Limits and Derivatives Class 11 NCERT Solutions | Exercise-wise Answers

NCERT Maths 11th 2023 ( Maths Ncert Solutions class 11th )

Pallavi Pathak
Updated on Jul 28, 2025 14:46 IST

By Pallavi Pathak, Assistant Manager Content

The subject matter experts at Shiksha created the Class 11 Limits and Derivatives NCERT solutions. The chapter introduces Calculus, a branch of mathematics that studies the change in the value of a function as the points in the domain change. Limits and Derivatives class 11 Chapter 12 Maths includes the definitions of limits and derivatives and some algebra involving these concepts.
The step-by-step solutions help students to understand the concepts clearly and also provide them the confidence to appear in their school exam, CBSE Board exam, and other entrance tests later on. These solutions are accurate, and students can depend on them for their exam preparation. If they practiced only from these solutions, after going through the NCERT chapter, it would be more than sufficient for their examination preparation.
For comprehensive subject-wise NCERT notes of Class 11 and Class 12, Check Here. You will get the reliable notes for quick revision.

Related Links

NCERT Solutions for Class 11 Maths NCERT Class 11 Notes NCERT Solutions Class 11 and 12 for Maths, Physics, Chemistry
Table of content
  • Class 11 Maths Chapter 12 Snapshot: Limits and Derivatives
  • Class 11 Math Limits and Derivatives: Key Topics, Weightage
  • Important Formulas of Class 11 Limits and Derivatives
  • Class 11 Maths Chapter 12 NCERT Solutions – Limits and Derivatives PDF Free
  • Class 11 Math Limits and Derivatives Exercise-wise Solution
  • Class 11 Math Limits and Derivatives Exercise 13.1 Solution
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Class 11 Maths Chapter 12 Snapshot: Limits and Derivatives

Here is a quick summary of the Limit and Derivatives Class 11 Maths:

  • According to Chapter 12 of Class 11 Maths, lim x a - f ( x ) is the expected value of f, which is called the left-hand limit of f at a, when x=a. Here, it is given that the values of f near x to the left of a.
  • To find the right-hand limit of f(x) at a, mathematically we show as - lim x a + f ( x ) . It is the expected value of f at x = a, given the values of f near x to the right of a.
  • There is also the concept of common value when the right and left-hand limits coincide. It is denoted by lim x a f ( x )

To get access to the chapter-wise Class 11 Maths NCERT Notes that offer solved examples and also provide the important topics, students must check here.

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Class 11 Math Limits and Derivatives: Key Topics, Weightage

Limits Class 11 Maths Chapter 12 is the part of the broader "Limits, Continuity & Differentiability" topic in the JEE Main exam. Hence, the students must focus on the core concepts of the chapter. Before starting the preparation of the chapter, it is wise to know the topics of this chapter. See below the topics covered in this chapter:

Exercise Topics Covered
12.1 Introduction
12.2 Intuitive Idea of Derivatives
12.3 Limits

Class 11 Limits and Derivatives Weightage in JEE Main Exam

Exam Number of Questions Weightage
JEE Main 1 to 3 questions 3.3% to 10%
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Important Formulas of Class 11 Limits and Derivatives

Important Formulae of Limits and Derivatives for CBSE and Competitive Exams

  • First Principle of Differentiation f ( x ) = lim h 0 f ( x + h ) f ( x ) h f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
  • Power Functions: d d x ( x n ) = n x n 1 \frac{d}{dx} (x^n) = n x^{n-1}
  • Trigonometric Functions:
    • d d x ( sin x ) = cos x \frac{d}{dx} (\sin x) = \cos x
    • d d x ( cos x ) = sin x \frac{d}{dx} (\cos x) = -\sin x
    • d d x ( tan x ) = sec 2 x \frac{d}{dx} (\tan x) = \sec^2 x
    • d d x ( cot x ) = cosec 2 x \frac{d}{dx} (\cot x) = -\csc^2 x
    • d d x ( sec x ) = sec x tan x \frac{d}{dx} (\sec x) = \sec x \tan x
    • d d x ( csc x ) = cosec x cot x \frac{d}{dx} (\csc x) = -\csc x \cot x
  • Exponential & Logarithmic Functions:
    • d d x ( e x ) = e x \frac{d}{dx} (e^x) = e^x
    • d d x ( a x ) = a x ln a \frac{d}{dx} (a^x) = a^x \ln a
    • d d x ( log a x ) = 1 x ln a \frac{d}{dx} (\log_a x) = \frac{1}{x \ln a}
  • Rules of Differentiation

    • Sum Rule: ( f + g ) = f + g (f + g)' = f' + g'
    • Difference Rule: ( f g ) = f g (f - g)' = f' - g'
    • Product Rule: ( u v ) = u v + u v (uv)' = u'v + uv'
    • Quotient Rule: ( u v ) = u v u v v 2 \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
    • Chain Rule: d d x f ( g ( x ) ) = f ( g ( x ) ) g ( x ) \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)
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Class 11 Maths Chapter 12 NCERT Solutions – Limits and Derivatives PDF Free

If students download the free Limits and Derivatives Class 11 PDF from the link given below, they will get the well-structured solutions of all the NCERT questions of chapter 12 Class 11 Maths. It will help them to do well in school exam, CBSE, and the JEE Main exam.

Chapter 5 Limits and Derivatives Class 11 NCERT Solutions: Download PDF for Free

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Class 11 Math Limits and Derivatives Exercise-wise Solution

Class 11th Math Limits and Derivatives helps students understand the fundamentals of calculus unit, including differentiation, application of derivatives, integral and more. There are various important topics in this chapter such as checking differentiability, continuity, derivatives of different functions, and more. Limits and Derivatives Class 11 Solutions will include solution of all the exercises in lucid language. Ex 13.1 Class 11 will include problems related to the basic concepts of Limits and Derivatives such as using limit, finding continuity and differentiability. Class 11 Math Ex 13.2 includes more concepts such as finding derivatives of various functions and using them as application. Miscellaneous exercise covers various topics covered throughout the chapter. Students can check all the solutions below;

 

 

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Class 11 Math Limits and Derivatives Exercise 13.1 Solution

Limits and Derivatives Exercise 13.1 focuses on fundamentals of the Limits. Student will be able to understand the application of limit to find first derivative of the given function for a given range. Ex 13.1 Class 11 Math in the Limits and Derivatives includes problems based on trigonometric functions, algebraic functions and exponential functions and finding their first derivative using the limits. Class 11 Math Ex 13.1 solution will contain solution of all 32 short and long answer type questions. Students can check the solutions below;

Class 11 Math Limits and Derivatives Exercise 13.1 NCERT Solution

Q1.  l i m x 3  x+ 3
A.1.  l i m x 3 x + 3 = 3 + 3 = 6.

 

Q2.  lim x π ( x 2 2 7 )
A.2.  l i m x π ( x 2 2 7 ) = π 2 2 7
Q3.  lim r 1 π r 2
A.3 l i m r 1 π r 2 = π · ( 1 ) 2 = π
Q4.  lim x 4 4 x + 3 x 2
A4.  l i m x 4 4 x + 3 x 2 = 4 × 4 + 3 4 2 = 1 6 + 3 2 = 1 9 2
Q&A Icon
Commonly asked questions
Q:  

40. Find the derivative of xnanxa for some constant a.

A: 

40. Given, f(x)= xnanxa.

So, f(x)=(xa)ddx(xna3)ddx(xa)(xnan)(xa)2=(xa)nxn1(xnan)(xa)2

=(xa)nxn1(xnan)(xa)2

=nxn1xnaxn1xn+an(xa)2

=nxnxxnaxn1+an(xa)2

Q:  

42. Find the derivative of the following functions:

(i)sin x cos x  (ii)secx (iii)5 sec x + 4 cosx

(iv) cosecx (v)3cot x + 5 cosec x

(vi) 5sinx6cosx+7 (vii) 2tanx7secx

Read more
A: 

(i) f(x)=sin x cos x

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?0sin(x+h)cos(x+h)?sinxcosxh

=limh?012h×[2sin(x+h)cos(x+h)?2sinxcosx]

=limh?012h[sin2(x+h)?sin2x]

=limh?012h[2cos2(x+h)+2x2sin2(x+h)?2x2]

=limh?01h[cos(2x+h)sinh]

=limh?0cos(2x+h)×limh?0sinhh

=cos(2x+0)

=cos2x

(ii) f(x)=secx

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?01h[sec(x+h)?secx]

=limh?01h[1cos(x+h)?1cosx]

=limh?01h[cosx?cos(x+h)cos(x+h)cosx]

=limh?01h[?2sin(x+x+h2)sin(x?(x+h)2)cos(x+h)cosx]

=limh?01h[?2sin(2x+h2)sin(?h/2)cos(x+h)cosx]

=limh?0(?1sin(2x+h2)cos(x+h)cosx×limh?0(?1)sinh/2h/2

=sinxcosx?cosx×1

=tanx?secx.

(iii) Given f(x)=5 sec x+4 cosx.

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?05sec(x+h)+4cos(x+h)?[5secx+4cosx]h.

=limh?05h[sec(x+h)?secx]+limh?04h[cos(x+h)?cosx]

=limh?05h[1cos(x+h)?1cosx]+limh?04h[?2sin(x+h+x2)sin(x+h?x2)]

=limh?05h[cosx?cos(x+h)cos(x+h)(cosx)]+limh?04h[?2sin(2x+h2)sinh2]

=limh?05h[?2sin(2x+h2)sin(?h/2)cos(x+h)cosx]?4limh?0sin(2x+h2)limh?0sinh/2h/2

=sin(2x+02)cos(x+0)cosx×1?4sin(2x2)

=5sinxcosx?1cosx?4sinx

=5tanx?secx?4sinx

(iv) Given f(x)=cosecx

f?(x)=limh?0f(x+h)?f(x)h

=limh?01h[cosec(x+h)?cosecx]

=limh?01h[1sin(x+h)?1sinx]

=limh?01h[sinx?sin(x+h)sin(x+h)sinx]

=limh?01h[2cos(x+x+h2)sin(x?(x+h)2)]sin(x+h)sinx]

=limh?01h[2cos(x+x+h2)sin(x?(x+h)2)sin(x+h)sinx]

=limh?01h[2cos(2x+h2)sin(?h/2)]sin(x+h)sinx]

=limh?0cos(2x+h2)sin(x+h)sinx×(?1)sin(2)h/2)

=cos(2x+02)sin(x+0)sinx×(?1)

=?cosxsinx×1sinx

=?cotx?cosecx

(v) Given,f(x)=3 cot x+5cosecx.

So, f?(x)=limh?0f(x+h)?f(x)h =2cos(x+0)cosx×1+7sin(2x+02)×(?1)

=limh?03cot(x+h)+5cosec(x+h)?[3cotx+5cosx)

h?03h[cot(x+h)?cotx]+limh?0

=?5h[cosec(x+h)?cosecx]

=limh?03h[cos(x+h)sin(x+h)?cosxsinx]+limh?05h[1sin(x+h)?1sinx]

=limh?03h[cos(x+h)sinx?cosxsin(x+h)sin(x+h)sinx]+limh?05h[sinx?sin(x+h)sin(x+h)sinx]

=limh?03h[sin(x?(x+h))sin(x+h)sinx+limh?05h[2cos(x+x+h2)sin(x?(x+h)2)sin(x+h)sinx

=limh?03sin(x+h)sinx×limh?0(?1)sinhh+limh?05h[2cos(2x+h2)sin(?h/2)sin(x+h)sinx

=3sinx?sinx×(?1)+5limh
Q:  

52.

Kindly consider the following

px2+qx+rax+b

A: 

52. Given, f (x) = px2+qx+rax+b

f(x)=(ax+b)ddx(px2+qx+r)(px2+qx+r)ddx(ax+b)(ax+b)2

=(ax+b)(2xp+q)(px2+qx+r)(a)(ax+b)2

=2apx2+2bpx+aqx+bqapx2aqxar(ax+b)2

=apx2+2bpx+bqar(ax+b)2

Q:  

59. 

Kindly consider the following

cosx1+sinx

A: 

59. Given, f (x) = cosx1+sinx

So, f?(x) = (1+sinx)ddx(cosx)cosxddx(1+sinx)(1+sinx)2

Putting (2) and (3) in (1) we get,

f(x)=(1+sinx)(sinx)cosx(cosx)(1+sinx)2

=sinxsin2xcos2x(1+sinx)2

=sinx(sin2x+cos2x)(1+sinx)2

=(sinx+1)(1+sinx)2=11+sinx.

Q:  

66.

Kindly consider the following

(x2 + 1) cos x

A: 

66. Given, f (x) = (x2 + 1) cos x

f? (x) = (x2 + 1) ddxcosx+cosxddx (x2+1)

= (x2+1) (sinx)+cosx (2x+0) [? ddxcosx=sinx]

= x2 sin x sin x + 2x cos x.

Q:  

Kindly consider the following

68. (x + cos x) (x tan x)

A: 

68. Given, f (x) = (x + cos x) (x tan x)

So, f?(x) = (x + cos x) ddxtanx)+ddx(x+cosx)·(xtanx)

=(x+cosx)(dxdxddxtanx)+(ddx+dcosxdx)(xtanx)

Let g (x) = tan x.

So, g?(x) =

 

=limh01h[sin(x+h)cos(x+h)sinxcosx]

=limh01h[sin(x+h)cosxsinxcos(x+h)cos(x+h)·cosx]

=limh01h[sin(x+hx)cosx·cos(x+h)]

=limh01cosx·cos(x+h)×limh0sinhh

=1cos2x×1

= sec2x ______ (2)

     Put (2) in (1) we get,

     f?(x) = (x + cos x) (1 - sec2x) +(1 - sin x) (x- tan x)

     We know that,

     1 + tan2x = sec2x

     Þ 1 - sec2x = - tan2x

     So, f?(x) = - tan2x(x + cos x) +(x- tan x) (1 - sin x).

Q:  

17. Kindly consider the following

A: 

17. Kindly go through the solution

==limx02sin2x2sin2x2

=limx0 (sinxx)2×x2limx0 (sinx2x2)2x22

= (1)2×x2×4 (1)2×x2

= 4

Q:  

30. If f(x)={|x|+1,x<00,x=0|x|1,x>0 For what value (s) of a does limxaf(x) exists?

A: 

30. Given, f(x)={|x|+1,x<00,x=0|x|1,x>0limxaf(x)=?

As | x | = {x,x<0x,x>0

We can rewrite f (x) = {x+1,x<00,x=0x1,x>0.

Case 1: when a<0,

limxaf(x)=limxa(x+1)=a+1

So, limxaf(x) = exist such that a< 0

Case II when a> 0,

limxaf(x)=limxa(x1)=a1

So, limxaf(x) exist such that a>0.

Case III when a = 0.

L.H.L = limxaf(x)=limxa(x+1)=limx0(x+1)=1

R.H.L = limxa+f(x)=limxa+(x1)=limx0+(x1)=1

Thus, limxaf(x)limxa+f(x)

So, limxaf(x) does not exist at a = 0.

Q:  

22. limxπ2tan2xxπ2

A: 

22. limxx2tan2xxπ2

Put y = x π2 . So that as y 0 cos π2

Q:  

12. limx21x+12x+2

A: 

Kindly go through the solution

=12×(2)=14.

Q:  

10 limz1z131z161

A: 

10.=limz1[z13113z1]÷limz1[z16116z1]

= l i m z 1 [ z 1 3 1 1 3 z 1 ] ÷ l i m z 1 [ z 1 6 1 1 6 z 1 ]

We know that,

limxaxnanxa=nan1

So,

= 1 3 × 6 = 2
Q:  

6. limx0(x+1)51x

A: 

6. limx0(x+1)51x=limx0x5+5x4+10x3+10x2+5x+11

Q6. limx0(x+1)51x

6. limx0(x+1)51x=limx0x5+5x4+10x3+10x2+5x+11

= l i m x 0 x 4 + 5 x 3 + 1 0 x 2 + 1 0 x + 5 .

(0)4 + 5(0)3 + 10(0)2 + 10(0) + 5

= 5.

Q:  

Kindly consider the following

71. x1+tanx

A: 

71. Given, f (x) x1+tanx

=x1+sinxcosx

=cosx·xcosx+sinx

So,f(x)=(cosx+sinx)ddx(xcosx)(xcosx)ddx(cosx+sinx)(cosx+sinx)2

 

=(cosx+sinx)(xsinx+cosx)(x·sinxcosx+xcos2x)(cosx+sinx)2

=xcosxsinx+cos2xxsin2x+sinxcosx+xsinxcosxxcos2x(cosx+sinx)2

=cos2x+sinxcosxx(sin2x+cos2x)(cosx+csinx)2

Dividing numerator and denominator by cos2x we get,

=cos2xcos2x+sinxcosx·cosxcosxx1cos2x(cosxcosx+sinxcosx)2

=1+tanxxsec2x(1+tanx)2

Q:  

1.  limx3 x+ 3

A: 

Exercise 13.1

1. 1. limx3x + 3 = 3 + 3 = 6.

Q:  

7. limx23x2x10x24

A: 

7.

=limx2x (3x+5)2 (3x+5) (x2) (x+2)

=limx2 (x2) (3x+5) (x2) (x+2)

=limx23x+5x+2

=3×2+52+2

=6+54

=114.

Q:  

9. Limx0ax+bcx+1

A: 

Kindly go through the solution

Q:  

18. limx0ax+xcosxbsinx

A: 

18. lim x0ax+xcosxbsinx=limx0x(a+cosx)bsinx

=1b×(a+cos0)1

=a+1b

Q:  

16. limx0cosxπx

A: 

16. limx0cosxπx=cos0π0=1π

Q:  

15. limxπsin(πx)π(πx)

A: 

15. limxπsin (πx)π (πx)=1π×limxπsin (πx)πx

=1π.

Q:  

8. limx3x4812x25x3

A: 

8.

=limx3 (x+3) (x2+9)2x+1.

= (3+3) (32+9)2×3+1

=6×186+1

=1087

Q:  

14. limx0sinaxsinbx,a,b0

A: 

14. limx0sinaxsinbx=limx0sinaxax×axsinbxbx×bx

=ab

Q:  

19. limx0xsecx

A: 

19. limx0xsecx=limx0xcosx

=0cos0

=01

= 0.

Q:  

4. limx44x+3x2

A: 

4. limx44x+3x2=4×4+342=16+32=192

Q:  

5. limx1x10+x5+1x1

A: 

5. limx1x10+x5+1x1= (1)10+ (1)5+1 (1)1=11+12=12

Q:  

21. limx0(cosecxcotx)

A: 

21. limx0(cosecxcotx)=limx0(1sinxcosxsinx)

=limx0(1cosxsinx)

=limx02sin2x22sinx2cosx2 {?cos2x=12sin2x1cos2x=2sin2x1cosx=2sin2x2?sin2x=2sinxcosxsinx=2sinx2cosx2{

=limx0x2

=tan02=0.

Q:  

2. limxπ(x227)

A: 

2. limxπ (x227)=π227

Q:  

3. limr1πr2

A: 

3limr1πr2=π· (1)2=π

Q:  

11. limx1ax2+bx+ccx2+bx+a=a+b+c  0

A: 

11. limx1ax2+bx+ccx2+bx+a=a+b+cc+b+a=1

Q:  

13. limx0axbx

A: 

13. limx0sinaxbx=1blimx0sinaxax×a

=ablimx0sinaxax 

=ab

Q:  

20. limx0sinax+bxax+sinbxa,b,a+b0

A: 

20. limx0sinax+bxax+sinbx=limx0sinaxax×ax+bxlimx0ax+sinbxbx×bx.

limx0sinaxax×limx0ax+limx0bxlimx0ax+limx0sinbxbx×limx0bx

=limx0ax+limx0bxlimx0ax+limx0bx

=limx0ax+bxax+bx

=limx01

= 1.

Q:  

23. Find limx0f(x) and limx1f(x) , where f (x) {2x+3,x03(x+1),x>0}

A: 

23. Given f (x) {2x+3,x03(x+1),x>0}

for limx0f(x),

left hand limit, L.H.S = limx0f(x) = limx0(2x+3)

= 2 0 + 3 = 3.

Right hand limit, R.H.L = limx0+f(x)=limx0+3(x+1)

= (0 + 1) = 3 1 = 3.

Thus, limx0f(x)=limx0+f(x)=limx0f(x)=3

For limx1f(x),

L.H.L = limx1f(x)=limx13(x+1)=3(1+1)=3×2=6

R.H.L = limx1+f(x)=limx1+3(x+1)=3(1+1)=3×2=6.

Thus, limx1f(x)=limx1+f(x)=limx1f(x)=6.

Q:  

24. Find limx1f(x) , where f(x)={x21,x1x21,x>1

A: 

24. Given f (x)= {x21, x1x21, x>1, limx1f (x)=?

Now, L.H.L = limx1f (x)=limx1 (x21)

12- 1 = 0

And R.H.L = limx1+f (x)=limx1+ (x21)  (1)2 1 = 1 = 2.

Thus,  limx1f (x)limx1+f (x)

So,  limx1f (x) does not exist.

Q:  

25. Evaluate limx0f(x), where f (x) = {|x|x,x00,x=0

A: 

25. Given f (x) = {|x|x,x00,x=0,limx0f(x)=?

We know that, |x|={x,x0x,x<0

Now,

L.H.L = limx0f(x)=limx0xx=limx01=1

and R.H.L = limx0+f(x)=limx0+xx=limx0+1=1

Thus, limx0f(x)limx0+f(x)

i e, limx0f(x) does not exist.

Q:  

26. Find limx0f(x), wheref (x) {x|x|,x00,x=0

A: 

26. Given, f (x) {x|x|, x00, x=0

L.H.S = limx0f (x)=limx0xx=limx01=1

R.H.L limx0+f (x)=limx0+xx=limx0+1=1

Thus,  limx0f (x)limx0+f (x)

i e,  limx0f (x) does not exist.

Q:  

27.Find limx5f(x),where f (x) =|x|5

A: 

27. limx5f (x)=limx5|x|5

= | 5 | 5

= 5 5

= 0

Q:  

28. Supposef (x) = {a+bx,x<14,x=1bax,x>1 and if limx1f(x) = f (1) what are possible values of a and b?

A: 

28. Given, f (x) =  {a+bx, x<14, x=1bax, x>1

Since we need limx1f (x) we need,

LHL =

limx1f (x)=limx1 (a+bx) = a + b × 1 = a + b

and RHL =

limx1+f (x)=limx1+ (bax) = b - a × 1 = b - a

Given,  limx1f (x)=f (1): we have the following equations

a + b = 4 ____ (1)

b - a = 4 ____ (2)

Adding (1) and (2) we get,

2b = 8

b = 4

Putting b = 4 in (1) we get

a + 4 = 4

a = 0

Q:  

31. If the function f(x) satisfies limx1f(x)2x21=π , evaluate limx1f(x).

A: 

31. Given, limx1f (x)2x21=π

limx1f (x)2limx1x21=π

limx1 [f (x)2]=πlimx1 (x21)

limx1f (x)limx12=π (121)

limx1f (x)2=0.

limx1f (x)=2

Q:  

39. For some constants aand b, find the derivative of

(i) (x- a) + (x - b) (ii) (ax2+b)2 (iii) x9xb

Read more
A: 

(c) Given, f(x)=(x-a)(x-b)

where a and b are constants.

So,

f(x)=(xa)ddx(xb)+(xb)ddx(xa)

=(x-a)+(x-b)

= 2x– a- b. 

(ii) Given f(x)= (ax2+b)2. where ab are constant

So, f(x)=ddx(ax2+b)2

=ddx(a2x4+b2+2ax2b)

=ddxa2x4+ddxb2+ddx2ax2b

4a2x3+0+4axb

4ax(ax2+b).

(iii) Given, f(x)= x9xb where a and bare constants

So, f(x)=(xb)ddx(xa)(xa)ddx(xb)(xb)2

=(xb)(xa)(xb)2

=ab(xb)2

Q:  

41. Find the derivative of

(i) 2x34 (ii) (5x3+3x1)(x1)

(iii) x3(5+3x) (iv) x5(36x9)

(v) x4(34x5) (vi) 2x+1x23x1

Read more
A: 

41. (i) f(x)=2x34

f(x)=ddx(2x34)

=2dxdx0

=2.

(ii) Given, f(x)= (5x3+3x1)(x1)

So, f(x)=(5x3+3x1)ddx(x1)+(x1)ddx(5x3+3x1)

=(5x3+3x1).1+(x1)(15x2+3)

5x3+3x1+i5x3+3x15x23

=20x315x2+6x4.

(iii) Given, f(x) = x3(5+3x)

So, f(x)=x3ddx(5+3x)+(5+3x)dx3dx=x33+(5+3x)(3)x4

=3x315x49x3

=15x46x3

=3x4(5+2x).

(iv) Given, f(x)= x5(36x9).

f(x)=x5ddx(36x9)+(36x9)ddxx5.

=x5(6x9x10)+(36x9)5x4

x5(54x10)+15x430x5

=54x530x5+15x4

=24x5+15x4

(v) Given, f(x)= x4(34x5).

So, f(x)=x4ddx(34x5)+(34x5)ddxx4

=x4(4x5×x6)+(34x5)(4x5)

=20x1012x5+16x10.

=36x1012x5.

=36x1012x5.

(vi) Given, f(x)= 2x+1x23x1

So, f(x)=ddx(2x+1)ddx(x23x1)

=(x+1)ddx2ddx(x+1)(x+1)2(3x1)dx2dxx2ddx(3x1)(3x1)2

=2(x+1)22x(3x1)3x2(3x1)2

=2(x+1)26x22x3x2(3x1)2

=2(x+1)23x22x(3x1)2

=2(x+1)2x(3x2)(3x1)2

Q:  

44. Find the derivative of the following functions from first principle:

(i) x (ii) (x)1 (iii) sin(x + 1) (iv) cos (xπ8)

Find the derivative of the following functions (it is to be understood that a, b, c, d,

p, q, r and s are fixed non-zero constants and m and n are integers):

Read more
A: 
44.
= l i m h 0 x h + x h
= l i m h 0 h h
= l i m h 0 1

= - 1.

(ii) Given, f(x) = (-x)-1

by first principle,

f(x) l i m h 0 [ ] ( x + h ) 1 ( x ) 1 h

(iii) Given, f(x) = sin(x + 1)

By first principle,

f’(x) = limh0f(x+h)f(x)h

=limh0sin(x+h+1)sin(x+1)h

=limh02h·cos(x+h+1+x+12)·sin(x+h+1(x+1)2)

=limh02hcos(2x+2+h2)sin(h2).

=limh0cos(2x+2+h2)limh0sinh2h2

=cos(2x+2+02)×1.

= cos (x + 1)

(iv) Given, f(x) = cos (xπ8)

By first principle,

f(x) = limh0f(x+h)f(x)h

=limh0cos(x+hπ8)cos(xπ8)h

 

Q:  

48. 

Kindly consider the following

1ax2+bx+c

A: 

48. Given, f (x) = 1ax2+bx+c

So, f? (x) =  (ax2+bx+c)ddx (1)1·ddx (ax2+bx+c) (ax2+bx+c)2

= (ax2+bx+c) (0) (2ax+b) (ax2+bx+c)2

= (2ax+b) (ax2+bx+c)2

Q:  

54.

Kindly consider the following

 

A: 
54. 

So f? (x) ddx (42)

=ddx (4)ddx (2)

=4ddx (x12)0=4×12x121=2x12=2

 

Q:  

60. 

Kindly consider the following

sinx+cosxsinxcosx

A: 

60. Given, f (x) = sinx+cosxsinxcosx

So, f?(x) = sinxcosxddx(sinx+cosx)(sinx+cos2)ddx(sinxcosx)(sinxcosx)2

Let g(x) = cos x and p(x) = sin x.

{from so g’(x) A ) (upto equation 3)

Let g(x) = cos2 and p(x) = sin x.

So, g?(x) = limh0g(x+h)g(x)h

=limh01h[cos(x+h)cosx]

=limh01h[2·sin(x+h+x2)sin(x+hx2)]

=limh01h[2sin(2x+h2)sin(h2)]

=sin(2x+02)×1

= -sin x ______ (2)

And p?(x) = limh0p(x+h)p(x)h

=limh0sin(x+h)sinxh

=limh01h2cos(2x+h2)sin(h2)

=limh0cos(2x+h2)×limh0sin(h2)(h2)

= cos x _____ (3)

Putting (2) and (3) in (1) we get,

f(x)=(sinxcosx)[cosxsinx](csinx+cosx)[cosx+sinx](sinxcosx)2

=(sinxcosx)2(sinx+cosx)2(sinxcosx)2

=(sin2x+cos2x)+2sinxcosx(sin2x+cos2x)2sinxcosx(sinxcosx)2

=11(sinxcosx)2=2(sinxcosx)2

Q:  

61.

Kindly consider the following

secx1secx+1

A: 

61. Given, f (x) = secx1secx+1

=1cosx11cosx+1

=1cosx1+cosx

So, f?(x) = (1+cosx)ddx(1cosx)(1cosx)ddx(1+cosx)(1+cosx)2

=(1+cosx)(1)ddx(cosx)(1cosx)ddx(cosx)(1+cosx)2

Let g(x) = cos x.

So, g?(x) =limh0g(x+h)g(x)h

=limh0cos(x+h)cosxh

=limh02hsin(x+h+x2)sin(x+hx2)

=limh02hsin(2x+h2)sin(h2)

=limh0sin(2x+h2)×limh0sinh2h2

=sin(2x+02)×1

= -sin x.

So, f?(x) (1+cosx)(1)(sinx)(1cosx)(sinx)(1+cosx)2

=sinx+cosxsinx+sinxsinxcosx(1+cosx)2

=2sinx(1+cosx)2

=2sinx(1+1secx)2=2sinx×sec2x(secx+1)2=2secx·sinx(sinx+1)2·cosx.

=2secx·tanx(sinx+1)2

Q:  

64.

Kindly consider the following

sin(x+a)cosx

A: 

64.  Given, f (x) = sin(x+a)cosx

f(x)=cosxddxsin(x+a)sin(x+a)ddxcosxcos2x

Let g?(x) = sin (x + a)

So, g?(x) = limh0g(x+h)g(x)h

= cos (x + a)

And P(x) = cos x

So, P?(x) = limh0p(x+h)p(x)h

Thus, f?(x) = cosx·cos(x+a)sin(x+a)(sinx)cos2x

=cosx·cos(x+a)+sin(x+a)sinxcos2x

=cos(x+ax)cos2x

=cosacos2x

Q:  

66.

Kindly consider the following

x4 (5 sin x 3 cos x)

A: 

65. Given, f (x) = x4. (5 sin x 3 cos x)

f(x)=x4ddx(5sinx3cosx)+(5sinx3cosx)dx4dx.

=x4[5ddxsinx3·dcosxdx]+[5sinx3cosx]·4x3

As ddxsinx=cosx

and ddxcosx=sinx

Thus,

f(x)=x4[5cosx+3sinx]+[5sinx3cosx]·4x3

=x3[5cosx+3xsinx+20sinx12cosx]

=x3[5xcosx+3xsinx+20sinx12cosx].

Q:  

Kindly consider the following

70. x2cos(π4)sinx

A: 

 70. Given, f (x) x2cos (π4)sinx

=x2sinx×cosπ4

So, f? (x) =

 

=cosπ4 [2·xsinxx2cosxsin2x]

=xcosπ4 [2sinxxcosxsin2x].

Q:  

Kindly consider the following

72. (x + sec x) (x- tan x

A: 

72. Given, f (x) = (x + sec x) (x tan x)

So, f?(x) = (x + see x) ddx(xtanx)+(xtanx)ddx(x+secx).

=(x+secx)(dxdxddxtanx)+(xtanx)(dxdx+ddxsecx)

Let g(x) = tan x.

So, g?(x) =

 

=limh01h[tan(x+h)tanx]

=limh01h[sin(x+h)cos(x+h)sinxcosx]

=limh01h[sin(x+h)cosxsinxcos(x+h)cos(x+h)cosx]

=limh01h[sin(x+hx)cos(x+h)cosx]

=limh01cos(x+h)cosx×limh0sinhh

 

=1cos2x=sec2x . _____ (2)

And P(x) = see x.

=limh0sec(x+h)secxh

=limh01h[1cos(x+h)1cosx]

=limh01h[cosxcos(x+h)cos(x+h)cosx]

=limh01h[2sin(x+x+h2)sin(x(x+h)2)cos(x+h)cosx]

 

=sinxcos2x = tan x sec x. ____ (3)

   Putting (2) and (3) in (1) we get,

   f?(x) = (x + sec x) (1 - sec2x) + (x- tan x) (1 + tan x sec x)

Q:  

Kindly consider the following

73. xsinnx

A: 

73. Given, f (x) = xsinnx

So, f?(x) = (sinxxdxdx)(xddxsinxx)(sinxx)2

= sinxxx·n(sinx)x1ddxsinx(sinx)2x

=(sinx)nx·x(sinx)n1·cosx(sinx)2n

=(sinx)n1[sinxnx·cosx](sinx)2n

=sinxn·x·cosx(sinx)2n(n1)

=sinxnxcosx(sinx)n+1

Q:  

32. If f (x) = {mx2+n,x<0nx+m,0x1nx3+m,x>1 . For what integers m and n does both limx0f(x) and limx1f(x) exist?

A: 

32. Given, f (x) = {mx2+n,x<0nx+m,0x1nx3+m,x>1.

For limx0f(x)lim,x0f(x)=limx0+f(x)

limx0(mx2+n)=limx0+(mx3+m)

n = m

So, limx0f(x) exist for n = m.

Again, limx1f(x)=limx1nx+m=n+m.

limx1+f(x)=limx1+nx3+m=n+m

So, limx1f(x)=limx1+f(x)=limx1f(x)=n+m, Thus, limx1f(x) exist for any integral value of m and n.

Q:  

33. Find the derivative

A: 

33. Given, f (x)=x2- 2 ., f (10)=?

We have,  

f (10)=limh0f (10+h)f (10)h

=limh0 [ (10+h)22] [1022]h

limh0102+h2+20h2102+2h

limh0h (h+20)h

limh0h+20

=20

Q:  

34. Find the derivative of x at x = 1.

A: 

34. Given, f (x)=x, f   (1)=?

We have,

f (1)=limh0f (1+h)f (1)h

 =limh01+h1h

=limh0hh

=limh01

=1

Q:  

35. Find the derivative of 99x at x = l00

A: 

35. Given, f (x)= 99x, f   (100)=?

So, f (100)= h0f (100+h)f (100)h

=limh099 (100+h)99 (100)h

=limh099×100+99×h99×100h

=limh099hh

=limh099

=99

Q:  

36. Find the derivative of the following functions from first principle.

A: 

36. (i) x327 (ii) (x -1)(x-2)

(iii) 1x2 (iv) x+1x1

A.4.(i) Given, f(x)=x327.

So, f(x)=limh0f(x+h)f(x)h

=limh0[(x+h)327][x327]h

=limh0x3+h3+3xh(x+h)27x3+27h

=limh0h(h2+3x(x+h))h

=limh0h2+3x(x+h)

=0+3 x(x+ 0)

=3x2

(ii) Given, f(x) =(x-1)(x-2)

=x2- 3x+2

So, f(x)=limh0f(x+h)f(x)h

limh0[(x+h)23(x+h)+2][x23x+2]h

limh0x2+h2+2xh3x3h+2x2+3x2h

limh0h(h+2x3)h

=limh0h+2x3

= 2x – 3.

(iii) Given, f(x)= 1x2

So, f(x)=limh0f(x+h)f(x)h

=limh01(x+h)21x2h

=limh0x2(x+h)2(x+h)2x2h

=limh0x2x2h22xhhx2(x+h)2 

=limh0h(h2x)hx2(x+h)2

=limh0h2xx2(x+h)2

=02xx2(x+0)2

=2xx4

=2x3

(iv) Given, f(x)= x+1x1

f(x)=limh0f(x+h)f(x)h

=limh01h[fx+h+1x+h1x+1x1]

=limh01h[(x+h+1)(x1)(x+1)(x+h1)(x+h1)(x1)]

=limh01h[x2x+hxh+x1x2hx+xxh+1(x1)(x+h1)]

=limh01h[2h(x1)(x+h1)]

=limh02(x1)(x+h1)

=2(x1)(x1)=2(x1)2

Q:  

37. For the function

f(x)=x100100+x9999++x22+x+1

Prove that f(1)=100f(0)

A: 

37. 

Given f(x)=x100100+x9999+?+x22+x+1

limh0(x+h)100x100100h+limh0(x+h)99x9999h++limx0(x+h)2x22h +limx0x+limx01

=100x99100+99x9899+?+2x2+0+1

=x99+x98+?+x+1

At x=0,

f(X)  =1.

and at x=1,

f(1)=199+198++12+1+1

=100 × 1

=100 ×f(0)

Hence, f(1)=100f(0)

Q:  

38.  Find the derivative of xn+axn1+a2xn2++an1x+an for some fixed real number a.

A: 

38. Given, f(x)=xn+axn1+a2xn2+?+an1x+an.

We know that,

ddx(xx)=nxx1

So,

f(x)=ddxxx+ddxaxx1+ddxa2xx2+?+ddx ax1x+ddxax=nxn1+a(n1)xn2+a2(n2)xn3+?+an1+0.

(?daxdx=adx dxand dadx=0whereaisconstant

Q:  

42. Find the derivative of cos x from first principle.

A: 

42. Given, f(x) = cos x

f(x+h)=cos(x+h)

By first principle,

f(x)=limxhf(x+h)f(x)h

=limxh1h[cos(x+h)cosx]

=limxh1h[2sin(x+h+x2)sin(x+hx2)]

=limxh1h[2sin(2x+h2)sin(h2)]

=limxh1h[2sin(2x+h2)sin(h2)]

=limxhsin(2x+h2)×limxhsinh/2h/2

=sin(2x+02)×1=sinx.

Q:  

47. (ax + b) (cx + d)2

A: 

47. Given, f (x) = (ax + b) (cx + d)2

So, f?(x) = (ax +b) ddx(cx+d)2+(cx+d)2ddx(ax+b)

=(ax+b)ddx[(c+d)(cx+d)]+(cx+d)2a

=(ax+b)[(c+d)ddx(x+d)+(c+d)ddx(c+d)]+a(cx+d)2

=(ax+b)[c(cx+d)+c(cx+d)]+a(cx+d)2

=(ax+b)·2·c(cx+d)+a(cx+d)2

Q:  

48. 

Kindly consider the following

 ax+bcx+d

A: 

48. Given, f (x) = ax+bcx+d

So, f (x) =  (cx+d)ddx (ax+b) (ax+b)ddx (cx+d) ( (x+d)2

= (cx+d)a (ax+b)c (cx+d)2

=acx+adacxcb (cx+d)2

=adbc (cx+d)2

Q:  

49. 

Kindly consider the following

1+1x11x

A: 

49. Given, f (x) = 1+1x11x=x+1xx1x=x+1x1

So, f (x) =  (x1)ddx (x+1) (x+1)ddx (x1) (x1)2

(x1) (x+1) (x1)2

=x1x1 (x1)2=2 (x1)2

Q:  

51. 

Kindly consider the following

(ax+b)px2+qx+r

A: 

51. Given, f (x) = (ax+b)px2+qx+r

So, f(x) = (px2+qx+r)ddx(ax+b)(ax+b)ddx(px2+qx+r)(px2+qx+r)2

 =(px2+qx+r)a(ax+b)(2xp+q)(px2+qx+r)2

=apx2+aqx+ar2apx2aqx2bpxb(px2+qx+9)2q.

=apx22bpx+arbq(px2+qx+r)2

Q:  

53. 

Kindly consider the following

ax4bx+cosx

A: 

53. Given, f (x) = =ax4bx+cosx

So, f(x) = limh0f(x+h)f(x)h

=limh01h[a(x+n)4b(x+h)2+cos(x+h)(ax4bx2+cosx)]

=limh0ah[1(x+h)41x4]limh0bh[1(x+h)21x2]+limh01h[cos(x+h)cosx]

=limh0ah[x4(x+h)4(x+h)4·x4]limh0bh[x2(x+h)2(x+h)2·x2]+

limh01h[2sin(x+h+x2)sin(x+hx2)]

=limh0ah[x4x44x3h6x2h24xh3h3x4·(x+h)4]limh0bh[x2x2h22xhx2(x+h)2]

+limh0ah[2sin(2x+h2)sinh2]

=limh0ah[h(4x36x2h4xh2h2)x4(x+h)4]limh0bh[h(h2x)x2(x+h)2]

limh0sin(2x+h2)limh0sinh2h2

=limh0a(4x36x2h4xh2h2)x4(x+h)4limh0b(h2x)x2(x+h)2

sin(2x+02)×1

=a(4x3)x4·x4b(2x)x2·x2sinx.

=4ax5+2bx3sinx

Q:  

55.

Kindly consider the following

(ax + b)n

 

A: 

55. Given, f (x) = (ax + b)n

Chain rule,  ddxu (x)n=nu (x)n1dudxu (x) where

u (x) is a function of x.

So, f (x) = ddx (ax+b)n

=n (ax+b)n1ddx (ax+b)

=na (ax+b)n1

Q:  

56. (ax + b)n (cx + d)m

A: 

56. Given, f (x) = (ax + b)n (cx + d)m

So, f’(x) = (ax+b)nddx(cx+d)n+(cx+d)mddx(ax+b)n

=(ax+b)x·mc(cx+d)m1+(cx+d)mna(ax+b)x1

=(ax+b)n1(cx+d)m1[(ax+b)·mc+(cx+d)na]

=(ax+b)n1(cx+d)m1[mc(ax+b)+na(cx+d)]

Q:  

57. 

Kindly consider the following

sin(x + a)

A: 

57. Given, f (x) = sin (x + a)

So, f (x) = limh0f (x+h)f (x)h

=limh0sin (x+h+a)sin (x+a)h

=limh01h·2cos (x+h+a+x+a2)sin (x+h+a (x+a)2)

=limh0cos (2x+2a+h2)limh0sin (h2)h2

=cos (2x+2a+0)2×1

= cos (x + a)

Q:  

58.

Kindly consider the following

cosec x. cot x

A: 

58. Given f (x) = cosec x. cot x.

By Leibnitz product rule,

So, g(x) = limh0g(x+h)g(x)h

=limh0cot(x+h)cotxh

=limh01h[cos(x+h)sin(x+h)cosxsinx]

=limh01h[sinxcos(x+h)cosxsin(x+h)sinxsin(x+h)]

=limh01h[sin(x(x+h))sinxsin(x+h)] [?csin(AB)=sinAcosBcosAsinB]

=limh01h[sin(h)sinx.sin(x+h)]

=limh01sinxsin(x+h)×(1)limh0sinhh

=1sinxsin(x+0)×(1)

= -cosec2x.______(2)

And hx) = limh0h(x+h)h(x)h

=limh0cosec(x+h)cosecxh

=limh01h[1sin(x+h)1sinx]

=limh01h[sinxsin(x+h)sinx·sin(x+h)]

=limh01h[2cos(x+x+h2)sin(x(x+h)2)sinxsin(x+h).]

Q:  

63.

Kindly consider the following

a+bsinxc+dcosx

A: 

63. Given, f (x) = a+bsinxc+dcosx

f?(x) = (c+dcosx)ddx(bsinx)(a+bsinx)ddx(c+dcosx)(c+dcosx)2

f(x)=(c+dcosx)·b·ddx(sinx)(a+bsinx)·d·ddx(cosx)(c+dcosx)2_____(1)

{Copy (A)}

So, g?(x) = limh0g(x+h)g(x)h

=limh01h[cos(x+h)cosx]

=limh01h[2·sin(x+h+x2)sin(x+hx2)]

=limh01h[2sin(2x+h2)sin(h2)]

=sin(2x+02)×1

= sin x ______ (2)

And p?(x) = limh0p(x+h)p(x)h

=limh0sin(x+h)sinxh

=limh01h2cos(2x+h2)sin(h2)

=limh0cos(2x+h2)×limh0sin(h2)(h2)

= cos x _____ (3)

So, put (2) and (3) in (1) we get,

f(x)=(c+dcosx)(b·cosx)(a+bsinx)(d·sinx)(c+dcosx)2

=beccosx+bdcos2x+adsinx+bdsin2x(c+dcosx)2

=bccosx+adsinx+bd(cos2x+sin2x)(c+dcosx)2

=bccosx+adsinx+bd(c+dcosx)2

Q:  

62.

Kindly consider the following

sinnx

A: 

62. Given, f (x) =sinnx

By chain rule,

f? (x) = n (sin x)n-1 ddh sin x

Let (gx) = sinx

So, g? (x) limh0g (x+h)g (x)h

=limh0sin (x+h)sinxh

=limh02hcos (2a+h2)sin (h2)

=limh0cos (2x+h2)×limh0sin (h2)h2

cos (2x+0)2×1

= cos x.

So, f? (x) = n (sin x)n-1 cos x.

Q:  

67.

Kindly consider the following

(ax2 + sin x) (p +q cos x)

A: 

67. Given, f (x) = (ax2 + sin x) (p +q cos x).

So, f? (x) = (ax2 + sin x) ddx  (p+qcosx)+ (p+qcosx)ddx (ax2+sinx)

= (ax2+sinx) (0+qddxcosx)+ (p+qcosx) (addx (x2)+ddxsinx)

= (ax2+sinx) (q (sinx)+ (p+qcosx) (a·2x+cosx)

= q sin x (ax2 + sin x) + (p + q cos x) (2ax + cos x)

Q:  

Kindly consider the following

69. 4x+5sinx3x+7cosx

A: 

69. Given, f (x) = 4x+5sinx3x+7cosx

So,

f(x)=(3x+7cosx)ddx(4x+5sinx)(4x+5sinx)ddx(3x+7cosx)(3x+7cosx)2

=(3x+7cosx)(4x+5cosx)(4x+5sinx)(37sinx)(3x+7cosx)2

=x+xx+x+xx+xxx+35sin2x_(3x+7cosx)2

=15xcosx+28cosx+28xsinx15sinx+35(cos2x+sin2x)(3x+7cosx)2

=35+15xcosx+28cosx+28xsinx15sinx(3x+7cosx)2

Q:  

29. Kindly consider the following

 width='1200' height=

A: 

29.

Given, f (x) = (x-a1) (x-a2)… (x-an)

So,  limxa1f (x)=limxa1 (xa1)limxa1 (xa2)? limxa1 (xan)

= (a1-a1) (a1-a2) … (a1-an)

= 0 (a1-a2) … (a1-an)

 = 0

And limxaf (x)=limxa (xa1)limxa (xa2)limxa (xan)

= (a-a1) (a-a2) … (a-an)

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