NCERT Solutions for Exercise 13.2 Class 11 Maths Chapter 13

NCERT Solutions for Exercise 13.2 Class 11 Maths Chapter 13 - Limits and Derivatives

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NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives Exercise 13.2- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 13: Limits and Derivatives Exercise 13.2-NCERT Solutions for Exercise 13.2 Class 11 Maths chapter 13 discuss derivatives and the questions related to them. The derivative is an important concept as the field of Science, Commerce and Engineering are concerned. So understanding the basics of derivatives is important. Exercise 13.2 Class 11 Maths helps to understand the basics of derivatives.

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Limits And Derivatives Class 11 Chapter 13 Exercise 13.2
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NCERT Solutions for Exercise 13.2 Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT solutions for Class 11 Maths Chapter 13 Exercise 13.2 solve questions from the definition of the derivative, algebra of derivative functions and Class 11 Maths chapter 13 exercise 13.2 presents questions from derivative of trigonometric and polynomial functions. The basic derivative ideas used to solve the Class 11th Maths chapter 13 exercise 13.2 are also helpful in NCERT syllabs Class 11 and 12 Physics and Chemistry also. With Class 11 Maths chapter 13 exercise 13.2, the following exercise is also present in the chapter limits and derivatives. All the solutions including ex 13.2 class 11 are written and prepared by subject experts at Careers360 in a very easy-to-understand language. Additionally, PDF formats of the solutions are available so that students can access them anytime without incurring any charges.

**In the CBSE Syllabus for the academic year 2023-24, this chapter has been renumbered as Chapter 12.

Download the PDF of NCERT Solutions for Class 11 Maths Chapter 13 – Limits and Derivatives Exercise 13.2

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Limits And Derivatives Class 11 Chapter 13 Exercise 13.2

Question:1 Find the derivative of $x ^ 2 -2 \: \: at \: \: x = 10$

Answer:

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.

Answer:

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.

Answer:

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:

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

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

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

$f'(100)=99$

Question:4 (i) Find the derivative of the following functions from first principle. $x ^3 -27$

Answer:

Given

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

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

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

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

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

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

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

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

Question:4.(ii) Find the derivative of the following function from first principle. $( x-1)(x-2)$

Answer:

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)

Question:4(iii) Find the derivative of the following functions from first principle. $1 / x ^2$

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)

Question:4(iv) Find the derivative of the following functions from first principle. $\frac{x +1}{x-1}$

Answer:

Given:

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

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

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

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

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

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

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

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

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

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

Question:5 For the function $f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1$ Prove that f '(1) =100 f '(0).

Answer:

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

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.

Question:6 Find the derivative of $x ^n + ax ^{n-1} + a ^ 2 x ^{n-2} + ....+ a ^ { n-1} x + a ^n$ for some fixed real number a.

Answer:

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

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

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

Question:7(i) For some constants a and b, find the derivative of $( x - a ) ( x -b )$

Answer:

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

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

Question:7(ii) For some constants a and b, find the derivative of $( ax ^2 + b)^2$

Answer:

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

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

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

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

Question:7(iii) For some constants a and b, find the derivative of $\frac{x - a }{x -b }$

Answer:

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

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

Question:8 Find the derivative of $\frac{x ^n - a ^n }{x - a }$ for some constant a.

Answer:

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

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

Question:9(i) Find the derivative of $2x - 3/4$

Answer:

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

$f'(x)=2-0$

$f'(x)=2$

Question:9(ii) Find the derivative of $( 5x^3 + 3x -1 ) ( x -1)$

Answer:

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

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

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

Question:9(iii) Find the derivative of $x ^{-3} ( 5 + 3x )$

Answer:

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

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

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

Question:9(iv) Find the derivative of $x ^5 ( 3 - 6 x ^{-9})$

Answer:

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

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

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

Question:9(v) Find the derivative of $x ^{-4} ( 3 - 4x ^{-5})$

Answer:

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

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

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

Question:9(vi) Find the derivative of $\frac{2}{x+1}- \frac{x^2 }{3 x-1}$

Answer:

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

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

Question:10 Find the derivative of $\cos x$ from first principle.

Answer:

Given,

f(x)= $\cos x$

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

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

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

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

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

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

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

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

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

Question:11(i) Find the derivative of the following functions: $\sin x \cos x$

Answer:

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,

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

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

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

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

Question:11(ii) Find the derivative of the following functions: $\sec x$

Answer:

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

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

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

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

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

Question:11 (iii) Find the derivative of the following functions: $5 \sec x + 4 \cos x$

Answer:

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

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

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

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

Question:11(iv) Find the derivative of the following functions: $\csc x$

Answer:

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

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

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

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

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

Question:11(v) Find the derivative of the following functions: $3 \cot x + 5 \csc x$

Answer:

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

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

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

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

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

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

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

Question:11(vi) Find the derivative of the following functions: $5 \sin x - 6 \cos x + 7$

Answer:

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,

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

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

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

Question:11(vii) Find the derivative of the following functions: $2 \tan x - 7 \sec x$

Answer:

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,

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

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

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

Benefits of NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2

As mentioned above derivatives are used in the subjects physics and chemistry and maths of Class 11 and also. So practising exercise 13.2 Class 11 Maths is crucial to get a good idea of concepts.
All the questions listed in the Class 11 Maths chapter 13 exercise 13.2 are important and may get questions from the exercise for Class 11 final exams.

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Key Features of NCERT Class 11 Maths Ex 13.2 Solution

Complete Exercise Coverage: The class 11 maths ex 13.2 solutions encompass all exercises and problems in Exercise 13.2 of the Class 11 Mathematics textbook.
Step-by-Step Solutions: The ex 13.2 class 11 solutions are presented in a clear, step-by-step format, aiding in understanding and application of problem-solving methods.
Clarity and Precision: Written with clarity and precision, ensuring easy comprehension of mathematical concepts and techniques.
Proper Mathematical Notation: Utilizes appropriate mathematical notations and terminology to enhance fluency in mathematical language.
Conceptual Understanding: Class 11 ex 13.2 solutions aim to deepen conceptual understanding rather than promoting rote memorization, encouraging critical thinking.
Free Accessibility: 11th class maths exercise 13.2 answers are typically available free of charge, providing an accessible resource for self-study.
Supplementary Learning Tool: Serves as a supplementary resource to complement classroom instruction and support exam preparation.
Homework and Practice: Students can use these class 11 ex 13.2 solutions to cross-verify their work, practice problem-solving, and enhance overall mathematical skills.

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Frequently Asked Questions (FAQs)

1. What is the value of f(x) =80x at x=100

The value of f(x) at 100 =8000

2. What is the value of derivative of f(x)=80x at x=100

The derivative of 80x is 80 which is a constant

3. Find the value of the derivative of 80x^2 at x=100

The derivative of 80x^2=160x and the value at 100=16000

4. Tanx is a trigonometric function. What is its derivative?

Sec^2x

5. What is the derivative of a constant ‘a’?

The derivative of ‘a’ =0 since ‘a’ is a constant

6. Why the derivative of a constant = 0?

Since the rate of change of constant =0, the derivative =0

7. What number of questions are solved in NCERT solutions for Class 11 Maths chapter 13 exercise 13.2?

11 questions are solved from exercise 13.2 Class 11 Maths

8. List out the topics covered in the Class 11th Maths chapter 13 exercise 13.2

Definition of derivative, derivative of polynomial and trigonometric functions.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 13.2 Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives Exercise 13.2- Download Free PDF

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Limits And Derivatives Class 11 Chapter 13 Exercise 13.2

Question:1 Find the derivative of

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Question:1 Find the derivative of $x ^ 2 -2 \: \: at \: \: x = 10$