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

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

Edited By Komal Miglani | Updated on Apr 24, 2025 12:21 PM IST

The concept of derivatives is explained in this exercise. In our daily life, we often observe situations where one quantity changes with respect to another, like the speed of a moving car, rise in water level in the oceans or reservoirs, or a change in prices of stocks with time. In mathematics, these concepts are explained using the concept of derivatives. Derivative is defined as the rate at which one quantity changes with respect to another or how fast or slow something is changing at a particular moment. The concept of derivatives forms the foundation for higher-level calculus.

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  1. Class 11 Maths Chapter 12 Exercise 12.2 Solutions - Download PDF
  2. Download PDF
  3. NCERT Solutions Class 11 Maths Chapter 12: Exercise 12.2
  4. Topic covered in Chapter 12 Limits and Derivatives Exercise 12.2
  5. NCERT Solutions of Class 11 Subject Wise
  6. Subject-Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 13.2 Class 11 Maths Chapter 13 - Limits and Derivatives
NCERT Solutions for Exercise 13.2 Class 11 Maths Chapter 13 - Limits and Derivatives

In NCERT, students will learn how to determine the derivatives of algebraic, polynomial, and trigonometric functions. NCERT Solutions of Exercise 12.2 are designed by subject experts in a very comprehensive and systematic manner that ensures the accuracy of the solutions. Derivatives help us to calculate how things change, which is useful in science, engineering, and economics. Check NCERT Solutions to get detailed solutions for Science and Maths from Class 6 to Class 12.

Class 11 Maths Chapter 12 Exercise 12.2 Solutions - Download PDF

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NCERT Solutions Class 11 Maths Chapter 12: Exercise 12.2

Question1: 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

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

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

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

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

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

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

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

$f'(10)=20+0$

$f'(10)=20$

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

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

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

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

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

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

$f'(1)=1$ (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

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

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

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

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

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

$f'(100)=99$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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,

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

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

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

Also read

Topic covered in Chapter 12 Limits and Derivatives Exercise 12.2

Exercise 12.2 introduces students to the derivative of a function. In this, students will find questions that primarily revolve around the following key topics:

1) Derivatives: Derivative is the measure of the rate at which the function is changing at a given point.

Mathematically derivative of the function f(x) at point x=a is

$f^{\prime}(a)=\lim\limits _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$

The instantaneous rate of change of a function at point a’ is shown by this formula.

2) Algebra of derivative of function : Rules used to find the differentiation of the combination of functions.

  • Addition and Subtraction rule : $\frac{d}{d x}(f(x) \pm g(x))=\frac{d}{d x} f(x) \pm \frac{d}{d x} g(x)$
  • Multiplication Rule : $\frac{d}{d x}(c \cdot f(x))=c \cdot \frac{d}{d x} f(x)$
  • Product Rule: $\frac{d}{d x}(f(x) \cdot g(x))=f^{\prime}(x) \cdot g(x)+f(x) \cdot g^{\prime}(x)$
  • Quotient Rule: $\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) \cdot f^{\prime}(x)-f(x) \cdot g^{\prime}(x)}{[g(x)]^2}$
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3) Derivative of Polynomial : $\frac{d}{d x}\left(x^n\right)=n \cdot x^{n-1}$

4) Derivatives of Trigonometric functions :

  • Sin function: $\frac{d}{d x} \sin x=\cos x$
  • Cos function: $\frac{d}{d x} \cos x=-\sin x$
  • Tan function: $\frac{d}{d x} \tan x=\sec ^2 x$
  • Cot function: $\frac{d}{d x} \cot x=-\csc ^2 x$
  • Sec function: $\frac{d}{d x} \sec x=\sec x \cdot \tan x$
  • Cosec function: $\frac{d}{d x} \csc x=-\csc x \cdot \cot x$

Also Read

NCERT Solutions of Class 11 Subject Wise

Students can refer to the subject-wise NCERT solutions. The links to solutions are given below:

Subject-Wise NCERT Exemplar Solutions

Students can access the NCERT exemplar solutions to enhance their deep understanding of the topic. These solutions are aligned with the CBSE syllabus and also help in competitive exams.

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 12 exercise 12.2?

11 questions are solved from exercise 12.2 Class 11 Maths

8. List out the topics covered in the Class 11th Maths chapter 12 exercise 12.2

Definition of derivative, derivative of polynomial and trigonometric functions. 

Articles

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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

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

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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