CBSE Class 11 Maths Revision Notes Chapter 13

Limits and Derivatives come under the field of calculus, which can be overwhelming and difficult for students to understand. Revision notes can help them to grasp core concepts clearly and build a strong base in this chapter.

Quick Links

Extramarks provides Revision Notes for Class 11 Mathematics Chapter 13 which contains lucid and concise explanations for every topic. It ensures an in-depth understanding of key concepts and formulas that will assist students in solving the chapter’s questions. Extramarks can help students learn this chapter with ease and score high marks.

Limits and Derivatives Class 11 Notes – Chapter Overview

Class 11 Revision Notes Limits and Derivatives

Extramarks Revision Notes for Class 11 Mathematics Chapter 13 Limits and Derivatives contain all important concepts of this chapter like Limits and the 7 Theorems, Limits of Polynomials and Rational functions, algebra of limits and all the respective formulas. These notes are easy to access and students can use them for self-study and for quick revisions.

Class 11 Mathematics Revision Notes Chapter 13

Consider a function f(x)=x2. Plotting it gives the value of x which approaches 0 as the value of function f(x) moves to 0.
If x→a, f(x)→l, then the limit of the function f(x) is represented by l. This is written symbolically as limx→af(x)=l.
The function should assume at a given point x=a irrespective of the limits.
X can approach a number in two ways, from left or right. This means that all the values of x near a could be less or greater than a.
- Right-hand limit – Value of f(x) which is dictated by values of f(x) when x tends to from the right which is written as limx→a+f(x).
- Left-hand limit – Value of f(x) which is dictated by values of f(x) when x tends to from the left which is written as limx→a−f(x).

The limit of f(x) as x tends to zero does not exist (even though the function is defined at 0) since left and right-hand limits are different.
The common value is the limit and is denoted by limx→af(x) if the right and left-hand limits coincide.

Standard Limits – Listed Below Are Some Standard Limits

limx→p (xn – pn)/x – p = npn-1
limx→0 (sin x)/x = 1
limx→p (sin (x – p))/(x – p) = 1
limx→0 (tan x)/x = 1
limx→p (tan (x – p))/(x – p) = 1
limx→0 (sin-1 x)/x = 1
limx→0 (tan-1 x)/x = 1
limx→0 (px – 1)/x = logep, p > 0 and p 1

Algebra of Limits:

If f and g are two functions such that both limx→af(x) and limx→ag(x) exist, then:

Limit of sum of two functions is sum of the limits of the functions, i.e. limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x).
Limit of difference of two functions is difference of the limits of the functions, i.e. limx→a[f(x)−g(x)]=limx→af(x)−limx→ag(x)
Limit of product of two functions is product of the limits of the functions, i.e., limx→a[f(x).g(x)]=limx→af(x).limx→ag(x)
Limit of quotient of two functions is the quotient of the limits of the functions (whenever the denominator is non zero), i.e., limx→af(x)g(x)=limx→af(x)limx→ag(x)
In a special case, when g is the constant function such that g(x)=λ, for some real number λ, we have limx→a[(λ.f)(x)]=λ.limx→af(x)

Polynomial Function:

A polynomial function is a function f(x) if it is a zero function or if f(x)=a0+a1x+a2x2+…+anxn where aiS is are real numbers such that an≠0 for some natural number n.

Rational Function:

A rational function is f(x)=g(x)/h(x) where g(x) and h(x) are polynomials such that h(x)≠0

limx→af(x)=limx→ag(x)/h(x)= limx→ag(x)/limx→ah(x) = g(a)/h(a).

Derivatives:

Derivatives – The derivative of a function y = f(x) is obtained by applying the equation change in y/change in x. Assume x goes from x to dx, and y changes from y to f(x) to f(x + dx), so,

Derivative of a function at point x is given by: f′(x)=limh→0f(x+h)−f(x)/h .
Derivative = change in y/change in x = dy/dx = f(x + dx) – f(x)/dx

Some Standard Derivatives – Mentioned Below Are Some Standard Derivatives

d(xi)/dx = ixi-1
d(sin x)/dx = cos x
d(cos x)/dx = -sin x
d(tan x)/dx = sec2x
d(cot x)/dx = -coses2x
d(sec x)/dx = secx . tanx
d(cosex x)/dx = -cosec x. cot x

Q.1

$\begin{array}{l} Find the limit : \\ \lim_{x \to 0} \frac{\sin x}{2 x^{2} - x} \end{array}$

Ans

$\begin{array}{l} We have, \\ \lim_{x \to 0} \frac{\sin x}{2 x^{2} - x} = \lim_{x \to 0} \frac{\sin x}{x (2 x - 1)} \\ = \lim_{x \to 0} \frac{\sin x}{x} \times \lim_{x \to 0} \frac{1}{2 x - 1} \\ = 1 \times \frac{1}{2 (0) - 1} [\lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = - 1 \end{array}$

Q.2

$\begin{array}{l} Free the limit: \\ \lim_{x \to 2} [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}] \end{array}$

Ans

$\begin{array}{l} We have, \\ \lim_{x \to 2} [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}] = \lim_{x \to 2} [\frac{x (x^{2} - 4 x + 4)}{x^{2} - 4}] \\ = \lim_{x \to 2} [\frac{x (x - 2) (x - 2)}{(x - 2) (x + 2)}] \\ = \lim_{x \to 2} [\frac{x (x - 2)}{(x + 2)}] \\ = \frac{2 (2 - 2)}{(2 + 2)} \\ = \frac{0}{4} = 0 \end{array}$

Q.3

$Compute : \lim_{x \to 7} (x + 7)$

Ans

$\lim_{x \to 7} (x + 7) = 7 + 7 = 14.$

Q.4

$Compute : \lim_{x \to 2} \frac{x^{3} + 8}{x^{2} - 1}$

Ans

$\begin{array}{l} \lim_{x \to 2} \frac{x^{3} + 8}{x^{2} - 1} = \frac{2^{3} + 8}{2^{2} - 1} \\ = \frac{8 + 8}{4 - 1} = \frac{16}{3} . \end{array}$

Q.5

$Find the limit: \lim_{x \to 3} \frac{x^{2} - 9}{x^{3} + 9 x - 6 x^{2}} .$

Ans

$\begin{array}{l} On substituting the value x = 3, the expression reduces to \frac{0}{0} . \\ Now, \\ \lim_{x \to 3} \frac{x^{2} - 9}{x^{3} + 9 x - 6 x^{2}} = \lim_{x \to 3} \frac{(x - 3) (x + 3)}{x (x^{2} + 9 - 6 x)} = \lim_{x \to 3} \frac{(x - 3) (x + 3)}{x {(x - 3)}^{2}} \\ = \lim_{x \to 3} \frac{(x + 3)}{x (x - 3)} = \frac{3 + 3}{3 (3 - 3)} = \frac{6}{0} = \infty \\ which is not defined. \end{array}$

Q.6

$Prove that for any positive integer n, \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1} .$

Ans

$\begin{array}{l} x^{n} - a^{n} = (x - a) (x^{n - 1} + x^{n - 2} a + x^{n - 3} a^{2} + \dots + {xa}^{n - 2} + a^{n - 1}) \\ \Rightarrow \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = \lim_{x \to a} (x^{n - 1} + x^{n - 2} a + x^{n - 3} a^{2} + \dots + {xa}^{n - 2} + a^{n - 1}) \\ = a^{n - 1} + a^{n - 2} a + a^{n - 3} a^{2} + + {aa}^{n - 2} + a^{n - 1} \\ = a^{n - 1} + a^{n - 1} + \dots + a^{n - 1} + a^{n - 1} \\ = {na}^{n - 1} \end{array}$

Q.7

$Compute: \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$

Ans

$\begin{array}{l} Put y = 1 + x, so that when x \to 0 then y \to 1 \\ \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} \lim_{y \to 1} \frac{\sqrt{y} - 1}{y - 1} = \lim_{y \to 1} \frac{y^{\frac{1}{2}} - 1^{\frac{1}{2}}}{y - 1} \\ = \frac{1}{2} {(1)}^{\frac{1}{2} - 1} = \frac{1}{2} (as \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1}) \end{array}$

Q.8

$Compute: \lim_{r \to 1} 2 πr$

Ans

$\lim_{r \to 1} 2 πr = 2 π \times 1 = 2 π$

Q.9

$Evaluate: \lim_{x \to 0} \frac{\sin 6 x}{\sin 3 x} .$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{\sin 6 x}{\sin 3 x} = \lim_{x \to 0} [\frac{\sin 6 x}{6 x} \times \frac{3 x}{\sin 3 x} \times 2] \\ = 2 \cdot \frac{\lim_{x \to 0} \frac{\sin 6 x}{6 x}}{\lim_{x \to 0} \frac{\sin 3 x}{3 x}} \\ = 2 \cdot \frac{\lim_{x \to 0} \frac{\sin 6 x}{6 x}}{\lim_{3 x \to 0} \frac{\sin 3 x}{3 x}} \\ = 2 \cdot \frac{1}{1} = 2 (as \lim_{x \to 0} \frac{\sin x}{x} = 1) \end{array}$

Q.10

$Evaluate: \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} .$

Ans

$\begin{array}{l} When we substitute (\frac{π}{2} - x) as y, then y \to 0 as x \to \frac{π}{2} \\ \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = \lim_{y \to 0} \frac{\tan 2 (\frac{π}{2} - y)}{- y} \\ = \lim_{y \to 0} \frac{\tan (π - 2 y)}{- y} = \lim_{y \to 0} \frac{- \tan (2 y)}{- y} = \lim_{y \to 0} \frac{\tan (2 y)}{y} \\ = \lim_{y \to 0} \frac{\sin (2 y)}{y \cos (2 y)} = \lim_{2 y \to 0} \frac{\sin (2 y)}{2 y} \times \frac{2}{\cos (2 y)} \\ = 1 \times \frac{2}{1} = 2 \end{array}$

Q.11

$Find \lim_{x \to 0} f(x)wheref(x) = {\frac{1}{| x |}, x \neq 0} .$

Ans

$\begin{array}{l} It is given that = \{\frac{1}{| x |}, x \neq 0\} \\ We observe that when x approaches to zero from left side i.e. \\ x is negative and close to zero then | x | is close to zero close to zero and positive. \\ Therefore 1 / | x | is large and positive. \\ \lim_{x \to 0^{-}} \frac{1}{| x |} \to \infty \\ Similarly if x approaches to zero from right side, i.e., \\ x is positive and close to zero then | x | is close to zero and positive. \\ Therefore 1 / | x | is large and positive. \\ \lim_{x \to 0^{+}} \frac{1}{| x |} \to \infty \\ Therefore limits exists and it tends to \infty . \end{array}$

Q.12

$Compute: \lim_{x \to 0} x \sec x .$

Ans

$\lim_{x \to 0} x \sec x = 0 \times \sec 0 = 0 \times 1 = 0$

Q.13

$\begin{array}{l} If f (x) = {\begin{matrix} {mx}^{2} + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ {nx}^{3} + m, & x > 1 \end{matrix} \\ For what values of m and n do \lim_{x \to 0} f (x) and \lim_{x \to 1} f (x) both exist. \end{array}$

Ans

$\begin{array}{l} It is given thatf (x) = \{\begin{matrix} {mx}^{2} + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ {nx}^{3} + m, & x > 1 \end{matrix} \\ \lim_{x \to 0} f(x)and \lim_{x \to 1} f(x)both exist. \\ \Leftrightarrow \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) and \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) \\ \Leftrightarrow \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (0 + h)and \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} f(1+ h) \\ \Leftrightarrow \lim_{h \to 0} \{m {(- h)}^{2} + n\} = \lim_{h \to 0} {nh + m} and \lim_{h \to 0} {n(1 - h) + m} = \lim_{h \to 0} \{{n (1 + h)}^{3} + m\} \\ \Leftrightarrow n = m and n + m = n + m \\ Hence \lim_{x \to 0} f(x)and \lim_{x \to 1} f(x)both exist for n = m . \end{array}$

Q.14 Compute the derivative of sin x w.r.t. x by first principle.

Ans

$\begin{array}{l} Let f (x) = sinx . \\ Then, \\ \frac{df (x)}{dx} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{\sin (x + h) - \sin (x)}{h} \\ = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + h}{2}) \sin (\frac{h}{2})}{h} \\ = \lim_{h \to 0} \cos (x + \frac{h}{2}) \cdot \lim_{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}} = \cos x \cdot 1 = \cos x \end{array}$

Q.15 Compute the derivative of cos²x.

Ans

$\begin{array}{l} \frac{df (x)}{dx} = \frac{d}{dx} (\cos x \cdot \cos x) \\ = (\cos x) ‘ \cos x + \cos x (\cos x) ‘ (By using product rule) \\ = (- \sin x) \cos x + \cos x (- \sin x) \\ = - 2 \sin x \cos x \\ = - \sin 2 x \end{array}$

Q.16

$\begin{array}{l} Find the derivative of : \\ (a) 3 x - 4 x^{2} \\ (b) (5 y^{4} - 2 y^{2} + 1) (y - 1) \end{array}$

Ans

$\begin{array}{l} (a) f (x) = 3 x - 4 x^{2} 3 \\ \Rightarrow \frac{df (x)}{dx} = (3 x - 4 x^{2}) ‘ \\ = 3 - 4 \cdot 2 x = 3 - 8 x \\ (b) f (y) = (5 y^{4} - 2 y^{2} + 1) (y - 1) \\ Using the product rule, \\ \frac{df (y)}{dy} = \frac{d}{dy} [(5 y^{4} - 2 y^{2} + 1) (y - 1)] \\ = (5 y^{4} - 2 y^{2} + 1) ‘ (y - 1) + (5 y^{4} - 2 y^{2} + 1) (y - 1) ‘ \\ = (20 y^{3} - 4 y) (y - 1) + (5 y^{4} - 2 y^{2} + 1) (1) \\ = 20 y^{4} - 20 y^{3} - 4 y^{2} + 4 y + 5 y^{4} - 2 y^{2} + 1 \\ = 25 y^{4} - 20 y^{3} - 6 y^{2} + 4 y + 1 \end{array}$

Q.17 Find the first derivative of (13t – 1) (5t³ + 2) w.r.t.t.

Ans

$\begin{array}{l} \frac{df (t)}{dt} = \frac{d}{dt} [(13 t - 1) (5 t^{3} + 2)] \\ = (13 t - 1) ‘ (5 t^{3} + 2) + (13 t - 1) (5 t^{3} + 2) ‘ (Using the product rule) \\ = 13 (5 t^{3} + 2) + (13 t - 1) (15 t^{2}) \\ = 65 t^{3} + 26 + 195 t^{3} - 15 t^{2} \\ = 260 t^{3} - 15 t^{2} + 26 \end{array}$

Q.18

$Find the first derivative of \frac{x + a}{x - a} by using First Principle.$

Ans

$\begin{array}{l} Let f (x) = \frac{x + a}{x - a} . \\ Then, \\ \frac{df (x)}{dx} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ \frac{df (x)}{dx} = \lim_{h \to 0} \frac{(\frac{x + h + a}{x + h - a}) - (\frac{x + a}{x - a})}{h} \\ = \lim_{h \to 0} \frac{(x + h + a) (x - a) - (x + a) (x + h - a)}{h (x + h - a) (x - a)} \\ = \lim_{h \to 0} \frac{(x^{2} + hx + ax - ax - ah - a^{2}) - (x^{2} + ax + hx + ah - ax - a^{2})}{h (x + h - a) (x - a)} \\ = \lim_{h \to 0} \frac{(x^{2} + hx + ax - ax - ah - a^{2}) - (x^{2} + ax + hx + ah - ax - a^{2})}{h (x + h - a) (x - a)} \\ = \lim_{h \to 0} \frac{x^{2} + hx + ax - ax - ah - a^{2} - x^{2} - ax - hx - ah + ax + a^{2}}{h (x + h - a) (x - a)} \\ = \lim_{h \to 0} \frac{- 2 ah}{h (x + h - a) (x - a)} \\ = \frac{- 2 a}{(x - a) (x - a)} = \frac{- 2 a}{x^{2} - 2 ax + a^{2}} \end{array}$

Q.19 Compute the derivative of √x from the first principle.

Ans

$\begin{array}{l} Let f (x) = \sqrt{x} and f (x + h) = \sqrt{(x + h)} \\ Then by first principle \\ f^{‘} (x) = \lim_{x \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{x \to 0} \frac{\sqrt{(x + h)} - \sqrt{x}}{h} \times \frac{\sqrt{(x + h)} + \sqrt{x}}{\sqrt{(x + h)} + \sqrt{x}} \\ = \lim_{x \to 0} \frac{(x + h) - x}{h (\sqrt{(x + h)} + \sqrt{x})} \\ = \lim_{x \to 0} \frac{h}{h (\sqrt{(x + h)} + \sqrt{x})} \\ = \frac{1}{\sqrt{x} + \sqrt{x}} \\ f^{‘} (x) = \frac{1}{2 \sqrt{x}} . \end{array}$

Q.20

$Compute the first derivative of \frac{x}{1 + tanx} .$

Ans

$\begin{array}{l} f (x) = \frac{x}{1 + tanx} \\ Using quotient rule for differentiation, \\ f^{‘} (x) = \frac{(1 + tanx) {(x)}^{‘} - x {(1 + tanx)}^{‘}}{{(1 + tanx)}^{2}} \\ = \frac{(1 + tanx) (1) - x (\sec^{2} x)}{{(1 + tanx)}^{2}} \\ = \frac{1 + tanx - {xsec}^{2} x}{{(1 + tanx)}^{2}} \end{array}$

Q.21

$Compute the first derivative of \frac{\sin x}{\cos (x + a)} .$

Ans

$\begin{array}{l} f (x) = \frac{\sin x}{\cos (x + a)} \\ Using the quotient rule for differentiation we get \\ f^{‘} (x) = \frac{\cos (x + a) {(sinx)}^{‘} - {\cos (x + a)}^{‘} \sin x}{\cos^{2} (x + a)} \\ = \frac{\cos (x + a) \cdot cosx - {- \sin (x + a)} \cdot sinx}{\cos^{2} (x + a)} \\ = \frac{cosx \cdot \cos (x + a) + sinx \cdot \sin (x + a)}{\cos^{2} (x + a)} \\ = \frac{\cos (x + a - x)}{\cos^{2} (x + a)} \\ = \frac{\cos a}{\cos^{2} (x + a)} \end{array}$

Q.22 Find the derivative of the function f(x) = 2x² + 3x – 3 at x = –2 by first principal method.

Ans

$\begin{array}{l} Wefindthederivativeofthefunctionat x = - 2 . \\ Wehave \\ f^{‘} (- 2) = \lim_{h \to 0} \frac{f (- 2 + h) - f (- 2)}{h} \\ = \lim_{h \to 0} \frac{[2 {(- 2 + h)}^{2} + 3 (- 2 + h) - 3] - [2 {(- 2)}^{2} + 3 (- 2) - 3]}{h} \\ = \lim_{h \to 0} \frac{[2 (4 + h^{2} - 4 h) - 6 + 3 h - 3] - [8 - 6 - 3]}{h} \\ = \lim_{h \to 0} \frac{[8 + 2 h^{2} - 8 h - 6 + 3 h - 3 - 8 + 6 + 3]}{h} \\ = \lim_{h \to 0} \frac{2 h^{2} - 5 h}{h} = \lim_{h \to 0} 2 h - 5 = 2 (0) - 5 = - 5 \end{array}$

Q.23

$Evaluate: \lim_{x \to 0} \frac{1 - \cos 2 x}{x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{1 - \cos 2 x}{x} = \lim_{x \to 0} \frac{2 \sin^{2} x}{x} \\ = \lim_{x \to 0} (2 \sin x \times \frac{\sin x}{x}) \\ = 2 \times 0 \times 1 = 0 \end{array}$

Q.24 Find the derivative of f(x) = x²cosx by first principle.

Ans

$\begin{array}{l} Let f (x) = x^{2} \cos x . Then, f (x + h) = {(x + h)}^{2} \cos (x + h) \\ ∴ \frac{d}{dx} f (x) = \lim_{x \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{x \to 0} \frac{{(x + h)}^{2} \cos (x + h) - x^{2} cosx}{h} \\ = \lim_{x \to 0} \frac{(x^{2} + 2 xh + h^{2}) \cos (x + h) - x^{2} \cos x}{h} \\ = \lim_{x \to 0} \frac{x^{2} {\cos (x + h) - cosx} + h (2 x + h) \cos (x + h)}{h} \\ = \lim_{x \to 0} \frac{x^{2} {\cos (x + h) - cosx}}{h} + \lim_{x \to 0} \frac{h (2 x + h) \cos (x + h)}{h} \\ = \lim_{x \to 0} \frac{x^{2} (- 2 \sin \frac{h}{2} \cdot \sin \frac{2 x + h}{2})}{h} + \lim_{x \to 0} (2 x + h) \cos (x + h) \\ = \lim_{x \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times - x^{2} \sin \frac{2 x + h}{2} + 2 x \cos x \\ = 1 \times - x^{2} \sin x + 2 x \cos x \\ = - x^{2} \sin x + 2 x \cos x \end{array}$

Q.25

$Compute the first derivative of \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} .$

Ans

$\begin{array}{l} f (x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} = \frac{\frac{x + 1}{x}}{\frac{x - 1}{x}} = \frac{x + 1}{x - 1} \\ Using quotient rule for differentiation, \\ f^{‘} (x) = \frac{(x - 1) {(x + 1)}^{‘} - {(x - 1)}^{‘} (x + 1)}{{(x - 1)}^{2}} \\ = \frac{(x - 1) (1) - (1) (x + 1)}{{(x - 1)}^{2}} \\ = \frac{x - 1 - x - 1}{{(x - 1)}^{2}} = \frac{- 2}{{(x - 1)}^{2}}, x \neq 1 \end{array}$

Q.26

$Evaluate: \lim_{x \to 0} \frac{sinx - 2 \sin (3 x) + \sin (5 x)}{x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{sinx - 2 \sin (3 x) + \sin (5 x)}{x} = \lim_{x \to 0} \frac{sinx - \sin (3 x) - \sin (3 x) + \sin (5 x)}{x} \\ = \lim_{x \to 0} \frac{[sinx - \sin (3 x)] + [\sin (5 x) - \sin (3 x)]}{x} \\ = \lim_{x \to 0} \frac{- 2 \sin x \cos (2 x) + 2 \sin x \cos (4 x)}{x} \\ = \lim_{x \to 0} \frac{2 \sin x [\cos (4 x) - \cos (2 x)]}{x} \\ = 2 \lim_{x \to 0} (\frac{\sin x}{x}) \cdot \lim_{x \to 0} [\cos (4 x) - \cos (2 x)] \\ = 2 \times 1 \times 0 = 0 \end{array}$

Q.27

$Prove that \frac{d}{dx} (x^{n}) = {nx}^{n - 1} .$

Ans

$\begin{array}{l} We can prove it with the help of first principle. \\ We have f (x) = x^{n} \\ \frac{df (x)}{dx} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{{(x + h)}^{n} - x^{n}}{h} = \lim_{h \to 0} \frac{{(x + h)}^{n} - x^{n}}{(x + h) - x} \\ = \lim_{z \to x} \frac{z^{n} - x^{n}}{z - x} …(i) \\ where z = x + h and z \to x as h \to 0 \\ Using \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1} in(i)we can conclude that \frac{df (x)}{dx} = {nx}^{n - 1} \\ Hence, \frac{d}{dx} (x^{n}) = {nx}^{n - 1} . \end{array}$

Q.28

$Differentiate with respect to x : \frac{{ax}^{2} + bx + c}{\sqrt{x}} .$

Ans

$\begin{array}{l} \frac{d}{dx} [\frac{{ax}^{2} + bx + c}{\sqrt{x}}] = \frac{d}{dx} [\frac{{ax}^{2}}{\sqrt{x}} + \frac{bx}{\sqrt{x}} + \frac{c}{\sqrt{x}}] \\ = \frac{d}{dx} [{ax}^{\frac{3}{2}} + {bx}^{\frac{1}{2}} + {cx}^{\frac{- 1}{2}}] \\ = \frac{d}{dx} [{ax}^{\frac{3}{2}}] + \frac{d}{dx} [{bx}^{\frac{1}{2}}] + \frac{d}{dx} [{cx}^{\frac{- 1}{2}}] \\ = a \frac{d}{dx} [x^{\frac{3}{2}}] + b \frac{d}{dx} [x^{\frac{1}{2}}] + c \frac{d}{dx} [x^{\frac{- 1}{2}}] \\ = a (\frac{3}{2} x^{\frac{1}{2}}) + b (\frac{1}{2} x^{\frac{- 1}{2}}) + c (- \frac{1}{2} x^{\frac{- 3}{2}}) \\ = \frac{3 a}{2} x^{\frac{1}{2}} + \frac{b}{2} x^{\frac{- 1}{2}} - \frac{c}{2} x^{\frac{- 3}{2}} . \end{array}$

Q.29 Find the first derivative of x³ – 4 at x = 2.

Ans

f(x) = x³ – 4
f'(x) = 3x²
f'(2) = 3 × 2² = 3 × 4 = 12

Q.30

$For the function, f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots \dots + \frac{x^{2}}{2} + x + 1 . Prove that f ‘ (1) = 100 f ‘ (0) .$

Ans

$\begin{array}{l} f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots \dots . + \frac{x^{2}}{2} + x + 1 \\ f^{‘} (x) = \frac{100 x^{99}}{100} + \frac{99 x^{98}}{99} + \dots \dots + \frac{2 x}{2} + 1 \\ = x^{99} + x^{98} + \dots \dots . + x + 1 \\ LHS- f^{‘} (1) = 1^{99} + 1^{98} + \dots \dots . + 1 + 1 (100 times) = 100 \\ RHS- f^{‘} (0) = 0^{99} + 0^{98} + \dots \dots . + 0 + 1 = 1 \\ Thus, f^{‘} (1) = 100 f^{‘} (0) \end{array}$

Q.31

$Evaluate : \lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} .$

Ans

$\begin{array}{l} On substituting the value x = - 2, the expression reduces to 0 / 0 . \\ Therefore, \\ \lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \to - 2} \frac{\frac{x + 2}{2 x}}{x + 2} \\ = \lim_{x \to - 2} \frac{1}{2 x} \\ = \frac{1}{2 (- 2)} = - \frac{1}{4} \end{array}$

Q.32

$Evaluate: \lim_{x \to 2} \frac{e^{x} - e^{2}}{x - 2} .$

Ans

$\begin{array}{l} \lim_{x \to 2} \frac{e^{x} - e^{2}}{x - 2} = \lim_{x - 2 \to 0} \frac{e^{2} (e^{x - 2} - 1)}{x - 2} \\ = e^{2} \lim_{y \to 0} \frac{e^{y} - 1}{y} [Let x - 2 = y] \\ = e^{2} \times 1 [\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = e^{2} \end{array}$

Q.33

$Evaluate: \lim_{x \to 2} \frac{e^{3 + x} - e^{5}}{x - 2}$

Ans

$\begin{array}{l} \lim_{x \to 2} \frac{e^{3 + x} - e^{5}}{x - 2} = \lim_{x - 2 \to 0} \frac{e^{5} (e^{x - 2} - 1)}{x - 2} \\ = e^{5} \lim_{y \to 0} \frac{e^{y} - 1}{y} [Let x - 2 = y] \\ = e^{5} \times 1 [\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = e^{5} \end{array}$

Q.34

$Evaluate : \lim_{x \to 5} \frac{e^{x} - e^{5}}{x - 5}$

Ans

$\begin{array}{l} \lim_{x \to 5} \frac{e^{x} - e^{5}}{x - 5} = \lim_{x - 5 \to 0} \frac{e^{5} (e^{x - 5} - 1)}{x - 5} \\ = e^{5} \lim_{y \to 0} \frac{e^{y} - 1}{y} [Let x - 5 = y] \\ = e^{5} \times 1 [\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = e^{5} \end{array}$

Q.35

$Evaluate : \lim_{x \to 0} \frac{e^{2 sinx} - 1}{x} .$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{e^{2 \sin x} - 1}{\sin x} = 2 \times \lim_{x \to 0} \frac{e^{2 \sin x} - 1}{2 \sin x} \\ = 2 \times \lim_{2 \sin x \to 0} \frac{e^{2 \sin x} - 1}{2 \sin x} [as x \to 0, 2 \sin x \to 0] \\ = 2 \times 1 [\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = 2 \end{array}$

Q.36

$Evaluate : \lim_{x \to 0} \frac{x (e^{2 x} - 1)}{1 - cosx} .$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{x (e^{2 x} - 1)}{1 - \cos x} = \lim_{x \to 0} \frac{e^{2 x} - 1}{x} \times \lim_{x \to 0} \frac{x^{2}}{1 - \cos x} \\ = 2 \lim_{2 x \to 0} \frac{e^{2 x} - 1}{2 x} \times \lim_{x \to 0} \frac{x^{2}}{2 \sin^{2} (\frac{x}{2})} \\ = 2 \times \lim_{2 x \to 0} \frac{e^{2 x} - 1}{2 x} \times \lim_{x \to 0} \frac{2 {(\frac{x}{2})}^{2}}{\sin^{2} (\frac{x}{2})} \\ = 4 \times \lim_{2 x \to 0} \frac{e^{2 x} - 1}{2 x} \times \lim_{x / 2 \to 0} \frac{{(\frac{x}{2})}^{2}}{\sin^{2} (\frac{x}{2})} \\ = 4 \times 1 \times 1 [\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 and \lim_{x \to 0} \frac{sinx}{x} = 1] \\ = 4 \end{array}$

Q.37

$Evaluate: \lim_{x \to 0} \frac{x (e^{2 x} - 1)}{\sin x} .$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{x (e^{2 x} - 1)}{\sin x} = \lim_{x \to 0} (\frac{x}{\sin x}) \times 2 \lim_{2 x \to 0} \frac{e^{2 x} - 1}{2 x} \\ = 1 \times 2 (1) [\lim_{x \to 0} \frac{\sin x}{x} = 1 and \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = 0 \end{array}$

Q.38

$Evaluate : \lim_{x \to 0} \frac{\sin x + (e^{5 x} - 1)}{2 x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{\sin x + (e^{5 x} - 1)}{2 x} = \lim_{x \to 0} \frac{\sin x}{2 x} + \lim_{x \to 0} \frac{e^{5 x} - 1}{2 x} \\ = \frac{1}{2} \lim_{x \to 0} \frac{\sin x}{x} + \frac{5}{2} \lim_{x \to 0} \frac{e^{5 x} - 1}{5 x} \\ = \frac{1}{2} \times 1 + \frac{5}{2} \times 1 [\lim_{x \to 0} \frac{sinx}{x} = 1 and \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = \frac{6}{2} \\ = 3 \end{array}$

Q.39

$Evaluate : \lim_{x \to 0} \frac{\sin x + \log_{e} (1 + x)}{2 x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{\sin x + \log_{e} (1 + x)}{2 x} = \lim_{x \to 0} \frac{\sin x}{2 x} + \lim_{x \to 0} \frac{\log_{e} (1 + x)}{2 x} \\ = \frac{1}{2} \lim_{x \to 0} \frac{\sin x}{x} + \frac{1}{2} \lim_{x \to 0} \frac{\log_{e} (1 + x)}{x} \\ = \frac{1}{2} \times 1 + \frac{1}{2} \times 1 [\begin{array}{l} \lim_{x \to 0} \frac{\sin x}{x} = 1 and \\ \lim_{x \to 0} \frac{\log_{e} (1 + x)}{x} = 1 \end{array}] \\ = 1 \end{array}$

Q.40

$\begin{array}{l} Find the following limit : \\ \lim_{x \to 0} \frac{\sqrt[3]{\sin x}}{\sqrt[3]{x}} \end{array}$

Ans

$\begin{array}{l} Wehave, \\ \lim_{x \to 0} \frac{\sqrt[3]{\sin x}}{\sqrt[3]{x}} = \lim_{x \to 0} \sqrt[3]{(\frac{\sin x}{x})} \\ = \sqrt[3]{\lim (\frac{\sin x}{x})} \\ = \sqrt[3]{1} [\lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = 1 \end{array}$

Q.41

$\begin{array}{l} Find the following limit : \\ \lim_{x \to 0} (\frac{x + \sin x}{x}) \end{array}$

Ans

$\begin{array}{l} Wehave, \\ \lim_{x \to 0} (\frac{x + \sin x}{x}) = \lim_{x \to 0} (\frac{x}{x} + \frac{\sin x}{x}) \\ = \lim_{x \to 0} 1 + \lim_{x \to 0} \frac{\sin x}{x} \\ = 1 + 1 [\lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = 2 \end{array}$

Q.42

$\begin{array}{l} Find the limit : \\ \lim_{x \to 3} \frac{x^{2} - 9}{x^{3} + 2 x - 15} \end{array}$

Ans

$\begin{array}{l} Wehave, \\ \lim_{x \to 3} \frac{x^{2} - 9}{x^{3} + 2 x - 15} = \lim_{x \to 3} \frac{(x - 3) (x + 3)}{(x + 5) (x - 3)} \\ = \lim_{x \to 3} \frac{(x + 3)}{(x + 5)} \\ = \frac{3 + 3}{3 + 5} \\ = \frac{6}{8} \\ = \frac{3}{4} \end{array}$

Q.43 Differentiate x log_ex with respect to x.

Ans

$\begin{array}{l} Let y = {xlog}_{e} x \\ Differentiating w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \{x \log_{e} x\} \\ = x \frac{d}{dx} \log_{e} x + \log_{e} x \frac{d}{dx} x \\ = x \cdot \frac{1}{x} + \log_{e} x . 1 \\ = 1 + \log_{e} x \end{array}$

Q.44 Differentiate sin x.e^x with respect to x.

Ans

$\begin{array}{l} Let y = \sin x \cdot e^{x} \\ Differentiating w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \{\sin x \cdot e^{x}\} = \sin x \frac{d}{dx} e^{x} + e^{x} \frac{d}{dx} \sin x \\ = \sin x \cdot e^{x} + e^{x} \cos x \\ = e^{x} (\sin x + \cos x) \end{array}$

Q.45 Differentiate sin²x with respect to x.

Ans

$\begin{array}{l} Let y = \sin^{2} x \\ Differentiating w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {\sin x \cdot \sin x} = \sin x \frac{d}{dx} \sin x + \sin x \frac{d}{dx} \sin x \\ = \sin x \cdot \cos x + \sin x \cdot \cos x \\ = 2 \sin x \cos x \end{array}$

Q.46 Find the derivative of cos x/log x.

Ans

$\begin{array}{l} Let y = \frac{\cos x}{\log x} \\ Differentiate w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \{\frac{\cos x}{\log x}\} \\ = \frac{\log x \frac{d}{dx} \cos x - \cos x \frac{d}{dx} \log x}{{(\log x)}^{2}} \\ = \frac{\log x {- sinx} - \cos x \frac{1}{x}}{{(\log x)}^{2}} \\ = \frac{- \sin x \cdot \log x - \frac{1}{x} \cdot \cos x}{{(\log x)}^{2}} \end{array}$

Q.47 Find the derivative of (x²+1)cos x.

Ans

$\begin{array}{l} Let y = (x^{2} - 1) \cos x \\ Differentiation w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \{(x^{2} - 1) \cos x\} \\ = (x^{2} - 1) \frac{d}{dx} \cos x + \cos x \frac{d}{dx} (x^{2} - 1) \\ = (x^{2} - 1) \times - \sin x + \cos x (2 x + 0) \\ = - (x^{2} - 1) \sin x + 2 x \cos x \\ = - x^{2} \sin x + \sin x + 2 x \cos x \end{array}$

Q.48 Find the derivative of (ax + b)(x+1)².

Ans

$\begin{array}{l} Let y = (ax + b) {(x + 1)}^{2} \\ Differentiating w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \{(ax + b) {(x + 1)}^{2}\} \\ = (ax + b) \frac{d}{dx} (x^{2} + 2 x + 1) + {(x + 1)}^{2} \frac{d}{dx} (ax + b) \\ = (ax + b) (2 x + 2) + {(x + 1)}^{2} (a + 0) \\ = 2 (x + 1) (ax + b) + a {(x + 1)}^{2} \\ = (x + 1) {2 (ax + b) + a (x + 1)} \end{array}$

Q.49

$\begin{array}{l} What is the value of \lim_{x \to 2} (5 x^{3} - 4 x^{2} + 5) ? \\ (a)29(b)24(c)12(d)60 \end{array}$

Ans

$\begin{array}{l} (a)29 \\ Explanation: \\ \lim_{x \to 2} (5 x^{3} - 4 x^{2} + 5) \\ = \lim_{x \to 2} (5 x^{3}) - \lim_{x \to 2} (4 x^{2}) + \lim_{x \to 2} (Addition Rule) \\ = 5 \lim_{x \to 2} x^{3} - 4 \lim_{x \to 2} x^{2} + 5 (Product Rule) \\ = 5 {(2)}^{3} - 4 {(2)}^{2} + 5 (Limit of Polynomials) \\ = 40 - 16 + 5 = 29 \end{array}$

Q.50

$\begin{array}{l} The value of \lim_{x \to 0} \frac{\sqrt{2 + x} - \sqrt{2}}{x} is \\ (a) \frac{1}{2} (b) \frac{1}{\sqrt{2}} \\ (c) \frac{1}{2 \sqrt{2}} (d) 2 \sqrt{2} \end{array}$

Ans

$\begin{array}{l} (c) \frac{1}{2 \sqrt{2}} \\ Explanation: \\ \lim_{x \to 0} \frac{\sqrt{2 + x} - \sqrt{2}}{x} \\ = \lim_{x \to 0} \frac{(\sqrt{2 + x} - \sqrt{2}) (\sqrt{2 + x} + \sqrt{2})}{x (\sqrt{2 + x} + \sqrt{2})} \\ = \lim_{x \to 0} \frac{2 + x - 2}{x (\sqrt{2 + x} + \sqrt{2})} \\ = \lim_{x \to 0} \frac{1}{(\sqrt{2 + x} + \sqrt{2})} = \frac{1}{2 \sqrt{2}} . \end{array}$

Q.51

$\begin{array}{l} If f (x) = x^{n} and if f^{‘} (1) = 10, the value of n is \\ (a) 9 (b) 10 (c) 89 (d) 90 \end{array}$

Ans

$\begin{array}{l} (b) 10 \\ Explanation: \\ We have, f (x) = x^{n} \\ Differentiating both sides w.r.t. x, we get f^{‘} (x) = {nx}^{n - 1} \\ Putting x = 1, we get \\ f^{‘} (1) = n [f^{‘} (1) = 10] \\ \Rightarrow n = 10 \end{array}$

Q.52

$Evaluate: \lim_{x \to \sqrt{10}} \frac{\sqrt{7 - 2 x} - (\sqrt{5} - \sqrt{2})}{x^{2} - 10}$

Ans

$\begin{array}{l} Wehave, \\ \lim_{x \to \sqrt{10}} \frac{\sqrt{7 - 2 x} - (\sqrt{5} - \sqrt{2})}{x^{2} - 10} \\ = \lim_{x \to \sqrt{10}} \frac{\sqrt{7 - 2 x} - \sqrt{{(\sqrt{5} - \sqrt{2})}^{2}}}{x^{2} - 10} \\ = \lim_{x \to \sqrt{10}} \frac{\sqrt{7 - 2 x} - \sqrt{(7 - 2 \sqrt{10})}}{x^{2} - 10} \\ = \lim_{x \to \sqrt{10}} \frac{\sqrt{7 - 2 x} - \sqrt{(7 - 2 \sqrt{10})}}{x^{2} - 10} \times \frac{\sqrt{7 - 2 x} + \sqrt{(7 - 2 \sqrt{10})}}{\sqrt{7 - 2 x} + \sqrt{(7 - 2 \sqrt{10})}} \\ = \lim_{x \to \sqrt{10}} \frac{(7 - 2 x) - (7 - 2 \sqrt{10})}{(x - \sqrt{10}) (x + \sqrt{10}) \{\sqrt{7 - 2 x} + \sqrt{(7 - 2 \sqrt{10})}\}} \\ = \lim_{x \to \sqrt{10}} \frac{- 2 x + 2 \sqrt{10}}{(x - \sqrt{10}) (x + \sqrt{10}) \{\sqrt{7 - 2 x} + \sqrt{(7 - 2 \sqrt{10})}\}} \\ = \lim_{x \to \sqrt{10}} \frac{- 2 (x - \sqrt{10})}{(x - \sqrt{10}) (x + \sqrt{10}) \{\sqrt{7 - 2 x} + \sqrt{(7 - 2 \sqrt{10})}\}} \\ = \lim_{x \to \sqrt{10}} \frac{- 2}{(x + \sqrt{10}) \{\sqrt{7 - 2 x} + \sqrt{(7 - 2 \sqrt{10})}\}} \\ = \lim_{x \to \sqrt{10}} \frac{- 2}{2 \sqrt{10} \{\sqrt{7 - 2 \sqrt{10}} + \sqrt{7 - 2 \sqrt{10}}\}} \\ = \frac{- 1}{\sqrt{10} \times 2 \times \sqrt{7 - 2 \sqrt{10}}} \\ = \frac{- 1}{2 \sqrt{10} (\sqrt{5} - \sqrt{2})} \\ = \frac{- 1}{2 \sqrt{10}} \times \frac{(\sqrt{5} + \sqrt{2})}{3} \\ = - \frac{(\sqrt{5} + \sqrt{2})}{6 \sqrt{10}} \end{array}$

Q.53

$\begin{array}{l} If f is a real valued function defined by f (x) = x^{2} + 4 x + 3, \\ then find f^{‘} (1) and f^{‘} (3) . \end{array}$

Ans

$\begin{array}{l} Wehave, \\ f (x) = x^{2} + 4 x + 3 \\ ∴ f^{‘} (1) = \lim_{h \to 0} \frac{f (1 + h) - f (1)}{h} \\ \Rightarrow f^{‘} (1) = \lim_{h \to 0} \frac{\{{(1 + h)}^{2} + 4 (1 + h) + 3\} - \{1^{2} + 4 \times 1 + 3\}}{h} \\ \Rightarrow f^{‘} (1) = \lim_{h \to 0} \frac{(h^{2} + 6 h + 8) - 8}{h} \\ \Rightarrow f^{‘} (1) = \lim_{h \to 0} \frac{h^{2} + 6 h}{h} \\ \Rightarrow f^{‘} (1) = \lim_{h \to 0} h + 6 \\ = 6 \\ and, \\ f^{‘} (3) = \lim_{h \to 0} \frac{f (3 + h) - f (3)}{h} \\ \Rightarrow f^{‘} (3) = \lim_{h \to 0} \frac{\{{(3 + h)}^{2} + 4 (3 + h) + 3\} - \{3^{2} + 4 \times 3 + 3\}}{h} \\ \Rightarrow f^{‘} (3) = \lim_{h \to 0} \frac{(h^{2} + 10 h + 24) - 24}{h} \\ \Rightarrow f^{‘} (3) = \lim_{h \to 0} \frac{h^{2} + 10 h}{h} \\ \Rightarrow f^{‘} (3) = \lim_{h \to 0} h + 10 \\ = 10 \end{array}$

Q.54

$Differentiate e^{\sqrt{tanx}} from first principal.$

Ans

$\begin{array}{l} Let f (x) = e^{\sqrt{\tan x}} \\ then, f (x + h) = e^{\sqrt{\tan (x + h)}} \\ \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{e^{\sqrt{\tan (x + h)}} - e^{\sqrt{tanx}}}{h} \\ \frac{d}{dx} (f (x)) = \lim_{h \to 0} e^{\sqrt{\tan x}} \{\frac{e^{\sqrt{\tan (x + h)} - \sqrt{tanx}} - 1}{h}\} \\ = e^{\sqrt{\tan x}} \lim_{h \to 0} \{\frac{e^{\sqrt{\tan (x + h)} - \sqrt{tanx}} - 1}{\sqrt{\tan (x + h)} - \sqrt{\tan x}} \times \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h}\} \\ = e^{\sqrt{\tan x}} \lim_{h \to 0} \{\frac{e^{\sqrt{\tan (x + h)} - \sqrt{\tan x}} - 1}{\sqrt{\tan (x + h)} - \sqrt{\tan x}}\} \cdot \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} \\ = e^{\sqrt{\tan x}} \times 1 \times \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} \times \frac{\sqrt{\tan (x + h)} + \sqrt{\tan x}}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} \\ = e^{\sqrt{tanx}} \times 1 \times \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h} \times \frac{1}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} \\ = e^{\sqrt{\tan x}} \times \lim_{h \to 0} \frac{\sin h}{h \cos (x + h) \cos x} \times \frac{1}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} \\ = e^{\sqrt{tanx}} \times \frac{1}{\cos^{2} h} \times \frac{1}{2 \sqrt{\tan x}} \\ = \frac{e^{\sqrt{\tan x}}}{2 \sqrt{\tan x}} \sec^{2} x \end{array}$

Q.55

$\begin{array}{l} Evaluate the following limits : \\ (i) \lim_{x \to 0} (\frac{\cos Ax - \cos Bx}{x^{2}}) \\ (ii) \lim_{x \to 0} \frac{\sin 2 x + \sin 3 x}{2 x + \sin 3 x} \\ (iii) \lim_{x \to 0} \frac{\sin 2 x + \sin 6 x}{\sin 5 x - \sin 3 x} \end{array}$

Ans

$\begin{array}{l} (i)Wehave, \\ \lim_{x \to 0} (\frac{cosAx - cosBx}{x^{2}}) \\ = \lim_{x \to 0} \frac{2 \sin (\frac{A + B}{2}) xsin (\frac{B - A}{2}) x}{x^{2}} [\begin{array}{l} cosC - cosD \\ = 2 \sin \frac{C + D}{2} \sin \frac{D - C}{2} \end{array}] \\ = 2 \lim_{x \to 0} \{\frac{\sin (\frac{A + B}{2}) x}{(\frac{A + B}{2}) x} \times (\frac{A + B}{2}) \frac{\sin (\frac{B - A}{2}) x}{(\frac{B - A}{2}) \times} \times (\frac{B - A}{2})\} \\ = 2 (\frac{B + A}{2}) (\frac{B - A}{2}) \lim_{x \to 0} \{\frac{\sin (\frac{A + B}{2}) x}{(\frac{A + B}{2}) x}\} \times \{\lim_{x \to 0} \frac{\sin (\frac{B - A}{2}) x}{2}\} \\ = (\frac{B^{2} - A^{2}}{2}) (1) (1) = \frac{B^{2} - A^{2}}{2} \\ (ii)Wehave, \\ \lim_{x \to 0} \frac{\sin 2 x + \sin 3 x}{2 x + \sin 3 x} \\ = \lim_{x \to 0} \frac{\frac{\sin 2 x}{x} + \frac{\sin 3 x}{x}}{2 + \frac{\sin 3 x}{x}} \\ = \lim_{x \to 0} \frac{2 (\frac{\sin 2 x}{2 x}) + 3 (\frac{\sin 3 x}{3 x})}{2 + 3 (\frac{\sin 3 x}{3 x})} \\ = \frac{2 \lim_{x \to 0} (\frac{\sin 2 x}{2 x}) + 3 \lim_{x \to 0} (\frac{\sin 3 x}{3 x})}{2 + 3 \lim_{x \to 0} (\frac{\sin 3 x}{3 x})} \\ = \frac{2 \times 1 + 3 \times 1}{2 + 3 \times 1} = \frac{5}{5} = 1 \\ (iii) Wehave, \\ \lim_{x \to 0} \frac{\sin 2 x + \sin 6 x}{\sin 5 x - \sin 3 x} \\ = \lim_{x \to 0} \frac{2 \sin 4 xcos 2 x}{2 sinxcos 4 x} \\ = \lim_{x \to 0} \{\frac{\frac{\sin 4 x}{4 x} \times 4 x \times \cos 2 x}{\frac{sinx}{x} \times x \times \cos 4 x}\} \\ = 4 \frac{\lim_{x \to 0} \frac{\sin 4 x}{4 x} \times \lim_{x \to 0} \cos 2 x}{\lim_{x \to 0} \frac{\sin x}{x} \times \lim_{x \to 0} \cos 4 x} \\ = 4 (\frac{1 \times 1}{1 \times 1}) = 4 \end{array}$

Q.56

$\begin{array}{l} Differentiate the following function with respect to x : \\ (i) \log_{x} x (ii) e^{3 logx} (iii) 2^{\log_{2} x} \\ (iv) 5 (2^{3 \log_{2} x}) (v) 5 e^{x} (vi) 9 \cdot (3^{x}) \end{array}$

Ans

$\begin{array}{l} (i) We have, \\ \frac{d}{dx} (\log_{x} x) = \frac{d}{dx} (1) = 0 . [∴ \log_{x} x = 1] \\ (ii) Wehave, \\ \frac{d}{dx} (e^{3 logx}) = \frac{d}{dx} (e^{{logx}^{3}}) \\ = \frac{d}{dx} (x^{3}) = 3 x^{2} [e^{logk} = k] \\ (iii) Wehave, \\ \frac{d}{dx} (2^{\log_{2} x}) = \frac{d}{dx} (x) = 1 [a^{\log_{a} n} = n] \\ (iv) Wehave, \\ \frac{d}{dx} (5 {.2}^{3 \log_{2} x}) \\ = 5 \cdot \frac{d}{dx} (2^{3 \log_{2} x}) \\ = 5 \cdot \frac{d}{dx} (2^{\log_{2} x^{3}}) = 5 \cdot \frac{d}{dx} (x^{3}) \\ = 5 (3 x^{2}) = 15 x^{2} . \\ (v) Wehave, \\ \frac{d}{dx} (5 e^{x}) = 5 \cdot \frac{d}{dx} (e^{x}) \\ = 5 e^{x} \\ (vi) Wehave, \\ \frac{d}{dx} (9 {.3}^{x}) = 9 \cdot \frac{d}{dx} (3^{x}) = 9 \cdot (3^{x} \log_{e} 3) . \end{array}$

Q.57

$\begin{array}{l} (i) The differentiation of cosec x with respect to x is - cosec x \cot x . \\ i.e., prove that \frac{d}{dx} (cosec x) = - cosec x \cot x \\ (ii) The differentiation of \log_{a} x (a > 0, a \neq 1) with respect to x is \frac{1}{x \log_{e} a} \\ i.e., prove that \frac{d}{dx} (\log_{a} x) = \frac{1}{{xlog}_{e} a} . \end{array}$

Ans

$\begin{array}{l} (i)Let f (x) = cosec x . \\ Then, \\ f (x + h) = cosec (x + h) \\ ∴ \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{cosec (x + h) - cosecx}{h} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{\frac{1}{\sin (x + h)} - \frac{1}{\sin x}}{h} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{sinx - \sin (x + h)}{h \sin x \sin (x + h)} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{2 \sin (\frac{x - x - h}{2}) \cos (\frac{x + x + h}{2})}{h \sin x \sin (x + h)} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{2 \sin (\frac{- h}{2}) \cos (x + h / 2)}{h \sin x \sin (x + h)} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{\sin (h / 2)}{h / 2} \times \lim_{h \to 0} \frac{\cos (x + h / 2)}{sinxsin (x + h)} \\ [\sin (- \frac{h}{2}) = - \sin \frac{h}{2}] \\ \Rightarrow \frac{d}{dx} (f (x)) = (- 1) \times \frac{cosx}{\sin x \sin x} \\ = - \cot x cosec x . \\ Hence, \frac{d}{dx} (cosec x) = - cosec x \cot x \\ (ii)Let f (x) = \log_{a} x \\ Then, \\ f (x + h) = \log_{a} (x + h) \\ ∴ \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{\log_{a} (x + h) - \log_{a} (x)}{h} \\ = \lim_{h \to 0} \frac{\log_{a} (\frac{x + h}{x})}{h} \\ \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{\log_{a} (1 + \frac{h}{x})}{h} \\ = \lim_{h \to 0} \frac{\log_{a} (1 + \frac{h}{x})}{(\log_{e} a) \cdot h} [\log_{a} λ = \frac{\log_{e} λ}{\log_{e} a}] \\ \Rightarrow \frac{d}{dx} (f (x)) = \frac{1}{\log_{e} a} \lim_{h \to 0} \frac{\log_{a} (1 + \frac{h}{x})}{x (\frac{h}{x})} \\ = \frac{1}{{xlog}_{e} a} [\lim_{h \to 0} \frac{\log_{a} (1 + \frac{h}{x})}{\frac{h}{x}} = 1] \\ Hence, \\ \frac{d}{dx} (\log_{a} x) = \frac{1}{{xlog}_{e} a} \end{array}$

Q.58

$The value of \lim_{x \to a} \frac{x^{m} - a^{m}}{x - a} is_____.$

Ans

$\begin{array}{l} x^{m} - a^{m} = (x - a) (x^{m - 1} + x^{m - 2} a + x^{m - 3} a^{2} + x^{m - 3} a^{3} + \dots + {xa}^{m - 2} + a^{m - 1}) \\ \lim_{x \to a} \frac{x^{m} - a^{m}}{x - a} = \lim_{x \to a} (x^{m - 1} + x^{m - 2} a + x^{m - 3} a^{2} + x^{m - 3} a^{3} + \dots + {xa}^{m - 2} + a^{m - 1}) \\ = x^{m - 1} + x^{m - 2} a + x^{m - 3} a^{2} + . .. + x a^{m - 2} + a^{m - 1} \\ = x^{m - 1} + x^{m - 1} + x^{m - 1} + x^{m - 1} + . .. + x^{m - 1} + x^{m - 1} \\ = m x^{m - 1} \end{array}$

Q.59 The derivative of x³(x – 3)² is ______.

Ans

$\begin{array}{l} {The given function is x}^{3} {(x - 3)}^{2} . \\ Let y = x^{3} {(x - 3)}^{2} \\ Differentiating w.r.t. x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {x^{3} {(x - 3)}^{2}} \\ = x^{3} \frac{d}{dx} {(x - 3)}^{2} + {(x - 3)}^{2} \frac{d}{dx} x^{3} \\ = x^{3} \times 2 (x - 3) + {(x - 3)}^{2} \times 3 x^{2} \\ = 2 x^{3} (x - 3) + 3 x^{2} {(x - 3)}^{2} \\ = x^{2} (x - 3) {2 x + 3 (x - 3)} \\ = x^{2} (x - 3) (2 x + 3 x - 9) \\ = x^{2} (x - 3) (5 x - 9) \\ {The derivative of x}^{3} {(x - 3)}^{2} is \underline{x^{2} (x - 3) (5 x - 9)} . \end{array}$

Please register to view this section

FAQs (Frequently Asked Questions)

1. What are some of the real-world applications of the concept of derivatives?

Examples of real-world applications of the derivative theory include:

Those who look after a reservoir need to know when it will overflow. This is accomplished by measuring the depth of the water at various points in time.
Rocket scientists can calculate the exact velocity with which a satellite should be launched from a rocket by knowing the height of the rocket at different points in time.
Knowing the present value of stock allows financial experts to forecast changes in its value in the stock market.

2. What is the sandwich theorem of limits?

According to the sandwich theorem of limits, if f, g, and z are real functions with f(x) <= g(x) <= z(x) for all x in the common domain, then for a real number a if limx→af(x)=l=limx→ag(x) , then limx→ag(x)=l.

3. How can one ace Chapter 13 “Limits and Derivatives”?

Calculus is the study of rates of change and the continuity of quantities. “Limits and Derivatives” accounts for around half of calculus. To ace this chapter, you must:

Pay attention in class and clarify any doubts directly with teachers as soon as possible. Students can also view videos to gain a better understanding.
The more one practises this chapter, the better one will become. It is essential to practise examples and question papers.
Extramarks revision notes are a good source to grasp the concepts mentioned in this chapter which is compiled in a concise form. It is written by subject matter experts according to the latest and updated CBSE syllabus.

4. Why should one learn about limits and derivatives?

Understanding the chapter and doing well on exams is aided by learning how mathematical concepts are applied in the actual world. Limits and derivatives play an important role not just in high-level mathematics but also in subjects such as physics and engineering. They can be used to calculate temperature variations, electric and magnetic fields, the rate of change of a quantity and other things.

5. What is the objective of studying limits and derivatives?

Learning real-world applications of principles in Mathematics aids in understanding the chapter, and as a result, scores well in exams. Limits and derivatives are significant not only in advanced mathematics but also in other fields such as physics and engineering. They can be used to calculate electric and magnetic fields, determine the rate of change of a quantity, and monitor temperature variations, among other things.

6. What is the Algebra of the Derivative of Functions?

Because the concept of derivatives directly involves limitations, we believe that the regulations for derivatives will closely follow those for limits. These are compiled in the following theorem.

5th Theorem:

Let f and g be two functions whose derivatives are specified in the same domain. Then,

The derivative of the sum of two functions is the sum of the functions’ derivatives.

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

The derivative of the 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) .

The following product rule gives the derivative of the product of two functions.

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

The following quotient rule gives the derivative of the quotient of two functions (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 .

CBSE Class 11 Maths Revision Notes Chapter 13

Class 11 Mathematics Revision Notes for Limits and Derivatives of Chapter 13

Limits and Derivatives Class 11 Notes – Chapter Overview

Class 11 Revision Notes Limits and Derivatives

Class 11 Mathematics Revision Notes Chapter 13

Standard Limits – Listed Below Are Some Standard Limits

Algebra of Limits:

Polynomial Function:

Rational Function:

Derivatives:

Some Standard Derivatives – Mentioned Below Are Some Standard Derivatives

Please register to view this section

FAQs (Frequently Asked Questions)

1. What are some of the real-world applications of the concept of derivatives?

2. What is the sandwich theorem of limits?

3. How can one ace Chapter 13 “Limits and Derivatives”?

4. Why should one learn about limits and derivatives?

5. What is the objective of studying limits and derivatives?

6. What is the Algebra of the Derivative of Functions?

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions

CBSE Class 11 Maths Revision Notes Chapter 13

Class 11 Mathematics Revision Notes for Limits and Derivatives of Chapter 13

Limits and Derivatives Class 11 Notes – Chapter Overview

Class 11 Revision Notes Limits and Derivatives

Class 11 Mathematics Revision Notes Chapter 13

Standard Limits – Listed Below Are Some Standard Limits

Algebra of Limits:

Polynomial Function:

Rational Function:

Derivatives:

Some Standard Derivatives – Mentioned Below Are Some Standard Derivatives

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. What are some of the real-world applications of the concept of derivatives?

2. What is the sandwich theorem of limits?

3. How can one ace Chapter 13 “Limits and Derivatives”?

4. Why should one learn about limits and derivatives?

5. What is the objective of studying limits and derivatives?

6. What is the Algebra of the Derivative of Functions?

Share

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions