frac{x }{2}+ 1+ 0$

$f '(x) = x^{99}+x^{98}+......x+1$

Now.

$f '(0) = 0^{99}+0^{98}+......0+1=1$

$f '(1) = 1^{99}+1^{98}+......1+1=100$

So,

Here

$1\times 100=100$

$f'(0)\times 100=f'(1)$

Hence Proved.

Given

$f(x)=x ^n + ax ^{n-1} + a ^ 2 x ^{n-2} + ....+ a ^ { n-1} x + a ^n$

As we know, the property,

$f'(x^n)=nx^{n-1}$

applying that property we get

$f'(x)=nx ^{n-1} + a(n-1)x ^{n-2} + a ^ 2(n-2) x ^{n-3} + ....+ a ^ { n-1} 1 + 0$

$f'(x)=nx ^{n-1} + a(n-1)x ^{n-2} + a ^ 2(n-2) x ^{n-3} + ....+ a ^ { n-1}$

Given

$f(x)=( x - a ) ( x -b )=x^2-ax-bx+ab$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=2x-a-b$

Given

$f(x)=( ax ^2 + b)^2=a^2x^4+2abx^2+b^2$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying those properties we get

$f'(x)=4a^2x^3+2(2)abx+0$

$f'(x)=4a^2x^3+4abx$

$f'(x)=4ax(ax^2+b)$

Given,

$f(x)=\frac{x - a }{x -b }$

Now As we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{(x-b)\frac{d(x-a)}{dx}-(x-a)\frac{d(x-b)}{dx}}{(x-b)^2}$

$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{(x-b)-(x-a)}{(x-b)^2}$

$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{a-b}{(x-b)^2}$

Hence

$f'(x)=\frac{a-b}{(x-b)^2}$

Given,

$f(x)=\frac{x ^n - a ^n }{x - a }$

Now As we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)\frac{d(x^n-a^n)}{dx}-(x^n-a^n)\frac{d(x-a)}{dx}}{(x-a)^2}$

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)nx^{n-1}-(x^n-a^n)}{(x-a)^2}$

$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$

Hence

$f'(x)=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$

Given:

$f(x)=2x - 3/4$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=2-0$

$f'(x)=2$

Given.

$f(x)=( 5x^3 + 3x -1 ) ( x -1)=5x^4+3x^2-x-5x^3-3x+1$

$f(x)=5x^4-5x^3+3x^2-4x+1$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=5(4)x^3-5(3)x^2+3(2)x-4+0$

$f'(x)=20x^3-15x^2+6x-4$

Given

$f(x)=x ^{-3} ( 5 + 3x )=5x^{-3}+3x^{-2}$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=(-3)5x^{-4}+3(-2)x^{-3}$

$f'(x)=-15x^{-4}-6x^{-3}$

Given

$f(x)=x ^5 ( 3 - 6 x ^{-9})=3x^5-6x^{-4}$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=(5)3x^4-6(-4)x^{-5}$

$f'(x)=15x^4+24x^{-5}$

Given

$f(x)=x ^{-4} ( 3 - 4x ^{-5})=3x^{-4}-4x^{-9}$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=(-4)3x^{-5}-(-9)4x^{-10}$

$f'(x)=-12x^{-5}+36x^{-10}$

Given

$f(x)=\frac{2}{x+1}- \frac{x^2 }{3 x-1}$

As we know the quotient rule of derivative:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

So applying this rule, we get

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{(x+1)\frac{d(2)}{dx}-2\frac{d(x+1)}{dx}}{(x+1)^2}-\frac{(3x-1)\frac{d(x^2)}{dx}-x^2\frac{d(3x-1)}{dx}}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{(3x-1)2x-x^23}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{6x^2-2x-3x^2}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$

Hence

$f'(x)=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$

Given,

f(x)=$\cos x$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\cos(x+h)-\cos(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\cos(x)\cos(h)-\cos(x)}{h}-\frac{\sin(x)\sin(h)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\cos(x)(\cos(h)-1)}{h}-\frac{\sin(x)\sin(h)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\cos(x)\frac{-2sin^2(h/2)}{h}\cdot -\sin(x)\frac{sinh}{h}$

$f'(x)=\cos(x)(0)-sinx(1)$

$f'(x)=-\sin(x)$

Given,

f(x)=$\sin x \cos x$

Now, As we know the product rule of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

So, applying the rule here,

$\frac{d(\sin x\cos x)}{dx}=\sin x\frac{d\cos x}{dx}+\cos x\frac{d\sin x}{dx}$

$\frac{d(\sin x\cos x)}{dx}=\sin x(-\sin x)+\cos x (\cos x)$

$\frac{d(\sin x\cos x)}{dx}=-\sin^2 x+\cos^2 x$

$\frac{d(\sin x\cos x)}{dx}=\cos 2x$

Given

$f(x)=\sec x=\frac{1}{\cos x}$

Now As we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(\frac{1}{\cos x})}{dx}=\frac{\cos x\frac{d(1)}{dx}-1\frac{d(\cos x)}{dx}}{\cos ^2x}$

$\frac{d(\frac{1}{\cos x})}{dx}=\frac{-1(-\sin x)}{\cos ^2x}$

$\frac{d(\frac{1}{\cos x})}{dx}=\frac{\sin x}{\cos ^2x}=\frac{\sin x}{\cos x}\frac{1}{\cos x}$

$\frac{d(\sec x)}{dx}=\tan x\sec x$

Given

$f(x)=5 \sec x + 4 \cos x$

As we know the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying the property, we get

$\frac{d(5\sec x+4\cos x)}{dx}=\frac{d(5\sec x)}{dx}+\frac{d(4\cos x)}{dx}$

$\frac{d(5\sec x+4\cos x)}{dx}=5\tan x\sec x+4(-\sin x)$

$\frac{d(5\sec x+4\cos x)}{dx}=5\tan x\sec x-4\sin x$

Given :

$f(x)=\csc x=\frac{1}{\sin x}$

Now As we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(\frac{1}{\sin x})}{dx}=\frac{(\sin x)\frac{d(1)}{dx}-1\frac{d(\sin x)}{dx}}{(\sin x)^2}$

$\frac{d(\frac{1}{\sin x})}{dx}=\frac{-1(\cos x)}{(\sin x)^2}$

$\frac{d(\frac{1}{\sin x})}{dx}=-\frac{(\cos x)}{(\sin x)}\frac{1}{\sin x}$

$\frac{d(\csc x)}{dx}=-\cot x \csc x$

Given,

$f(x)=3 \cot x + 5 \csc x$

As we know  the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying the property,

$\frac{d(3\cot x+5 \csc x)}{dx}=\frac{d(3\cot x)}{dx}+\frac{d(5\csc x)}{dx}$

$\frac{d(3\cot x+5 \csc x)}{dx}=3\frac{d(\frac{\cos x}{\sin x})}{dx}+\frac{d(5\csc x)}{dx}$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\frac{d(\frac{\cos x}{\sin x})}{dx}$

Now As we know the quotient rule of derivative,

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

So applying this rule, we get

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x\frac{d(\cos x)}{dx}-\cos x(\frac{d(\sin x)}{dx})}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{-\sin^2 x-\cos^2 x}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\left[\frac{\sin^2 x+\cos^2 x}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\left[\frac{1}{\sin^2x}\right]$

$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\csc^2x$

Given,

$f(x)=5 \sin x - 6 \cos x + 7$

Now as we know the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

So, applying the property,

$f'(x)=5 \cos x - 6 (-\sin x ) + 0$

$f'(x)=5 \cos x + 6 (\sin x )$

$f'(x)=5 \cos x + 6 \sin x$

Given

$f(x)=2 \tan x - 7 \sec x$

As we know  the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying this property,

$\frac{d(2\tan x+7\sec x)}{dx}=2\frac{d\tan x}{dx}+7\frac{d\sec x}{dx}$

$\frac{d(2\tan x+7\sec x)}{dx}=2\sec^2x +7(-\sec x\tan x)$

$\frac{d(2\tan x+7\sec x)}{dx}=2\sec^2x -7\sec x\tan x$

CBSE NCERT solutions for class 11 maths chapter 13 limits and derivatives-Miscellaneous Exercise

Given.

f(x)=-x

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-(x+h)-(-x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-h}{h}$

$f'(x)=-1$

Given.

f(x)= $( - x ) ^{-1}$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-(x+h)^{-1}-(-x)^{-1}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-\frac{1}{x+h}+\frac{1}{x}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{-x+x+h}{x(x+h)}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{h}{(x+h)(x)}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{x(x+h)}$

$f'(x)=\frac{1}{x^2}$

Given.

$f(x)=\sin ( x+1)$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\sin(x+h+1)-\sin(x+1)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2\cos(\frac{x+h+1+x+1}{2})\sin(\frac{x+h+1-x-1}{2})}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2\cos(\frac{2x+h+2}{2})\sin(\frac{h}{2})}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\cos(\frac{2x+h+2}{2})\sin(\frac{h}{2})}{\frac{h}{2}}$

$f'(x)=\cos(\frac{2x+0+2}{2})\times 1$

$f'(x)=\cos(x+1)$

Given.

$f(x)=\cos ( x - \pi /8 )$

Now, As we know, The derivative of any function at x is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\cos(x+h-\pi/8)-cos(x-\pi/8)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-2\sin\left ( \frac{x+h-\pi/8+x-\pi/8 }{2}\right )\sin\left ( \frac{x+h-\pi/8-x+\pi/8}{2} \right )}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-2\sin\left ( \frac{2x+h-\pi/4 }{2}\right )\sin\left ( \frac{h}{2} \right )}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-\sin\left ( \frac{2x+h-\pi/4 }{2}\right )\sin\left ( \frac{h}{2} \right )}{\frac{h}{2}}$

$f'(x)=\sin\left ( \frac{2x+0-\pi/4}{2} \right )\times 1$

$f'(x)=-\sin\left (x-\pi/8 \right )$

Given

f(x)= x + a

As we know, the property,

$f'(x^n)=nx^{n-1}$

applying that property we get

$f'(x)=1+0$

$f'(x)=1$

Given

$f(x)=\left(px + q\right) \left ( \frac{r}{x}+ s \right )$

$f(x)=pr+psx+\frac{qr}{x}+qs$

As we know, the property,

$f'(x^n)=nx^{n-1}$

applying that property we get

$f'(x)=0 + ps +\frac{-qr}{x^2}+0$

$f'(x)=ps + q \left ( \frac{-r}{x^2} \right )$

$f'(x)=ps - \left ( \frac{qr}{x^2} \right )$

Given,

$f(x)=( ax + b ) ( cx + d )^2$

$f(x)=( ax + b ) ( c^2x^2+2cdx+d^2)$

$f(x)=ac^2x^3+2acdx^2+ad^2x+bc^2x^2+2bcdx+bd^2$

Now,

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=3ac^2x^2+4acdx+ad^2+2bc^2x+2bcd+0$

$f'(x)=3ac^2x^2+4acdx+ad^2+2bc^2x+2bcd$

$f'(x)=3ac^2x^2+(4acd+2bc^2)x+ad^2+2bcd$

Given,

$f(x)=\frac{ax+b}{cx+d}$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, The derivative of f(x) is

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{(cx+d)d(\frac{d(ax+b)}{dx})-(ax+b)(\frac{d(cx+d)}{dx})}{(cx+d)^2}$

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{(cx+d)a-(ax+b)c}{(cx+d)^2}$

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{acx+ad-acx-bc}{(cx+d)^2}$

$\frac{d(\frac{ax+b}{cx+d})}{dx}=\frac{ad-bc}{(cx+d)^2}$

Hence Derivative of the function is

$\frac{ad-bc}{(cx+d)^2}$.

$\frac{1 + \frac{1}{x}}{1- \frac{1}{x}}$

Given,

$f(x)=\frac{1 + \frac{1}{x}}{1- \frac{1}{x}}$

Also can be written as

$f(x)=\frac{x+1}{x-1}$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, The derivative of f(x) is

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{(x-1)d(\frac{d(x+1)}{dx})-(x+1)(\frac{d(x-1)}{dx})}{(x-1)^2}$

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{(x-1)1-(x+1)1}{(x-1)^2}$

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{x-1-x-1}{(x-1)^2}$

$\frac{d(\frac{x+1}{x-1})}{dx}=\frac{-2}{(x-1)^2}$

Hence Derivative of the function is

$\frac{-2}{(x-1)^2}$

$\frac{1 }{ax ^2 + bx + c}$

Given,

$f(x)=\frac{1 }{ax ^2 + bx + c}$

Now, As we know the derivative of any such  function  is given by

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, The derivative of f(x) is

$\frac{d(\frac{1}{ax^2+bx+c})}{dx}=\frac{(ax^2+bx+c)d(\frac{d(1)}{dx})-1(\frac{d(ax^2+bx+c)}{dx})}{(ax^2+bx+c)^2}$

$\frac{d(\frac{1}{ax^2+bx+c})}{dx}=\frac{0-(2ax+b)}{(ax^2+bx+c)^2}$

$\frac{d(\frac{1}{ax^2+bx+c})}{dx}=-\frac{(2ax+b)}{(ax^2+bx+c)^2}$

$\frac{ax + b }{px^2 + qx + r }$

Given,

$f(x)=\frac{ax + b }{px^2 + qx + r }$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, The derivative of f(x) is

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{(px^2+qx+r)d(\frac{d(ax+b)}{dx})-(ax+b)(\frac{d(px^2+qx+r)}{dx})}{(px^2+qx+r)^2}$

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{(px^2+qx+r)a-(ax+b)(2px+q)}{(px^2+qx+r)^2}$

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{apx^2+aqx+ar-2apx^2-aqx-2bpx-bq}{(px^2+qx+r)^2}$

$\frac{d(\frac{ax+b}{px^2+qx+r})}{dx}=\frac{-apx^2+ar-2bpx-bq}{(px^2+qx+r)^2}$

$\frac{px^2 + qx + r }{ax +b }$

Given,

$f(x)=\frac{px^2 + qx + r }{ax +b }$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence, The derivative of f(x) is

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{(ax+b)d(\frac{d(px^2+qx+r)}{dx})-(px^2+qx+r)(\frac{d(ax+b)}{dx})}{(ax+b)^2}$

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{(ax+b)(2px+q)-(px^2+qx+r)(a)}{(ax+b)^2}$

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{2apx^2+aqx+2bpx+bq-apx^2-aqx-ar}{(ax+b)^2}$

$\frac{d(\frac{px^2+qx+r}{ax+b})}{dx}=\frac{apx^2+2bpx+bq-ar}{(ax+b)^2}$

$\frac{a}{x^4} - \frac{b}{x^2 } + \cos x$

Given

$f(x)=\frac{a}{x^4} - \frac{b}{x^2 } + \cos x$

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=\frac{-4a}{x^5} - (\frac{-2b}{x^3 }) +( -\sin x)$

$f'(x)=\frac{-4a}{x^5} +\frac{2b}{x^3 } -\sin x$

Given

$f(x)=4 \sqrt x - 2$

It can also be written as

$f(x)=4 x^{\frac{1}{2}} - 2$

Now,

As we know, the property,

$f'(x^n)=nx^{n-1}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying that property we get

$f'(x)=4 (\frac{1}{2})x^{-\frac{1}{2}} - 0$

$f'(x)=2x^{-\frac{1}{2}}$

$f'(x)=\frac{2}{\sqrt{x}}$

$( ax + b ) ^ n$

Given

$f(x)=( ax + b ) ^ n$

Now, As we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

And, the property,

$f'(x^n)=nx^{n-1}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

applying those properties we get,

$f'(x)=n(ax+b)^{n-1}\times a$

$f'(x)=an(ax+b)^{n-1}$

Given

$f(x)=( ax + b ) ^ n ( cx + d ) ^ m$

Now, As we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

And the Multiplication property of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

And, the property,

$f'(x^n)=nx^{n-1}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties we get,

$f'(x)=(ax+b)^n(m(cx+d)^{m-1})+(cx+d)(n(ax+b)^{n-1})$

$f'(x)=m(ax+b)^n(cx+d)^{m-1}+n(cx+d)(ax+b)^{n-1}$

Given,

$f(x)=\sin ( x + a )$

Now, As we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

Applying this property we get,

$f'(x)=\cos ( x + a )\times 1$

$f'(x)=\cos ( x + a )$

Given,

$f(x)=\csc x \cot x$

the Multiplication property of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Applying the property

$\frac{d(\csc x)(\cot x))}{dx}=\csc x\frac{d\cot x}{dx}+\cot x\frac{d\csc x}{dx}$

$\frac{d(\csc x)(\cot x))}{dx}=\csc x(-\csc^2x)+\cot x(-\csc x \cot x)$

$\frac{d(\csc x)(\cot x))}{dx}=-\csc^3x-\cot^2 x\csc x$

Hence derivative of the function is $-\csc^3x-\cot^2 x\csc x$.

$\frac{\cos x }{1+ \sin x }$

Given,

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

Now, As we know the derivative of any function

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{(1+\sin x )d(\frac{d\cos x}{dx})-\cos x(\frac{d(1+\sin x)}{dx})}{(1+\sin x)^2}$

Hence, The derivative of f(x) is

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{(1+\sin x )(-\sin x)-\cos x(\cos x)}{(1+\sin x)^2}$

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{-\sin x-\sin^2 x -\cos^2 x}{(1+\sin x)^2}$

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=\frac{-\sin x-1}{(1+\sin x)^2}$

$\frac{d(\frac{\cos x}{1+\sin x})}{dx}=-\frac{1}{(1+\sin x)}$

Given

$f(x)=\frac{\sin x + \cos x }{\sin x - \cos x }$

Also can be written as

$f(x)=\frac{\tan x + 1 }{\tan x - 1 }$

which further can be written as

$f(x)=-\frac{\tan x + tan(\pi/4) }{1-\tan(\pi/4)\tan x }$

$f(x)=-\tan(x-\pi/4)$

Now,

$f'(x)=-\sec^2(x-\pi/4)$

$f'(x)=-\frac{1}{\cos^2(x-\pi/4)}$

$\frac{\sec x -1}{\sec x +1}$

Given,

$f(x)=\frac{\sec x -1}{\sec x +1}$

which also can be written as

$f(x)=\frac{1-\cos x}{1+\cos x}$

Now,

As we know the derivative of such function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

So, The derivative of the function is,

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{(1+\cos x)d(\frac{d(1-\cos x)}{dx})-(1-\cos x)(\frac{d(1+\cos x)}{dx})}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{(1+\cos x)(-(-\sin x))-(1-\cos x)(-\sin x)}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{\sin x+\sin x\cos x+\sin x-\cos x\sin x}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{2\sin x}{(1+\cos x)^2}$

Which can also be written as

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{2\sec x\tan x}{(1+\sec x)^2}$.

Given,

$f(x)=\sin^ n x$

Now, As we know the chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

And, the property,

$f'(x^n)=nx^{n-1}$

Applying those properties, we get

$f'(x)=n\sin^ {n-1} x \cos x$

Hence Derivative of the given function is $n\sin^ {n-1} x \cos x$

$\frac{a + b \sin x }{c+ d \cos x }$

Given Function

$f(x)=\frac{a + b \sin x }{c+ d \cos x }$

Now, As we know the derivative of any function  of this type is:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence derivative of the given function will be:

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{(c+d\cos x)(\frac{d(a+b\sin x)}{dx})-(a+b\sin x)(\frac{d(c+d\cos x x)}{dx})}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{(c+d\cos x)(b\cos x)-(a+b\sin x)(d(-\sin x))}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+bd\cos^2 x+ad\sin x+bd\sin^2 x}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+ad\sin x+bd(\sin^2 x+\cos^2 x)}{(c+d\cos x)^2}$

$\frac{d(\frac{a+b\sin x}{c+d\cos x})}{dx}=\frac{cb\cos x+ad\sin x+bd}{(c+d\cos x)^2}$

$\frac{\sin ( x+a )}{ \cos x }$

Given,

$f(x)=\frac{\sin ( x+a )}{ \cos x }$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence the derivative of the given function is:

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\frac{d(\sin (x+a))}{dx})-\sin(x+a)(\frac{d(\cos x)}{dx})}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\cos(x+a))-\sin(x+a)(-\sin (x))}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{(\cos x)(\cos(x+a))+\sin(x+a)(\sin (x))}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{\cos (x+a-x)}{(\cos x)^2}$

$\frac{d(\frac{\sin(x+a)}{\cos x})}{dx}=\frac{\cos (a)}{(\cos x)^2}$

Given

$f(x)=x ^ 4 ( 5 \sin x - 3 \cos x )$

Now, As we know, the Multiplication property of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Hence derivative of the given function is:

$\frac{d(x ^ 4 ( 5 \sin x - 3 \cos x ))}{dx}=x^4\frac{d ( 5 \sin x - 3 \cos x )}{dx}+ ( 5 \sin x - 3 \cos x )\frac{dx^4}{dx}$

$\frac{d(x ^ 4 ( 5 \sin x - 3 \cos x ))}{dx}=x^4(5\cos x+3\sin x)+ ( 5 \sin x - 3 \cos x )4x^3$

$\frac{d(x ^ 4 ( 5 \sin x - 3 \cos x ))}{dx}=5x^4\cos x+3x^4\sin x+ 20x^3 \sin x - 12x^3 \cos x$

Given

$f(x)=( x^2 +1 ) \cos x^{}$

Now, As we know the product rule  of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

The derivative of the given function is

$\frac{d(( x^2 +1 ) \cos x)}{dx}=( x^2 +1 ) \frac{d\cos x}{dx}+\cos x\frac{d( x^2 +1 ) }{dx}$

$\frac{d(( x^2 +1 ) \cos x)}{dx}=( x^2 +1 ) (-\sin x)+\cos x(2x)$

$\frac{d(( x^2 +1 ) \cos x)}{dx}= -x^2 \sin x-\sin x+2x\cos x$

Given,

$f(x)=( ax ^2 + \sin x ) ( p + q \cos x )$

Now As we know the Multiplication property of derivative,(the product rule)

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

And also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties we get,

$\frac{d(( ax ^2 + \sin x ) ( p + q \cos x ))}{dx}=(ax^2+\sin x)\frac{d(p+q\cos x)}{dx}+(p+qx)\frac{d(ax^2+sinx)}{dx}$

$\\\frac{d((ax ^2+\sin x ) ( p + q \cos x ))}{dx}=(ax^2+\sin x)(-q\sin x)+(p+qx)(2ax+\cos x)$

$f'(x)=-aqx^2\sin x-q\sin^2 x+2apx+p\cos x+2aqx^2+qx\cos x$

$f'(x)=x^2(-aq\sin x+2aq) +x(2ap+q\cos x)+p\cos x-q\sin^2 x$

Given,

$f(x)=( x+ \cos x ) ( x - \tan x )$

And the Multiplication property of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties we get,

$\frac{d(( x+ \cos x ) ( x - \tan x ))}{dx}=(x+\cos x)\frac{d(x-\tan x)}{dx}+(x-\tan x)\frac{d(x+\cos x)}{dx}$

$=(x+\cos x)(1-\sec^2x)+(x-\tan x)(1-\sin x)$

$=(x+\cos x)(-\tan^2x)+(x-\tan x)(1-\sin x)$

$=(-\tan^2x)(x+\cos x)+(x-\tan x)(1-\sin x)$

$\frac{4x + 5 \sin x }{3x+ 7 \cos x }$

Given,

$f(x)=\frac{4x + 5 \sin x }{3x+ 7 \cos x }$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties,we get

$\frac{d(\frac{4x+5\sin x}{3x+7\cos x})}{dx}=\frac{(3x+7\cos x)d(\frac{d(4x+5\sin x)}{dx})-(4x+5\sin x)(\frac{d(3x+7\cos x)}{dx})}{(3x+7\cos x)^2}$

$\frac{d(\frac{4x+5\sin x}{3x+7\cos x})}{dx}=\frac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}$

$=\frac{12x+28\cos x+15x\cos x+35\cos^2x-12x-15\sin x+28x\sin x+35\sin^2 x}{(3x+7\cos x)^2}$

$=\frac{12x+28\cos x+15x\cos x-12x-15\sin x+28x\sin x+35(\sin^2 x+\cos^2x)}{(3x+7\cos x)^2}$

$=\frac{28\cos x+15x\cos x-15\sin x+28x\sin x+35}{(3x+7\cos x)^2}$

$\frac{x ^2 \cos ( \pi /4 )}{\sin x }$

Given,

$f(x)=\frac{x ^2 \cos ( \pi /4 )}{\sin x }$

Now, As we know the derivative of any function

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Hence the derivative of the given function is

$\frac{d(\frac{x^2\cos(\pi/4)}{\sin x})}{dx}=\frac{(\sin x)d(\frac{d(x^2\cos(\pi/4))}{dx})-(x^2\cos(\pi/4))(\frac{d\sin x}{dx})}{\sin^2x}$

$\frac{d(\frac{x^2\cos(\pi/4)}{\sin x})}{dx}=\frac{(\sin x)(2x\cos (\pi/4))-(x^2\cos(\pi/4))(\cos x)}{\sin^2x}$

$\frac{d(\frac{x^2\cos(\pi/4)}{\sin x})}{dx}=\frac{2x\sin x\cos (\pi/4)-x^2\cos x\cos(\pi/4)}{\sin^2x}$

$\frac{x }{ 1+ \tan x }$

Given

$f(x)=\frac{x }{ 1+ \tan x }$

Now, As we know the derivative of any function

$\frac{d(\frac{x}{1+\tan x})}{dx}=\frac{(1+\tan x)d(\frac{dx}{dx})-x(\frac{d(1+\tan x)}{dx})}{(1+\tan x)^2}$

$\frac{d(\frac{x}{1+\tan x})}{dx}=\frac{(1+\tan x)(1)-x(\sec^2x)}{(1+\tan x)^2}$

$\frac{d(\frac{x}{1+\tan x})}{dx}=\frac{1+\tan x-x\sec^2x}{(1+\tan x)^2}$

Given

$f(x)=( x + \sec x ) ( x - \tan x )$

Now, As we know  the Multiplication property of derivative,

$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$

Also the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

Applying those properties we get,

the derivative of the given function is,

$\frac{d(( x + \sec x ) ( x - \tan x )}{dx}=(x+\sec x)\frac{d(x-\tan x)}{dx}+(x-\tan x)\frac{d(x+\sec x)}{dx}$

$\frac{d(( x + \sec x ) ( x - \tan x )}{dx}=(x+\sec x)(1-\sec^2x)+(x-\tan x)(1+\sec x\tan x)$

$\frac{x }{\sin ^ n x }$

Given,

$f(x)=\frac{x }{\sin ^ n x }$

Now, As we know the derivative of any function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

Also chain rule of derivative,

$[f(g(x))]'=f'(g(x))\times g'(x)$

Hence the derivative of the given function is

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nxd(\frac{dx}{dx})-x(\frac{d(\sin^nx)}{dx})}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nx(1)-x(n\sin^{n-1}x\times \cos x)}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nx-x\cos xn\sin^{n-1}x}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^nx-x\cos xn\sin^{n-1}x}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{\sin^{n-1}x(\sin x-nx\cos x)}{sin^{2n}x}$

$\frac{d(\frac{x}{\sin^nx})}{dx}=\frac{(\sin x-nx\cos x)}{sin^{n+1}x}$

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13