NCERT Solutions Class 11 Maths Chapter 13

NCERT Solutions for Class 11 Mathematics Chapter 13 – Limits and Derivatives

Mathematics is the subject which helps students to learn the art of patience and the art of perseverance. It develops the strong thinking abilities of the students. As a result, they become capable of coping with the problems more rationally.

Limits and derivatives are an introduction to calculus. Once students have a sound conceptual understanding of this chapter, they will be able to solve all the other chapters of calculus like differentiation, indefinite and definite integration, applications of derivatives and differential equations with ease. The topics like intuitive ideas of derivatives, limits, algebra of limits, limits of trigonometric functions, derivatives, algebra of derivatives of functions, derivatives of polynomials, and trigonometric functions are covered in the NCERT Solutions for Class 11 Mathematics Chapter 13.

The NCERT Solutions have chapter notes related to the entire chapter making it easier for the students to study during their preparation. It is designed while adhering to the latest CBSE syllabus, ensuring they get access to the latest resources. They can also find questions covered from all the segments of the chapter, helping them prepare everything in detail. Extramarks leaves no stones unturned when it comes to providing the best learning material with unmatchable speed and accuracy for students irrespective of the class and subject.Extramarks, is one of the best  online educational platforms, is trusted by   students and teachers  for its good quality resources. Students can find NCERT textbooks, NCERT Exemplar, NCERT solutions, NCERT revision notes, NCERT-related additional questions, CBSE past year papers and mock tests on the Extramarks’ website.


Key Topics Covered in NCERT Solutions for Class 11 Mathematics Chapter 13

Calculus is the essential part of Class 11 and Class 12 Mathematics as it is interrelated with all the other chapters of Mathematics. Limits and derivatives help in a better understanding of topics like integration and differentiation. The limits and derivatives will be applied while solving problems of differentiation and integration, thereby aiding in making the calculations easier for the students.

NCERT Solutions for Class 11 Mathematics Chapter 13 covers topics like intuitive ideas of derivatives, limits, algebra of limits, limits of trigonometric functions, derivatives, algebra of derivatives of functions, derivatives of polynomials and trigonometric functions. Students can get the NCERT solutions from the Extramarks’ website. After completing this chapter, students will be able to recall all the conditions of limits and be able to  find their derivatives without much difficulty. 



Limits and derivatives are the initial chapters of Calculus. The chapter begins with the intuitive idea of the derivatives without actually giving their definition. Furthermore, it gives a naive definition of limits and helps students to learn the algebraic sum of the limits. As you proceed, you will learn about the derivatives and their algebraic sum. You will also get to know about derivatives for polynomials and trigonometric functions.

You can learn about this chapter’s concepts and their applications in the NCERT Solutions for Class 11 Mathematics Chapter 13, available on the Extramarks’ website.


The Intuitive Idea of the Derivatives 

Here, you will learn to find the object’s velocity with experiments and a few cases and examples. Let us study each of them one by one:

In the first set of computations,

The Average velocity between t = t 1 and t = t2 equals the distance travelled between t = t1 and t = t2 seconds divided by (t 2 – t 1 ).


The average velocity in the first two seconds = (Distance travelled between t2 = 2, t1 = 0) / Time interval (t2 – t1)

In the second set of computations,

 The average velocities for various time intervals starting at t = 2 seconds.

The average velocity ‘v’ between t = 2 seconds and t = t2  seconds is 

= (Distance travelled between 2 seconds and t2 seconds) / (t2 – 2) 

= [(Distance travelled in t2 seconds) – (Distance travelled in 2 seconds)] / (t2 – 2)


In the first set of computations, we calculated average velocities in increasing time intervals ending at t = 2 and then hoped that nothing changed just before t = 2. 

In the second set of computations, we calculated the average velocities decreasing in time intervals ending at t = 2 and then hoped that nothing changes after t = 2. 

Purely on the physical grounds, these sequences of average velocities must approach a common limit.



The above discussion clearly states the limiting process. Now, let us learn some more about limits with the help of some observations:

We say  limx→a  f (x) is the expected value of f at x = a given the values of f near x to the left of a. This value is called the left-hand limit of f at a. 

We say  limx→a f (x) is the expected value of f at x = a given the values of f near x to the right of a. This value is called the right-hand limit of f (x) at a. 

If the right and left-hand limits coincide, we call that common value the limit of f (x) at x = a and denote it by limx→a f (x).  


Algebra of Limits 

Let us learn about the algebra of limits through a theorem without proofs.

Theorem 1:

Let f and g be two functions such that both limx→a f (x) and limx→a g(x) exist. 


(i) Limit of the sum of two functions is the sum of the limits of the functions, i.e., 

 limx→a [f (x) + g(x)] = limx→a f (x) +  limx→a g(x)


(ii) Limit of difference between two functions is the difference of the limits of the functions, i.e., 

 limx→a [f (x)g(x)] = limx→a f (x) limx→a g(x)


(iii) Limit of the product of two functions is the product of the limits of the functions, i.e., 

 limx→a [f (x). g(x)] = limx→a f (x).  limx→a g(x)


(iv) Limit of the quotient of two functions is the quotient of the limits of the functions (whenever the denominator is non-zero), i.e., 

 limx→a [f (x) / g(x)] =  limx→a f (x) / limx→ag(x)


Limits of polynomials and relational functions  

Let us learn about the limits of polynomials through a theorem and draw observations.

The function f is said to be a polynomial function of degree n f(x) = a0 + a1 x + a2 x2 +. . . + an. x n , where a is a real number such that an ≠ 0 for some natural number n. We know that  limx→a x = a. 


Limx→ax2 = (x-x) = limx→a x . limx→a x = a.a = a2


Theorem 2:

For any positive integer n,

Limx→a(xn-an)/(x-x) = nan-1


Limits of Trigonometric Functions 

You can get a clear-cut understanding of the limits of trigonometric functions with the theorems listed below:

Theorem 3:

Let f and g be two real-valued functions with the same domain such that f (x)g( x) for all x in the domain of definition, For some a, if both  Limx→a f(x) and  Limx→a g(x) exist, then  Limx→a f(x) Limx→ag(x).


Theorem 4 (Sandwich theorem):

Let f, g and h be real functions such that f(x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if  Limx→af(x) = l =  Limx→ah(x), then  Limx→ag(x) = l


Theorem 5 :

The following are two important limits. 

(i) Limx→0(sin x)/x = 1

(ii)Limx→0(1 – cos x)/x = 0



Derivatives are the rate of change of a function with respect to the variable. It is applied on all the fundamental units and is quite beneficial in solving the problems of calculus and differential equations.


Definition 1:

Suppose f is a real-valued function and a is a point in its domain of definition. The derivative of f at a is defined by

Limh→0 [f(a+h) − f(a)] / h 


Provided this limit exists. Derivative of f (x) at a is denoted by f′(a)

The detailed information about the derivatives is provided in the NCERT Solutions for Class 11 Mathematics Chapter 13, available on the Extramarks’ website.


Algebra of derivatives of functions:

The Algebra of derivatives of functions can be concluded with the help of the following theorem:


Theorem 5:

Let f and g be two functions such that their derivatives are defined in a common domain. 



(i) Derivative of the sum of two functions is the sum of the derivatives of the functions 

d/dx.[ f(x) + g(x) ] = d/dx.f(x) + d/dx.g(x)


(ii) Derivative of difference of two functions is the difference of the derivatives of the functions

d/dx.[ f(x) – g(x) ] = d/dx.f(x) – d/dx.g(x)


(iii) Derivative of the product of two functions is given by the following product rule

d/dx.[ f(x). g(x) ] = d/dx.f(x).g(x) + f(x).d/dx.g(x) 


(iv) Derivative of the quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero)

d/dx.[ f(x) / g(x) ] = [ d/dx.f(x).g(x) –  f(x).d/dx.g(x) ] / (g(x))2

You can find more information on this topic in the NCERT Solutions for Class 11 Mathematics Chapter 13 available on the Extramarks’ website.


Theorem 6:

Derivative of f(x) = xn is nxn-1 for any positive integer n. 


Derivatives of polynomials and trigonometric functions 

You can find the derivatives of polynomials and trigonometric functions using the following theorem:


Theorem 7:  

Let f(x) = an xn + an-1 xn-1 +…..+ a1 x + a0 be a polynomial function, where ai s are all real numbers and an ≠ 0. Then, the derivative function is given by 

df(x) / d(X) = nan xn-1 + (n-1)an-1 xn-2 +….+ 2a2 x + a1


NCERT Solutions for Class 11 Mathematics Chapter 13 Exercise &  Solutions

You can test your understanding by solving the questions in the NCERT textbook and look for its detailed and in-depth solutions and methodologies in the NCERT Solutions for Class 11 Mathematics Chapter 13. In this way, you can prepare yourselves and be confident for your upcoming examinations. The right solutions and answers provided in it will help you rectify your mistakes and shortcomings. As a result, you will turn into a smart learner. You can look for multiple ways to solve a question and choose the best solution to tackle those tricky and difficult questions in the NCERT solutions.

You can avail of NCERT Solutions for Class 11 Mathematics Chapter 13 from the Extramarks’ website. Click on the  links given below  to view exercise-specific questions and solutions for NCERT Solutions for Class 11 Mathematics Chapter 13:

  • Chapter 13 Class 11 Mathematics: Exercise13.1
  • Chapter 13 Class 11 Mathematics: Exercise13.2
  • Chapter 13 Class 11 Mathematics: Miscellaneous Exercise


Along with Class 11 Mathematics solutions, you can explore NCERT solutions on our Extramarks’ website for all primary and secondary classes 

  • NCERT Solutions Class 1
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NCERT Exemplar Class 11 Mathematics

NCERT Exemplar Class 11 Mathematics book has a collection of all NCERT-related questions. Students can find basic to advanced level questions in it. This makes them capable of solving different types of questions. As a result, they will acquire more analytical and critical thinking skills.

The book is specially designed by subject matter experts and gives insights into all the topics from the NCERT Class 11 Mathematics textbook. The benefits provided in this book will overall make a difference not only in-class assignments, tests as well as in the competitive exams later. The simple reason is CBSE itself prescribes NCERT books for board exams and is the best option for competitive exams as well. Thus, they will ensure that even the minutest doubt is resolved and the students develop an interest in learning and mastering the topic with ease.

It helps in laying the foundation for all the basic as well as advanced concepts in the manner required for different competitive examinations. Thus, help to boost the confidence level of the students. They can access  NCERT Exemplar Class 11 Mathematics easily from the Extramarks’ website.


Key Features for NCERT Solutions for Class 11 Mathematics Chapter 13

Practice makes you perfect. The more you practice, the easier it will get. Extramarks provides a repository of resources with solutions for students to step up their learning and be confident. . Hence, NCERT Solutions for Class 11 Mathematics Chapter 13 provides you with a lot of questions to practice. The key features are as follows: 

  • Students can find questions from the NCERT textbook, NCERT Exemplar, and other reference books in our NCERT Solutions for Class 11 Mathematics Chapter 13. They will be able to grasp the concepts, think and apply these concepts to solve those tricky questions easily after going through these solutions.
  • Experienced subject matter experts have provided answers to all the questions after thoroughly checking and verifying them while strictly adhering to the CBSE guidelines.
  • After completing the NCERT Solutions for Class 11 Mathematics Chapter 13, students will be able to solve the chapters like differentiation and integration accurately in tests and exams. It encourages the students to master the topic and achieve high grades..

Q.1 Evaluate the following limits in Exercises 1 to 22.




















limz1  z131z161




limx2  1x+12x+2




limx0  sin a xsin b x,a,b0


limxπ  sin(πx)π(πx)


limx0  cos x(πx)


limx0  cos2x1cosx1


limx0  ax+x cos xb sin x


limx0x sec x


limx0  sin ax+bxax+sin bx,   a,b,a+b0


limx0(cosec xcotx)


limxπ2  tan 2xxπ2







limr1πr2=π(1)2    =π


limx4(4x+3x2)=4(4)+342     =192


limx1(x10+x5+1x1)=(1)10+(1)5+1(1)1      =11+111      =12


limx0(x+1)51xLet x+1=yx=y1Then,limy1y51y1=limy1y515y1=5(1)51[∵limxaxnanxa=nan1] =5limx0(x+1)51x=5.


limx2(3x2x10x24)=limx2{3x26x+5x10(x2)(x+2)}      =limx2{3x(x2)+5(x2)(x2)(x+2)}      =limx2{(x2)(3x+5)(x2)(x+2)}      =limx2{(3x+5)(x+2)}      =3(2)+52+2      =114


limx3x4812x25x3=limx3(x2)2922x26x+x3  =limx3(x29)(x2+9)2x(x3)+1(x3) =limx3(x3)(x+3)(x2+9)(x3)(2x+1)  =limx3(x+3)(x2+9)(2x+1)  =limx3(3+3)(32+9)(2×3+1)  =6×187  =1087


limx0(ax+bcx+1)=a(0)+bc(0)+1     =b1     =b


limz1  z131z161=limz1  z1621z161 =2121∵limxaxnanxa=nan1 =2


limx1(ax2+bx+ccx2+bx+a)=a(1)2+b(1)+cc(1)2+b(1)+a       =a+b+cc+b+a       =1


limx2  1x+12x+2=limx22+x2xx+2=limx2x+22x(x+2)limx2  1x+12x+2=12(2)=14


limx0sin axbx=1blimx0sin axax×a=ablimax0(sin axax)=ab×1 [∵limx0sin xx=1]=ab


limx0  sinaxsinbx=alimax0(sinaxax)blimbx0(sinbxbx)=a×1b×1 [∵limx0sinxx=1]=ab


limxπ  sin(πx)π(πx)=1π{limπx0  sin(πx)(πx)}=1π(1)=1π


limx0  cosx(πx)=cos0(π0)    =1π [∵cos0=1]


limx0  cos2x1cosx1=limx0  12sin2x112sin2(x2)1=limx0  2sin2x2sin2(x2) =limx0  (sin2xx2){sin2(x2)4(x2)2}=4limx0  (sinxx)2{sin(x2)(x2)}2=4(1)2(1)2limx0  cos2x1cosx1=4


limx0  ax+x cos xb sin x=limx0  x(a+cosx)b sin x    =limx0(a+cos x)blimx0(sin xx)    =(a+cos 0)b(1)limx0  ax+x cos xb sin x=a+1b


limx0x sec x=0.sec0=0×1=0


limx0  sin ax+bxax+sin bx=limx0  x(a sin axax+b)x(a+b sin bxbx) =limx0  (a sin axax+b)(a+b sin bxbx)=(a×1+b)(a+b×1)=a+ba+blimx0  sin ax+bxax+sin bx=1


limx0(cosec xcotx)=limx0(1sinxcosxsinx)   =limx0(1cosxsinx)   =limx0(2sin2x22sinx2cosx2)   =limx0(sinx2cosx2) =limx0(tanx2) =tan0limx0(cosec xcotx)=0


Given:  limxπ2  tan 2xxπ2Let xπ2=y x=π2+yIf xπ2 then y0limxπ2  tan 2xxπ2=limy0  tan2(π2+y)π2+yπ2       =limy0  tan(π+2y)y       =limy0  tan 2yy [∵ tan (π+x)=tan x]       =2limy0  tan 2y2y =2×1[limx0tan xx=1]limxπ2  tan 2xxπ2=2


Find limx0f(x) and limx1f(x),wheref (x)={2x+3,x03(x+1),x>0


The given function is:f(x)={2x+3,x03(x+1),x>0At x=0,L.H.L.=limx0f(x)  =limx0(2x+3)  =2(0)+3  =3R.H.L.=limx0+f(x)  =limx03(x+1)  =3(0+1)  =3limx0f(x)=limx0+f(x)=limx0f(x)=3Atx=1,L.H.L.=limx1f(x)   =limx13(x+1)  =3(1+1)  =6R.H.L.=limx1+f(x)  =limx13(x+1)  =3(1+1)  =6limx1f(x)=limx1+f(x)=limx1f(x)=6


Find limx1 f(x), where f(x)={x21,x1x21,x>1


Thegivenfunctionis;fx={x21,x1x21,x>1Atx=1,L.H.L.=limx1fx =imx1x21 =121=0R.H.L.=limx1+fx =limx1x21 =121=2limx1fxlimx1+fxThus,limx1fxdoesnotexist.


Findlimx0f(x), where f(x)={x|x|,x00,x=0




Suppose f(x)={a+bx,x<14,x=1bax,x>1and iflimx1 f(x)=f(1) what are possible values of a and b?


The given function is:f(x)={a+bx,x<14,x=1bax,x>1 For x=1,     L.H.L.=limx1f(x)         =limx1(a+bx)         =a+b   R.H.L.=limx1+f(x)         =limx1(bax)          =baandf(1)=4According to given condition,lim x1f(x)=f(1)limx1f(x)=limx1+f(x)=f(1)    a+b=ba=4    a+b=4 and ba=4Solving both equations, we get   a=0 and b=4Thus, the possible values of a and b are 0 and 4 respectively.


Find limx5 f(x),where f(x)=|x|5


The given function is,f(x)=|x|5={x5,x<05,x=0x5,x>0For x=5,L.H.L.=limx5f(x)  =limx5(x5)[When x>0, |x|=x]   =55  =0R.H.L.=limx5+f(x)  =limx5(x5)[When x>0, |x|=x]  =55  =0limx5f(x)=limx5+f(x)=limx5f(x)=0Thus, limx5f(x)=0

Q.7 Let a1, a2, …, an be fixed real numbers and define a function f(x) = (x – a1) (x – a2)… (x – an).

What islimxa1f(x)? For some aa1,a2,...,an,compute limxaf(x).


The given function is:f(x)=(xa1)(xa2)(xan)     limxa1f(x)=limxa1{(xa1)(xa2)(xan)}         ={limxa1(xa1)}{limxa1(xa2)}{limxa1(xan)}         =(a1a1)(a1a2)(a1an)         =(0)(a1a2)(a1an)         =0 And    limxaf(x)=limxa{(xa1)(xa2)(xan)} ={limxa(xa1)}{limxa(xa2)}{limxa(xan)}=(aa1)(aa2)(aan)


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


The given function is:f(x)={|x|+1,x<00,x=0|x|1,x>0For existence of limxaf(x), CaseI:Whena=0limx0fx=limx0x+1=limx0x+1Whenx<0,x=x=0+1=1limx0+fx=limx0+x1 =limx0x1Whenx>0,x=x =01 =1Here,limx0fxlimx0+fxlimx0fx does not exist.CaseII:Whena<0limxafx=limxax+1 =limxax+1Whenx<0,x=x =a+1limxa+fx=limxa+x+1 =limxax+1Whena<x<0,x=x =a+1limxafx=limxa+fx=a+1 Thus, limxaf(x) exists at x=a, where a<0.CaseIII:When a>0limxaf(x)=limxa(|x|1) =limxa(x1)[When0<x<a,|x|=x] =a1limxa+f(x)=limxa+(|x|1) =limxa(x1)[Whena<x<0,|x|=x] =a1limxaf(x)=limxa+f(x)=a1Thus, limxaf(x) exists at x=a, where a>0.Therefore, limxaf(x)  exists for all a0.




Given:  limx1f(x)2x21=π    limx1{f(x)2}=limx1π(x21)limx1f(x)limx12=π(121)limx1f(x)limx12=π(0)              limx1f(x)=limx12             limx1f(x)=2.




We have,f(x)={mx2+n,x<0nx+m,0x1nx3+m,x>0For x=0,limx0f(x)=limx0(mx2+n)         =nlimx0+f(x)=limx0(nx3+m)         =mSince, limx0f(x) exists. So,limx0f(x)=limx0+f(x)    n=mTherefore, limx0f(x) exists if m=n.For x=1, limx1f(x)=limx1(nx+m)        =n+mlimx1+f(x)=limx1(nx3+m)        =n+mSince, limx0f(x) exists. So,limx1f(x)=limx1+f(x)=limx1f(x)Therefore, limx1f(x) exists for any value of m and n.

Q.11 Find the derivative of x2 – 2 at x = 10.


We are given:  f(x)=x22Since,        f(10)=limh0f(10+h)f(10)h                 f(10)=limh0{(10+h)22}{(10)22}h=limh0(100+20h+h22)(1002)h=limh098+20h+h298h =limh0h(20+h)hf(10)=20Thus, the derivative of the function x22 at x=10 is 20.

Q.12 Find the derivative of 99x at x = l00.



Q.13 Find the derivative of x at x = 1.


Wearegiven:fx=xSince,f1=limh0f1+hf1h f1=limh01+h1h=limh01+h1h=limh0hh f1=1Thus,thederivativeofthefunctionxatx=1is1.

Q.14 Find the derivative of the following functions from first principle.

(i) x327 (ii) (x1)(x2)(iii) 1x2 (iv) x+1x1


Since,f(x)=limh0f(x+h)f(x)h(i)f(x)=x327  f(x)=limh0{(x+h)327}(x327)h   =limh0x3+3x2h+3xh2+h327x3+27h   =limh03x2h+3xh2+h3h   =limh0(3x2+3xh+h2)hh   =3x2+3x(0)+(0)2  f(x)=3x2 (ii)    f(x)=(x1)(x2)     =x23x+2  f(x)=limh0(x+h)23(x+h)+2(x23x+2)h    =limh0x2+2xh+h23x3h+2x2+3x2h    =limh02xh+h23hh    =limh0(2x+h3)hh   f(x)=2x3(iii)    f(x)=1x2  f(x)=limh01(x+h)21x2h    =limh0x2(x+h)2h(x+h)2x2    =limh0x2x22xhh2h(x+h)2x2    =limh02xhh2h(x+h)2x2    =limh0h(2x+h)h(x+h)2x2     =(2x+0)(x+0)2x2    =2xx4f(x)=2x3(iv)f(x)=x+1x1  f(x)=limh0x+h+1(x+h1)x+1(x1)h    =limh0(x+h+1)(x1)(x+1)(x+h1)h(x+h1)(x1)    =limh0(x2+xh+xxh1)(x2+xhx+x+h1)h(x+h1)(x1)    =limh0x2+xh+xxh1x2xh+xxh+1h(x+h1)(x1)    =limh02hh(x+h1)(x1)    =2(x+01)(x1)      f(x)=2(x1)2


For the function f(x)=x100100+x9999+...+x22+x+1 Prove that f(1)=100 f(0).


We have,f(x)=x100100+x9999+...+x22+x+1Differentiating w.r.t. x, we getf(x)=ddx(x100100+x9999+...+x22+x+1)         =ddxx100100+ddxx9999+...+ddxx22+ddxx+ddx1         =100x99100+99x9899+...+2x12+1+0[ddxxn=nxn1]f(x)=x99+x98+...+x+1Substituting x=1, we getL.H.S.:    f(1)=(1)99+(1)98+...+(1)+1        =100Substituting x=0, we getf(0)=(0)99+(0)98+...+(0)+1        =1R.H.S.:100f(0)=100(1)        =100 Thus,f(1)=100 f(0).Therefore, it is proved.

Q.16 Find the derivative of xn + axn−1 + a2xn−2 + . . .+ an−1x + an for some fixed real number a.


We have,f(x)=xn+axn1+a2xn2+. . .+an1x+anDifferentiating w.r.t. x, we getf(x)=ddx(xn)+ddx(axn1)+ddx(a2xn2)+. . .+ddx(an1x)+ddxan=nxn1+a(n1)xn2+a2(n2)xn3+...+an1+0=nxn1+a(n1)xn2+a2(n2)xn3+...+an1




The given function is, f(x)={x|x|,x00,x=0At x=0,L.H.L.=limx0f(x)       =limx0(x|x|)        =limx0(xx)[whenx<0,|x|=x]        =1R.H.L.=limx0+f(x)      =limx0+(x|x|)        =limx0(xx)[whenx<0,|x|=x]         =1Since,limx0f(x)limx0+f(x)So, limx0f(x) does not exist.


For some constants a and b, find the derivative of(i)(xa)(xb)(ii)(ax2+b)2(iii)xaxb


(i) We have,f(x)=(xa)(xb)Differentiating w.r.t. x, we getf(x)=ddx{(xa)(xb)} =(xa)ddx(xb)+(xb)ddx(xa) [By Leibnitz rule]=(xa)(ddxxddxb)+(xb)(ddxxddxa)=(xa)(10)+(xb)(10)=xa+xb       f(x)=2xab(ii) We have,f(x)=(ax2+b)2        =(ax2+b)(ax2+b)Differentiating w.r.t. x, we getf(x)=ddx{(ax2+b)(ax2+b)}[[By Leibnitz rule]]f(x)={ddx(ax2+b)}(ax2+b)+(ax2+b){ddx(ax2+b)}          =(addxx2+ddxb)(ax2+b)+(ax2+b)(addxx2+ddxb)          =(2ax+0)(ax2+b)+(ax2+b)(2ax+0)          =2ax(ax2+b)+2ax(ax2+b)          =4ax(ax2+b)(iii) We have,f(x)=xaxbDifferentiating w.r.t. x, we get f’x=ddxxaxb=xbddxxaxaddxxbxb2Byquotientrule=xb10xa10xb2=xbx+axb2f’x=abxb2






Find the derivative of(i) 2x34 (ii) (5x3+3x1)(x1)(iii) x3(5+3x) (iv) x5(36x9)(v) x4(34x5) (vi) 2x+1x23x1


iWehave,fx=2x34Differentiatingw.r.t.x,wegetfx=ddx2x34 =ddx2xddx34 =2-0 =2iiWehave,fx=5x3+3x-1x-1 Differentiatingw.r.t.x,wegetf’x=ddx5x3+3x-1x-1 =ddx5x3+3x-1x-1+5x3+3x-1ddxx-1  ByLeibnitz’sproductrule =5ddxx3+3ddxx-ddx1x-1+5x3+3x-1ddxx-ddx1 =15x2+3-0x-1+5x3+3x-11-0 =15x3+3x-15x2-3+5x3+3x-1 =20x3-15x2+6x-4iiiWehave,fx= x-35+3xDifferentiatingw.r.t.x,wegetf’x=ddxx-35+3x=ddxx-35+3x+x-3ddx5+3x  By Leibnitz’s product rule=-3x-45+3x+x-30+3 =15x49x3+3x3 =15x46x3 =3x45+2xivWehave,fx=x536x9Differentiatingw.r.t.x,weget fx=ddxx536x9=ddxx536x9+x5ddx36x9By Leibnitz’s product rule=5x436x9+x50+54x10=15x430x5+54x5=15x4+24x5=15x4+24x5vWe have,fx=x434x5Differentiating w.r.t. x, we get fx=ddxx434x5 =ddxx434x5+x4ddx34x5 ByLeibnitz’sproductrule=4x534x5+x40+20x6=12x5+16x10+20x10=12x5+36x10=12x5+36x10viWehave,fx=2x+1x23x1Differentiatingw.r.t.x,weget fx=ddx2x+1x23x1=x+1ddx22ddxx+1x+123x1ddxx2x2ddx3x13x12 ByQuotientRule=x+1021+0x+123x12xx2303x12=2x+126x22x3x23x12 =–2x+12x3x-23x-12

Q.21 Find the derivative of cos x from first principle.


Since, f(x)=cos x   f(x)=limh0f(x+h)f(x)h     =limh0cos(x+h)cos(x)h     =limh02sin(x+hx2)sin(x+h+x2)h     =limh20sin(h2)(h2)×limh0sin(2x+h2)     =1×sin(2x+02) f(x)=sinx

Q.22 Find the derivative of the following functions:

(i) sin x cos x
(ii) sec x
(iii) 5sec x + 4cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x − 6cos x + 7
(vii) 2 tan x − 7sec x


(i) We have,f(x)=sinxcosxDifferentiating w.r.t. x, we getf(x)=ddx{sinxcosx}=(ddxsin x)cos x+sin x(ddxcos x)[By Leibnitz’s product rule]=cos x cos x+sin x(sin x)=cos2xsin2x=cos 2x(ii) We have,f(x)=secxDifferentiating w.r.t. x, we getf(x)=ddx(sec x)=sec x tan x(iii) We have,f(x)=5sec x+4cos xDifferentiating w.r.t. x, we getf(x)=ddx(5sec x+4cos x)=5ddxsec x+4ddxcos x=5(sec x tan x)+4(sin x)=5sec x tan x4sin x (iv) We have,f(x)=cosec xDifferentiating w.r.t. x, we get         f(x)=ddx(cosec x)=cosec x cot x(v) We have,f(x)=3cot x+5cosec xDifferentiating w.r.t. x, we getf(x)=ddx(3cot x+5cosec x)=3ddxcot x+5ddxcosec x=3(cosec2x)5cosec x cot x=3cosec2x5cosec x cot x(vi) We have,f(x)=5sin x6cos x+7Differentiating w.r.t. x, we getf(x)=ddx(5sin x6cos x+7)=5ddxsin x6ddxcos x+ddx7=5cos x+6sin x+0=5cos x+6sin x(vii) We have,f(x)=2tan x7sec x Differentiating w.r.t. x, we getf(x)=ddx(2tan x7sec x)=2ddxtan x7ddxsec x=2sec2x7secx tan x




limx0e4x1x=lim4x0e4x14x×4(x04x0)    =1×4    =4




limx0e2+xe2x=limx0e2(ex1)x      =e2limx0(ex1)x      =e2×1 [∵limx0ex1x=1]      =e2




limx0exe5x5=limx0e5(ex51)x5   =e5limx0ex51x5   =e5×1 [∵limx0ex1x=1]   =e5




limx0esinx1x=limsinx0esinx1sinx×limx0sinxx(x0sinx0)          =1×1 [∵limx0ex1x=1 and limx0sinxx=1] limx0esinx1x=1




limx0exe3x3=limx0e3(ex31)x3    =e3limx0ex31x3   =e3×1 [∵limx0ex1x=1]   =e3




limx0x(ex1)1cosx=limx0ex1x×limx0x21cosx      =1×limx0x22sin2(x2)      =limx02(x2)2{sin(x2)}2      =limx02{sin(x2)(x2)}2      =2(1)2      =2




limx0loge(1+2x)x=lim2x0loge(1+2x)2x×2[x02x0]   =1×2 [∵limx0loge(1+x)x=1]   =2




limx0log(1+x3)sin3x=limx30log(1+x3)x3×limx30x3sin3x  [∵x0x30] =1×1 [∵limx0log(1+x)x=1 and limx0xsinx=1]=1

Q.31 Find the derivative of the following functions from first principle:

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


(i) Since, f(x)=x    f(x)=limh0f(x+h)f(x)h      =limh0(x+h)(x)h     =limh0xh+xh     =limh0hh     =1(ii) Since, f(x)=(x)1=1x   f(x)=limh0f(x+h)f(x)h     =limh01(x+h)(1x)h     =limh01(x+h)+1xh     =limh0x+x+hx(x+h)h     =limh0hhx(x+h)     =1x(x+0)     =x2(iii) Since, f(x)=sin(x+1)    f(x)=limh0f(x+h)f(x)h     =limh0sin(x+h+1)sin(x+1)h     =limh02cos(x+h+1+x+12)sin(x+h+1x12)h     =limh02cos(2x+h+22)sin(h2)h     =limh0cos(2x+h+22)×limh20sin(h2)(h2)[h0h20]     =cos(2x+0+22)×1     =cos(x+1)(iv) Since, f(x)=cos(xπ8)   f(x)=limh0f(x+h)f(x)h     =limh0cos(x+hπ8)cos(xπ8)h     =limh02sin(x+hπ8+xπ82)sin(x+hπ8x+π82)h =limh0sin(2x+h2π82)×limh20sin(h2)(h2)   [h0h20]     =sin(2x+02π82)×1     =sin(xπ8)

Q.32 Find the derivative of the function (x + a).


Let f(x)=x+aDifferentiating w.r.t. x, we getf’(x)=ddx(x+a)        =ddxx+ddxa        =1+0        =1


Find the derivative of the function (px+q)(rx+s).


Let f(x)=(px+q)(rx+s)Differentiating w.r.t. x, we get f’(x)=ddx{(px+q)(rx+s)}        =(px+q){ddx(rx+s)}+(rx+s){ddx(px+q)}        =(px+q)(rddxx1+ddxs)+(rx+s)(pddxx+ddxq)        =(px+q)(rx2+0)+(rx+s)(p×1+0)        =prxqrx2+prx+ps        =qrx2+ps        =qrx2+ps    f(x)=qrx2+ps

Q.34 Find the derivative of the function (ax + b)(cx + d)2.


Let f(x)=(ax + b)(cx + d)2Differentiating w.r.t. x, we getf’(x)=ddx{(ax + b)(cx + d)2}[By Leibnitz product rule]        =(ax + b){ddx(cx + d)2}+(cx + d)2{ddx(ax + b)}        =(ax + b){ddx(cx + d)(cx + d)}+(cx + d)2(a+0)         =(ax + b){(cx + d)ddx(cx + d)+(cx + d)ddx(cx + d)} +a(cx + d)2        =(ax + b){(cx + d)(c+0)+(cx + d)(c+0)} +a(cx + d)2        =(ax + b)2c(cx + d)+a(cx + d)2        =2c(ax + b)(cx + d)+a(cx + d)2


Find the derivative of the function ax+bcx+d.


Let f(x)=ax+bcx+dDifferentiating w.r.t. x, we getf’(x)=ddx{ax+bcx+d}        =(cx+d)ddx(ax+b)(ax+b)ddx(cx+d)(cx+d)2        =(cx+d)(a+0)(ax+b)(c+0)(cx+d)2        =a(cx+d)c(ax+b)(cx+d)2        =acx+adacxbc(cx+d)2=adbc(cx+d)2


Find the derivative of the function 1+1x11x.


Let f(x)=1+1x11x         =x+1x1Differentiating w.r.t. x, we getf’(x)=ddx(x+1x1)        =(x1)ddx(x+1)(x+1)ddx(x1)(x1)2          =(x1)(1+0)(x+1)(10)(x1)2         =x1x1(x1)2f(x)=2(x1)2


Find the derivative of the function 1ax2+ bx + c.


Let f(x)=1ax2+bx+cDifferentiating w.r.t. x, we get f’(x)=ddx(1ax2+bx+c)        =(ax2+bx+c)ddx(1)(1)ddx(ax2+bx+c)(ax2+bx+c)2        =(ax2+bx+c)(0)(1)(2ax+b+0)(ax2+bx+c)2        =(2ax+b)(ax2+bx+c)2


Find the derivative of the function ax + bpx2+ qx + r.


Let f(x)=ax+bpx2+qx+rDifferentiating w.r.t. x, we getf’(x)=ddx(ax+bpx2+qx+r)        =(px2+qx+r)ddx(ax+b)(ax+b)ddx(px2+qx+r)(px2+qx+r)2        =(px2+qx+r)(a)(ax+b)(2px+q+0)(px2+qx+r)2         =a(px2+qx+r)(ax+b)(2px+q)(px2+qx+r)2        =apx2+aqx+ar(2apx2+axq+2bpx+bq)(px2+qx+r)2        =apx2+aqx+ar2apx2axq2bpxbq(px2+qx+r)2f’(x)=apx22bpx+arbq(px2+qx+r)2


Find the derivative of the function px2+ qx + rax + b.


Let f(x)=px2+qx+rax+bDifferentiating w.r.t. x, we getf’(x)=ddx(px2+qx+rax+b)        =(ax+b)ddx(px2+qx+r)(px2+qx+r)ddx(ax+b)(ax+b)2        =(ax+b)(2px+q+0)(px2+qx+r)(a)(ax+b)2         =(ax+b)(2px+q)a(px2+qx+r)(ax+b)2        =2apx2+axq+2bpx+bq(apx2+aqx+ar)(ax+b)2        =2apx2+axq+2bpx+bqapx2aqxar(ax+b)2      f’(x)=apx2+2bpx+bqar(ax+b)2


Find the derivative of the function ax4bx2+cos x.


Let f(x)=ax4bx2+cosx        =ax4bx2+cosxDifferentiating w.r.t. x, we get      f’(x)=ddx(ax4bx2+cosx)          =addxx4bddxx2+ddxcos x         =4ax5+2bx3sin x         =4ax5+2bx3sin x


Find the derivative of the function 4x2.


Let f(x)=4x2        =4x122 Differentiating w.r.t. x, we get f’(x)=ddx(4x122)        =4ddxx12ddx (2)        =4×12x120        =2x12        =2x

Q.42 Find the derivative of the function (ax + b)n.


Let fx=ax+bn and fx+h=ax+h+bn =ax+ah+bnBy first Principle,  f’x=limh0fx+hfxhddxax+bn=limh0ax+ah+bnax+bnh=limh0ax+bnax+ah+bax+bn1h=limh0ax+bnax+bax+b+ahax+bn1h=limh0ax+bn1+ahax+bn1h=limh0ax+bn1+nahax+b+nn1a2h22!ax+b+...1hUsing binomial theorem=limh0ax+bnnaax+b+nn1a2h2!ax+b+...=ax+bnnaax+b+0+...=naax+b×ax+bnddxax+bn=naax+bn1

Q.43 Find the derivative of the function (ax + b)n (cx + d)m.


Let f(x)=(ax+b)n(cx+d)mDifferentiating w.r.t. x, we getf’(x)=ddx{(ax+b)n(cx+d)m}By Leibnitz product rule:f’(x)=(ax+b)nddx(cx+d)m+(cx+d)mddx(ax+b)n =(ax+b)nddxy2+(cx+d)mddxy1   ...(i) [Let y1=(ax+b)n,y2=(cx+d)m]By first principle:      ddxy1=limh0(ax+ah+b)n(ax+b)nh=limh0(ax+b)n{(ax+ah+bax+b)n1}h=limh0(ax+b)n{(ax+bax+b+ahax+b)n1}h=limh0(ax+b)n{(1+ahax+b)n1}h=limh0(ax+b)n{1+nahax+b+n(n1)a2h22!(ax+b)+...1}h[Using binomial theorem]=limh0(ax+b)n{naax+b+n(n1)a2h2!(ax+b)+...}=(ax+b)n{naax+b+0+...}=naax+b×(ax+b)nddxy1=na(ax+b)n1 andddxy2=limh0(cx+ch+d)m(cx+d)mh=limh0(cx+d)m{(cx+ch+dcx+d)m1}h=limh0(cx+d)m{(cx+dcx+d+chcx+d)m1}h=limh0(cx+d)m{(1+chcx+d)m1}h=limh0(cx+d)m{1+mchcx+d+m(m1)c2h22!(cx+d)+...1}h[Using binomial theorem]=limh0(cx+d)m{mccx+d+m(m1)c2h2!(cx+d)+...}=(cx+d)m{mccx+d+0+...}=mccx+d×(cx+d)mddxy2=mc(cx+d)m1Substituting the values of ddxy1and ddxy2 in equation(i),we get x=ax+bnmccx+dm1+cx+dmnaax+bn1=ax+bn1cx+dm1mcax+b+nacx+d

Q.44 Find the derivative of the function sin(x + a).


Let f(x)=sin(x+a) and f(x+h)=sin(x+h+a)By first Principle,f’(x)=limh0f(x+h)f(x)h=limh0sin(x+h+a)sin(x+a)h=limh02cos(x+h+a+x+a2)sin(x+h+axa2)h=limh0cos(2x+h+2a2)×limh20sin(h2)(h2)[h0h20] =cos(2x+0+2a2)×1=cos(x+a)×1=cos(x+a)Therefore,ddxsin(x+a)=cos(x+a)

Q.45 Find the derivative of the function cosec x cot x.


Let f(x)=cosecx cot x Differentiating w.r.t. x, we get f’(x)=ddx(cosecx cot x)By Leibnitz product rule: f’(x)=cosecxddxcot x+cot xddxcosecx=cosecx(cosec2x)+cot x(cosecx cot x)=cosec3xcosecx cot2x


Find the derivative of the function cos x1+sin x.


Let f(x)=cosx1+sinxDifferentiating w.r.t. x, we getf’(x)=ddx(cosx1+sinx)=(1+sin x)ddxcos xcos xddx(1+sin x)(1+sin x)2=(1+sin x)(sin x)cos x(0+cos x)(1+sin x)2=sin xsin2xcos2x(1+sin x)2=sin x(sin2x+cos2x)(1+sinx)2 =sin x1(1+sin x)2=(1+sin x)(1+sin x)2f(x)=1(1+sin x)


Find the derivative of the function sinx + cosxsinxcosx.


Let f(x)=sin x+cos xsin xcos xDifferentiating w.r.t. x, we getf’(x)=ddx(sin x+cos xsin xcos x)={(sin xcos x)ddx(sin x+cos x)(sin x+cos x)ddx(sin xcos x)}(sin xcos x)2=(sin xcos x)(cos xsin x)(sin x+cos x)(cos x+sin x)(sin xcos x)2=(cos2x2cos x sin x+sin2x)(sin2x+2sin x cos x+cos2x)(sinxcosx)2 =(12cos x sin x)(1+2sin x cos x)(sin xcos x)2=1+2cos x sin x12sin x cos x(sin xcos x)2         f’(x)=2(sin xcos x)2


Find the derivative of the function secx1secx+1.


Let f(x)=sec x1sec x+1Differentiating w.r.t. x, we getf’(x)=ddx(sec x1sec x+1) ={(sec x+1)ddx(sec x1)(sec x1)ddx(sec x+1)}(sec x+1)2=(sec x+1)(sec x tan x0)(sec x1)(sec x tan x+0)(sec x+1)2=(sec2x tan x+sec x tan x)(sec2x tan xsec x tan x)(sec x+1)2=sec2x tan x+sec x tan xsec2x tan x+sec x tan x(sec x+1)2f’(x)=2 sec x tan x(sec x+1)2

Q.49 Find the derivative of the function sinn x.


Let y=sinnxFor n=1 y=sin xDifferentiating w.r.t. x, we get      dydx=ddxsinx   =cosxFor n=2y=sin2xDifferentiating w.r.t. x, we get      dydx=ddxsin2x   =ddx(sinx.sinx)   =sinx(ddxsin x)+sin x(ddxsin x)[By Leibnitz product rule]   =sin x cos x+sin x cos x      dydx=2sin x cos x ...(i)For n=3y=sin3xDifferentiating w.r.t. x, we get      dydx=ddxsin3x   =ddx(sinx.sin2x)    =sin x(ddxsin2x)+sin2x(ddxsin x)[By Leibnitz product rule]   =sin x(2sinxcosx)+sin2x cos x     [From equation (i)]   =2sin2x cos x+sin2x cos x      dydx=3sin2x cos xAccording to this pattern, we can sayddx(sinnx)=n sin(n1)x cos xLet this relation should be true for n=kThen, ddx(sinkx)=k sin(k1)x cos x ...(ii)Now,ddx(sink+1x)=ddx(sinkx sin x) =sinkx(ddxsinx)+(ddxsinkx)sin x=sinkx(cosx)+(k sin(k1)x cos x)sin x[From equation (ii)]=sinkx cos x+k sinkx cos x=(k+1)sinkx cos xSo, the declaration of equation (ii) is true for n=k+1.Hence, by Mathematical inducation,ddx(sinnx)=n sinn1x cos x


Find the derivative of the function a + bsinxc + dcosx.


Let f(x)=a+b sin xc+d cos xDifferentiating w.r.t. x, we getf’(x)=ddx(a+b sin xc+d cos x)={(c+d cos x)ddx(a+b sin x)(a+b sin x)ddx(c+d cos x)}(c+d cos x)2=(c+d cos x)(0+b cos x)(a+b sin x)(0d sin x)(c+d cos x)2=bc cos x+bd cos2x+ad sin x+bd sin2x(c+d cos x)2 =bc cos x+bd(cos2x+sin2x)+ad sin x(c+d cos x)2=bc cos x+bd(1)+ad sin x(c+d cos x)2f’(x)=bc cos x+ad sin x+bd(c+d cos x)2


Find the derivative of the function sin(x+a)cosx.


Let fx=sinx+acosx=sinxcosa+cosxsinacosxDifferentiatingw.r.t.x,wegetx=ddxsinxcosa+cosxsinacosx=cosxddxsinxcosa+cosxsinasinxcosa+cosxsinaddxcosxcosx2=cosxcosxcosasinxsinasinxcosa+cosxsinasinxcos2x=cos2xcosasinxcosxsina+sin2xcosa+sinxcosxsinacos2x=cos2xcosa+sin2xcosacos2x=cos2x+sin2xcosacos2x=1×cosacos2xx=cosacos2x

Q.52 Find the derivative of the function x4 (5sin x – 3cos x).


Let f(x)=x4(5sinx3cosx)Differentiating w.r.t. x, we get f’(x)=ddx{x4(5sinx3cosx)}By Leibnitz product rule: f’(x)=x4ddx(5sinx3cosx)+(5sinx3cosx)ddxx4 =x4(5cosx+3sinx)+(5sinx3cosx)×4x3 =x3(5xcosx+3xsinx+20sinx12cosx)

Q.53 Find the derivative of the function (x2 + 1) cosx.


Let f(x)=(x2+1)cosxDifferentiating w.r.t. x, we get f’(x)=ddx{(x2+1)cosx}By Leibnitz product rule: f’(x)=(x2+1)ddxcosx+cosxddx(x2+1)=(x2+1)(sinx)+cosx(2x+0)=x2sinxsinx+2xcosx

Q.54 Find the derivative of the function (ax2 + sin x)(p + q cos x).


Let f(x)=(ax2+sinx)(p+qcosx)Differentiating w.r.t. x, we get f’(x)=ddx{(ax2+sinx)(p+qcosx)}By Leibnitz product rule: f’(x)={(ax2+sinx)ddx(p+qcosx)+(p+qcosx)ddx(ax2+sinx)} ={(ax2+sinx)(0qsinx)+(p+q cosx)(2ax+cosx)}=q sin x(ax2+sinx)+(p+q cos x)(2ax+cosx)

Q.55 Find the derivative of the function (x + cos x)(x – tan x).


Let f(x)=(x+cosx)(xtan x)Differentiating w.r.t. x, we get f’(x)=ddx{(x+cosx)(xtan x)} f’(x)={(x+cosx)ddx(xtan x)+(xtan x)ddx(x+cosx)} By Leibnitz product rule: f’(x)={(x+cosx)ddx(xtan x)+(xtan x)ddx(x+cosx)}=(x+cosx)(1sec2x)+(xtan x)(1sinx)=(x+cosx)(sec2x1)+(xtanx)(1sinx)=tan2x(x+cosx)+(xtanx)(1sinx)


Find the derivative of the function 4x+5sinx3x+7cosx.


Let f(x)=4x+5sinx3x+7cosxDifferentiating w.r.t. x, we getf’(x)=ddx(4x+5sinx3x+7cosx) ={(3x+7cosx)ddx(4x+5sinx)(4x+5sinx)ddx(3x+7cosx)}(3x+7cosx)2=(3x+7cosx)(4+5cosx)(4x+5sinx)(37sinx)(3x+7cosx)2=(12x+28cosx+15xcosx+35cos2x12x15sinx+28xsinx+35sin2x)(3x+7cosx)2=(28cosx+15xcosx+35cos2x15sinx+28xsinx+35(1cos2x))(3x+7cosx)2 =(28cosx+15xcosx+35cos2x15sinx+28xsinx+3535cos2x)(3x+7cosx)2f’(x)=(35+15xcosx+28cosx+28xsinx15sinx)(3x+7cosx)2


Find the derivative of the function x2 cos(π4)sinx.


Let f(x)=x2cos(π4)sinxDifferentiating w.r.t. x, we get f’(x)=ddx(x2cos(π4)sinx) =sinx{ddxx2cos(π4)}x2cos(π4)(ddxsinx)(sinx)2 =sinx{2xcos(π4)}x2cos(π4)(cosx)sin2x =2xsinxcos(π4)x2cos(π4)(cosx)sin2x =xcos(π4)(2sinxxcosx)sin2x


Find the derivative of the function x1+tanx.


Let fx=x1+tanxDifferentiating w.r.t. x, we getf’x=ddxx1+tanx=1+tanxddxxxddx1+tanx1+tanx2=1+tanx1x0+sec2x1+tanx2=1+tanxxsec2x1+tanx2

Q.59 Find the derivative of the function (x + sec x)(x – tan x).


Let f(x)=(x+secx)(xtanx)Differentiating w.r.t. x, we get f’(x)=ddx{(x+secx)(xtanx)}By Leibnitz product rule: f’(x)={(x+secx)ddx(xtan x)+(xtan x)ddx(x+secx)} f’(x)=(x+secx)(1sec2x)+(xtan x)(1+sec x tan x)


Find the derivative of the function xsinnx.


Let fx=xsinnxDifferentiating w.r.t. x, we get     f’x=ddxxsinnx     =sinnxddxxxddxsinnxsinnx2     =sinnx1xnsinn1xcosxsinnx2∵ddxsinnx=nsinn1xcosx     =sinn1xsinxnxcosxsin2nxddxxsinnx=sinxnxcosxsinn+1x

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FAQs (Frequently Asked Questions)

1. What are the topics covered in the NCERT Solutions for Class 11 Mathematics Chapter 13?

The NCERT Solutions for Class 11 Mathematics Chapter 13 has complete information about the limits and derivatives. The topics like intuitive ideas of derivatives, limits, algebra of limits, limits of trigonometric functions, derivatives, algebra of derivatives of functions, derivatives of polynomials and trigonometric functions are explained with examples. The students will get detailed and authentic solutions without having to look anywhere else. 

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