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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 Vishal kumar | Updated on Nov 14, 2023 02:17 PM IST

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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  1. NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives Exercise 13.2- Download Free PDF
  2. Download the PDF of NCERT Solutions for Class 11 Maths Chapter 13 – Limits and Derivatives Exercise 13.2
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  4. Limits And Derivatives Class 11 Chapter 13 Exercise 13.2
  5. Question:1 Find the derivative of
  6. More About NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2
  7. Benefits of NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2
  8. Key Features of NCERT Class 11 Maths Ex 13.2 Solution
  9. NCERT Solutions of Class 11 Subject Wise
  10. Subject Wise NCERT Exampler 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

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

More About NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2

Most of the questions in exercise 13.2 Class 11 Maths are related to the derivative of polynomial functions. There are eleven questions solved under NCERT solutions for Class 11 Maths chapter 13 exercise 13.2. Questions 10 and 11 of Class 11 Maths chapter 13 exercise 13.2 asks to find the derivative of some trigonometric functions. Derivatives are such concepts that require good practice since these are frequently used in the field of science and technology, engineering, economics etc.

Also Read| Limits And Derivatives Class 11 Notes

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

  1. 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.

  2. 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.

  3. Clarity and Precision: Written with clarity and precision, ensuring easy comprehension of mathematical concepts and techniques.

  4. Proper Mathematical Notation: Utilizes appropriate mathematical notations and terminology to enhance fluency in mathematical language.

  5. Conceptual Understanding: Class 11 ex 13.2 solutions aim to deepen conceptual understanding rather than promoting rote memorization, encouraging critical thinking.

  6. Free Accessibility: 11th class maths exercise 13.2 answers are typically available free of charge, providing an accessible resource for self-study.

  7. Supplementary Learning Tool: Serves as a supplementary resource to complement classroom instruction and support exam preparation.

  8. 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.

Also, see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Frequently Asked Question (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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Popular Questions

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
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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

 
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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
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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
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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 .
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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
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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.
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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.
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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
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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
Read More

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