Chapter 12 Limits and Derivatives

NCERT Solutions Class 11 Maths Chapter 12 – Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 12 (Limits and Derivatives) introduce students to the world of Calculus, the branch of mathematics that deals with change. While algebra and geometry often focus on static objects, Calculus allows us to study quantities as they approach a specific value (Limits) and how they change at any given instant (Derivatives). This is arguably the most important chapter for anyone pursuing Physics, Engineering, or Economics.

NCERT Solutions for Class 11 Maths Chapter 12 by ALLEN are developed by expert faculty to bridge the gap between intuitive understanding and mathematical rigour. The solutions deal with the basic definitions of limits, the idea of the left-hand and right-hand limits and the passage to derivatives by the First Principle.

Calculus accounts for around 35-40% of the mathematics syllabus in competitive exams hence, it is a must to master Limits and Derivatives for JEE Main and JEE Advanced. These solutions are a good base for class 12 topics such as Continuity , Differentiability and Integration.

1.0Class 11 Maths Chapter 12 : Key Concepts

This chapter focuses on understanding the behavior of functions near a point and the rate of change of those functions. Key lessons include:

Intuitive Idea of Limits: Understanding what happens to f(x) as x gets arbitrarily close to a number a.
Algebra of Limits: Learning the rules for the limit of the sum, difference, product, and quotient of two functions.
Limits of Polynomial and Rational Functions: Techniques for solving 0/0 forms using factorization and rationalization.
Limits of Trigonometric Functions: Mastering the fundamental limit: $lim_{x \to 0} \frac{s i n x}{x} = 1$
Derivatives: The measure of how a function changes as its input changes.
Derivative at a Point: Defined as the slope of the tangent to the curve at that point.
First Principle of Derivatives: The formal definition used to find the derivative of any function: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$
Algebra of Derivatives:
- Sum/Difference Rule: $(u \pm v)^{'} = u^{'} \pm v^{'}$
- Product Rule (Leibniz Rule): (uv)' = u'v + uv'
- Quotient Rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 12

EXERCISE - 12.1 Evaluate the following limits in Q. 1 to 22.

$lim_{x \to 3} x + 3$ Sol. $lim_{x \to 3} x + 3 = 3 + 3 = 6$
$lim_{x \to π} (x - \frac{22}{7})$ Sol. $lim_{x \to π} (x - \frac{22}{7}) = (π - \frac{22}{7})$
$lim_{r \to 1} π r^{2}$ Sol. $lim_{r \to 1} π r^{2} = π (1)^{2} = π$
$lim_{x \to 4} \frac{4 x + 3}{x - 2}$ Sol. $lim_{x \to 4} \frac{4 x + 3}{x - 2} = \frac{4 ( 4 ) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2}$
$lim_{x \to - 1} \frac{x ^{10} + x ^{5} + 1}{x - 1}$ Sol. $lim_{x \to - 1} \frac{x ^{10} + x ^{5} + 1}{x - 1}$

$= \frac{( - 1 ) ^{10} + ( - 1 ) ^{5} + 1}{- 1 - 1} = \frac{1 - 1 + 1}{- 2} = - \frac{1}{2}$

$lim_{x \to 0} \frac{( x + 1 ) ^{5} - 1}{x}$ Sol. $lim_{x \to 0} \frac{( x + 1 ) ^{5} - 1}{x}$ Put $x + 1 = y$ so that $y \to 1$ as $x \to 0$ Accordingly,

$x \to 0 lim \frac{( x + 1 ) ^{5} - 1}{x} = y \to 1 lim \frac{y ^{5} - 1}{y - 1} = y \to 1 lim \frac{y ^{5} - 1 ^{5}}{y - 1} = 5. 1^{5 - 1} [x \to a lim \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}] = 5$

$∴ lim_{x \to 0} \frac{( x + 5 ) ^{5} - 1}{x} = 5$

$lim_{x \to 2} \frac{3 x ^{2} - x - 10}{x ^{2} - 4}$ Sol. At $x = 2$ , the value of the given rational function takes the form. $\frac{0}{0}$

$∴ x \to 2 lim \frac{3 x ^{2} - x - 10}{x ^{2} - 4} = x \to 2 lim \frac{( x - 2 ) ( 3 x + 5 )}{( x - 2 ) ( x + 2 )} = x \to 2 lim \frac{3 x + 5}{x + 2} = \frac{3 ( 2 ) + 5}{2 + 2} = \frac{11}{4}$

$lim_{x \to 3} \frac{x ^{4} - 81}{2 x ^{2} - 5 x - 3}$ Sol. At $x = 3$ , the value of the given rational function takes the form. $\frac{0}{0}$ $∴ lim_{x \to 3} \frac{x ^{4} - 81}{2 x ^{2} - 5 x - 3}$

$= lim_{x \to 3} \frac{( x - 3 ) ( x + 3 ) ( x ^{2} + 9 )}{( x - 3 ) ( 2 x + 1 )}$

$= x \to 3 lim \frac{( x + 3 ) ( x ^{2} + 9 )}{2 x + 1} = \frac{( 3 + 3 ) ( 3 ^{2} + 9 )}{2 ( 3 ) + 1} = \frac{6 \times 18}{7} = \frac{108}{7}$

$lim_{x \to 0} \frac{ax + b}{cx + 1}$ Sol. $lim_{x \to 0} \frac{ax + b}{cx + 1} = \frac{a ( 0 ) + b}{c ( 0 ) + 1} = b$
$lim_{z \to 1} \frac{z ^{\frac{1}{3}} - 1}{z ^{\frac{1}{6}} - 1}$ Sol. $lim_{z \to 1} \frac{z ^{\frac{1}{3}} - 1}{z ^{\frac{1}{6}} - 1}$

At $z = 1$ , the value of the given function takes the form. $\frac{0}{0}$ Put $z^{\frac{1}{6}} = x$ so that $z \to 1$ as $x \to 1$

Accordingly,

$z \to 1 lim \frac{z ^{\frac{1}{3}} - 1}{z ^{\frac{1}{6}} - 1} = x \to 1 lim \frac{x ^{2} - 1}{x - 1} = x \to 1 lim \frac{x ^{2} - 1 ^{2}}{x - 1} = 2. 1^{2 - 1} [x \to a lim \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}] = 2$

$∴ lim_{z \to 1} \frac{z ^{\frac{1}{3}} - 1}{z ^{\frac{1}{6}} - 1} = 2$

$lim_{x \to 1} \frac{ax ^{2} + bx + c}{cx ^{2} + bx + a}, a + b + c \neq = 0$ Sol. $lim_{x \to 1} \frac{ax ^{2} + bx + c}{cx ^{2} + bx + a} = \frac{a ( 1 ) ^{2} + b ( 1 ) + c}{c ( 1 ) ^{2} + b ( 1 ) + a}$

$= \frac{a + b + c}{a + b + c} = 1 [a + b + c \neq = 0]$

$lim_{x \to 0} \frac{s i n ax}{bx}$ Sol. $lim_{x \to 0} \frac{s i n ax}{bx}$ At $x = 0$ , the value of the given function takes the form $\frac{0}{0}$ .

Now, $lim_{x \to 0} \frac{s i n ax}{bx} = lim_{x \to 0} \frac{s i n ax}{ax} \times \frac{ax}{bx}$

$= x \to 0 lim (\frac{sin a x}{a x}) \times (\frac{a}{b}) = \frac{a}{b} a x \to 0 lim (\frac{sin a x}{a x}) [x \to 0 \Rightarrow a x \to 0] = \frac{a}{b} \times 1 [y \to 0 lim \frac{sin y}{y} = 1] = \frac{a}{b}$

$lim_{x \to 0} \frac{s i n a x}{s i n b x}, a, b \neq = 0$ Sol. $lim_{x \to 0} \frac{s i n ax}{s i n bx}, a, b \neq = 0$ At $x = 0$ , the value of the given function takes the form. $\frac{0}{0}$ Now, $lim_{x \to 0} \frac{s i n ax}{s i n bx}$

$= x \to 0 lim \frac{( \frac{s i n a x}{a x} ) \times a x}{( \frac{s i n b x}{b x} ) \times b x} = (\frac{a}{b}) \times \frac{lim _{a x \to 0} ( \frac{s i n a x}{a x} )}{lim _{b x \to 0} ( \frac{s i n b x}{b x} )} [x \to 0 \Rightarrow a x \to 0 and x \to 0 \Rightarrow b x \to 0] = (\frac{a}{b}) \times \frac{1}{1} [y \to 0 lim \frac{sin y}{y} = 1] = \frac{a}{b}$

$lim_{x \to π} \frac{s i n ( π - x )}{π ( π - x )}$ Sol. $lim_{x \to π} \frac{s i n ( π - x )}{π ( π - x )}$ It is seen that $x \to π \Rightarrow (π - x) \to 0$

$∴ lim_{x \to π} \frac{s i n ( π - x )}{π ( π - x )} = \frac{1}{π} lim_{(π \to x) \to 0} \frac{s i n ( π - x )}{( π - x )}$

$= \frac{1}{π} \times 1 [y \to 0 lim \frac{sin y}{y} = 1] = \frac{1}{π}$

$lim_{x \to 0} \frac{c o s x}{π - x}$ Sol. $lim_{x \to 0} \frac{c o s x}{π - x} = \frac{c o s 0}{π - 0} = \frac{1}{π}$

$lim_{x \to 0} \frac{c o s 2 x - 1}{c o s x - 1}$ Sol. $lim_{x \to 0} \frac{c o s 2 x - 1}{c o s x - 1}$ At $x = 0$ , the value of the given function takes the form. $\frac{0}{0}$ Now, $lim_{x \to 0} \frac{c o s 2 x - 1}{c o s x - 1}$

$= lim_{x \to 0} \frac{1 - 2 s i n ^{2} x - 1}{1 - 2 s i n ^{2} \frac{x}{2} - 1} [cos x = 1 - 2 sin^{2} \frac{x}{2}]$

$= lim_{x \to 0} \frac{s i n ^{2} x}{s i n ^{2} \frac{x}{2}}$

$[x \to 0 \Rightarrow \frac{x}{2} \to 0]$

$[lim_{y \to 0} \frac{s i n y}{y} = 1]$

$= 4$

$lim_{x \to 0} \frac{ax + x c o s x}{b s i n x}$ Sol. $lim_{x \to 0} \frac{ax + x c o s x}{b s i n x}$ At $x = 0$ , the value of the given function takes the form. $\frac{0}{0}$

$x \to 0 lim \frac{a x + x cos x}{b sin x} = \frac{1}{b} x \to 0 lim \frac{x ( a + cos x )}{sin x} = \frac{1}{b} \times \frac{1}{( lim _{x \to 0} \frac{s i n x}{x} )} \times x \to 0 lim (a + cos x) = \frac{1}{b} \times (a + cos 0) [x \to 0 lim \frac{sin x}{x} = 1] = \frac{a + 1}{b}$

$lim_{x \to 0} x sec x$ Sol. $lim_{x \to 0} x sec x = lim_{x \to 0} \frac{x}{c o s x} = \frac{0}{c o s x} = \frac{0}{1} = 0$

$lim_{x \to 0} \frac{s i n ax + bx}{ax + s i n bx}, a, b, a + b \neq = 0$ Sol. At $x = 0$ , the value of the given function takes the form. $\frac{0}{0}$ Now, $lim_{x \to 0} \frac{s i n ax + bx}{ax + s i n bx}$

$= lim_{x \to 0} \frac{( \frac{s i n a x}{a x} ) a x + b x}{a x + b x ( \frac{s i n b x}{b x} )}$

$= \frac{lim _{x \to 0} ( a x ) + lim _{x \to 0} b x}{lim _{x \to 0} a x + lim _{x \to 0} b x} [x \to 0 lim \frac{sin x}{x} = 1] = \frac{lim _{x \to 0} ( a x + b x )}{lim _{x \to 0} ( a x + b x )} = x \to 0 lim \frac{( a x + b x )}{( a x + b x )} = x \to 0 lim (1) = 1$

$lim_{x \to 0} (cosec x - cot x)$ Sol. At $x = 0$ , the value of the given function takes the form. $\infty - \infty$ Now, $lim_{x \to 0} (cosec x - cot x)$

$= x \to 0 lim (\frac{1}{sin x} - \frac{cos x}{sin x}) = x \to 0 lim (\frac{1 - cos x}{sin x}) = x \to 0 lim \frac{( \frac{1 - c o s x}{x} )}{( \frac{s i n x}{x} )} = \frac{lim _{x \to 0} \frac{1 - c o s x}{x}}{lim _{x \to 0} \frac{s i n x}{x}} = x \to 0 lim \frac{( a x + b x )}{( a x + b x )} = \frac{0}{1} [x \to 0 lim \frac{1 - cos x}{x} = 0 and x \to 0 lim \frac{sin x}{x} = 1] = 0$

$lim_{x \to \frac{π}{2}} \frac{t a n 2 x}{x - \frac{π}{2}}$ Sol. $lim_{x \to \frac{π}{2}} \frac{t a n 2 x}{x - \frac{π}{2}}$ At $x = \frac{π}{2}$ , the value of the given function takes the form. $\frac{0}{0}$ Now, but $x - \frac{π}{2} = y$ so that $x \to \frac{π}{2}, y \to 0$

$∴ x \to \frac{π}{2} lim \frac{tan 2 x}{x - \frac{π}{2}} = y \to 0 lim \frac{tan 2 \times ( y + \frac{π}{2} )}{y} = y \to 0 lim \frac{tan ( π + 2 y )}{y} = y \to 0 lim \frac{tan 2 y}{y} [tan (π + 2 y) = tan 2 y] = y \to 0 lim \frac{sin 2 y}{y cos 2 y} = y \to 0 lim (\frac{sin 2 y}{2 y} \times \frac{2}{cos 2 y}) = (2 y \to 0 lim \frac{sin 2 y}{2 y}) \times 2 y \to 0 lim (\frac{2}{cos 2 y}) = 1 \times \frac{2}{cos 0} [y \to 0 \Rightarrow 2 y \to 0] = 1 \times \frac{2}{1} = 2$

Find $lim_{x \to 0} f (x)$ and $lim_{x \to 1} f (x)$ where

$f (x) = {2 x + 3, 3 (x + 1), x \leq 0 x > 0$

Sol. The given function is $f (x) = {2 x + 3, 3 (x + 1), x \leq 0 x > 0$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} [2 x + 3] = 2 (0) + 3 = 3$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 3 (x + 1) = 3 (0 + 1) = 3$ $∴ lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) = lim_{x \to 0} f (x) = 3$

$x \to 1^{-} lim f (x) = x \to 1 lim 3 (x + 1) = 3 (1 + 1) = 6 x \to 1^{+} lim f (x) = x \to 1 lim 3 (x + 1) = 3 (1 + 1) = 6$

$∴ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x) = lim_{x \to 1} f (x) = 6$

Find $lim_{x \to 1} f (x)$ where $f (x) = {x^{2} - 1, - x^{2} - 1, x \leq 1 x > 1$ Sol. The given function is

$f (x) = {x^{2} - 1, - x^{2} - 1, x \leq 1 x > 1 x \to 1^{-} lim f (x) = x \to 1 lim [x^{2} - 1] = 1^{2} - 1 = 1 - 1 = 0 x \to 1^{+} lim f (x) = x \to 1 lim [- x^{2} - 1] = - (1^{2}) - 1 = - 1 - 1 = - 2$

It is observed that $lim_{x \to 1^{-}} f (x) \neq = lim_{x \to 1^{+}} f (x)$ Hence, $lim_{x \to 1} f (x)$ does not exist.

Evaluate $lim_{x \to 0} f (x)$ , where $f (x) = {\frac{∣ x ∣}{x}, 0, x \neq = 0 x = 0$ Sol. The given function is $f (x) = {\frac{∣ x ∣}{x}, 0, x \neq = 0 x = 0$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} [\frac{∣ x ∣}{x}] = lim_{x \to 0} (\frac{- x}{x})$

[When x is negative, $∣ x ∣ = - x$ ]

$= lim_{x \to 0} (- 1) = - 1$

$x \to 0^{+} lim f (x) = x \to 0^{+} lim [\frac{∣ x ∣}{x}] = x \to 0 lim [\frac{x}{x}] [When x is positive, ∣ x ∣ = x] = x \to 0 lim (1) = 1$

It is observed that $lim_{x \to 0^{-}} f (x) \neq = lim_{x \to 0^{+}} f (x)$ Hence, $lim_{x \to 0} f (x)$ does not exist.

Evaluate $lim_{x \to 0} f (x)$ , where $f (x) = {\frac{x}{∣ x ∣}, 0, x \neq = 0 x = 0$ Sol. The given function is $f (x) = {\frac{x}{∣ x ∣}, 0, x \neq = 0 x = 0$

$x \to 0^{-} lim f (x) = x \to 0^{-} lim [\frac{x}{∣ x ∣}] = x \to 0 lim (- 1) = x \to 0 lim [\frac{x}{- x}] [When x < 0, ∣ x ∣ = - x] = - 1 x \to 0^{+} lim f (x) = x \to 0^{+} lim [\frac{x}{∣ x ∣}] = x \to 0 lim [\frac{x}{x}] [When x > 0, ∣ x ∣ = x] = x \to 0 lim (1) = 1$

It is observed that $lim_{x \to 0^{-}} f (x) \neq = lim_{x \to 0^{+}} f (x)$ Hence, $lim_{x \to 0} f (x)$ does not exist.

Evaluate $lim_{x \to 5} f (x)$ , where $f (x) = ∣ x ∣ - 5$ Sol. The given function is $f (x) = ∣ x ∣ - 5$

$lim_{x \to 5^{-}} f (x) = lim_{x \to 5^{-}} [∣ x ∣ - 5] = lim_{x \to 5} (x - 5)$

[When $x > 0, ∣ x ∣ = x$ ]

$x \to 5^{+} lim f (x) = 5 - 5 = 0 = x \to 5^{+} lim (∣ x ∣ - 5) = x \to 5 lim (x - 5)$

[When $x > 0, ∣ x ∣ = x$ ]

$= 5 - 5 = 0$

$∴ lim_{x \to 5^{-}} f (x) = lim_{x \to 5^{+}} f (x) = 0$ Hence, $lim_{x \to 5} f (x) = 0$

Suppose $f (x) = ⎩ ⎨ ⎧ a + bx, 4, b - ax, if x < 1 if x = 1 if x > 1$ and $lim_{x \to 1} f (x) = f (1)$ what are possible values of a and b? Sol. The given function is $f (x) = ⎩ ⎨ ⎧ a + bx, 4, b - ax, if x < 1 if x = 1 if x > 1$

$x \to 1^{-} lim f (x) = x \to 1 lim (a + b x) = a + b x \to 1^{+} lim f (x) = x \to 1 lim (b - a x) = b - a f (1) = 4$

It is given that $lim_{x \to 1} f (x) = f (1)$ $∴ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x) = lim_{x \to 1} f (x) = f (1)$ $\Rightarrow a + b = 4$ and $b - a = 4$

On solving these two equations, we obtain $a = 0$ and $b = 4$ . Thus, the respective possible values of a and b are 0 and 4.

Let $a_{1}, a_{2} \dots ., a_{n}$ be fixed real number and define a function

$f (x) = (x - a_{1}) (x - a_{2}) \dots\dots (x - a_{n})$

What is $lim_{x \to a_{1}} f (x)$ ? For some $a \neq = a_{1}, a_{2} \dots a_{n}$ compute $lim_{x \to a} f (x)$ Sol. The given function

$f (x) = (x - a_{1}) (x - a_{2}) \dots\dots (x - a_{n}) x \to a_{1} lim f (x) = x \to a_{1} lim [(x - a_{1}) (x - a_{2}) \dots .. (x - a_{n})] x \to a_{1} lim (x - a_{1})] [x \to a_{1} lim (x - a_{2})] \dots .. [x \to a_{1} lim (x - a_{n})] a_{1} - a_{1}) (a_{1} - a_{2}) \dots .. (a_{1} - a_{n}) = 0$

$∴ lim_{x \to a_{1}} f (x) = 0$

Now,

$= = ∴ x \to a lim f (x) = x \to a lim [(x - a_{1}) (x - a_{2}) \dots\dots (x - a_{n})] [x \to a lim (x - a_{1})] [x \to a lim (x - a_{2})] \dots\dots [x \to a lim (x - a_{n})] (a - a_{1}) (a - a_{2}) \dots .. (a - a_{n}) x \to a lim f (x) = (a - a_{1}) (a - a_{2}) \dots .. (a - a_{n})$

If $f (x) = ⎩ ⎨ ⎧ ∣ x ∣ + 1, 0, ∣ x ∣ - 1, x < 0 x = 0 x > 0$ . For what value(s) of a does $lim_{x \to a} f (x)$ exists? Sol. The given function is $f (x) = ⎩ ⎨ ⎧ ∣ x ∣ + 1, 0, ∣ x ∣ - 1, x < 0 x = 0 x > 0$ When $a = 0$ ,

$x \to 0^{-} lim f (x) = x \to 0^{-} lim (∣ x ∣ + 1) = x \to 0 lim (- x + 1) [If x < 0, ∣ x ∣ = - x] = - 0 + 1 = 1$

$x \to 0^{+} lim f (x) = x \to 0^{+} lim (∣ x ∣ - 1) = x \to 0 lim (x - 1) [If x > 0, ∣ x ∣ = x] = 0 + 1 = - 1$

Here, it is observed that $lim_{x \to 0^{-}} f (x) \neq = lim_{x \to 0^{+}} f (x)$ $∴ lim_{x \to 0} f (x)$ does not exist When a < 0

$x \to a^{-} lim f (x) = x \to a^{-} lim (∣ x ∣ + 1) = x \to a lim (- x + 1) [x < a < 0 \Rightarrow ∣ x ∣ = - x] = - a + 1$

$x \to a^{+} lim f (x) = x \to a^{+} lim (∣ x ∣ + 1) = x \to a lim (- x + 1) [a < x < 0 \Rightarrow ∣ x ∣ = - x] = - a + 1$

$∴ lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = - a + 1$

Thus, limit of $f (x)$ exists at $x = a$ , where $a < 0$ . When $a > 0$

$= = lim_{x \to a^{-}} f (x) = lim_{x \to a^{-}} (∣ x ∣ - 1) lim_{x \to a} (x - 1) a - 1$

$x \to a^{+} lim f (x) = x \to a^{+} lim (∣ x ∣ - 1) = x \to a lim (x - 1) [0 < a < x \Rightarrow ∣ x ∣ = x] = a - 1$

Thus, limit of $f (x)$ exists at $x = a$ , where $a > 0$ . Thus, $lim_{x \to a} f (x)$ exists for all $a \neq = 0$ .

If the function $f (x)$ satisfies, $lim_{x \to 1} \frac{f ( x ) - 2}{x ^{2} - 1} = π$ evaluate $lim_{x \to 1} f (x)$ Sol. $lim_{x \to 1} \frac{f ( x ) - 2}{x ^{2} - 1} = π$ $\Rightarrow \frac{l i m _{x \to 1} ( f ( x ) - 2 )}{l i m _{x \to 1} ( x ^{2} - 1 )} = π$ $\Rightarrow lim_{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$ $\Rightarrow lim_{x \to 1} (f (x) - 2) = π (1^{2} - 1)$ $\Rightarrow lim_{x \to 1} (f (x) - 2) = 0$ $\Rightarrow lim_{x \to 1} f (x) - lim_{x \to 1} 2 = 0$ $\Rightarrow lim_{x \to 1} f (x) - 2 = 0$ $∴ lim_{x \to 1} f (x) = 2$
If $f (x) = ⎩ ⎨ ⎧ mx^{2} + n, nx + m, nx^{3} + m, x < 0 0 \leq x \leq 1. x > 1$ For what integers m and n does $lim_{x \to 0} f (x)$ and $lim_{x \to 1} f (x)$ exist? Sol. The given function is

$f (x) = ⎩ ⎨ ⎧ mx^{2} + n, nx + m, nx^{3} + m, x < 0 0 \leq x \leq 1 x > 1$

$x \to 0^{-} lim f (x) x \to 0^{+} lim f (x) = x \to 0 lim (m x^{2} + n) = m (0)^{2} + n = n = x \to 0 lim (n x + m) = n (0) + m = m$

Thus, $lim_{x \to 0} f (x)$ exists if $m = n$ .

$x \to 1^{-} lim f (x) x \to 1^{+} lim f (x) = x \to 1 lim (n x + m) = n (1)^{3} + m = m + n = x \to 1 lim (n x^{3} + m) = n (1)^{3} + m = m + n$

$∴ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x) = lim_{x \to 1} f (x)$ Thus, $lim_{x \to 1} f (x)$ exist for any integral value of m and n

EXERCISE - 12.2

Find the derivative of $x^{2} - 2$ at $x = 10$ . Sol. Let $f (x) = x^{2} - 2$ . Accordingly,

$f^{'} (10) = h \to 0 lim \frac{f ( 10 + h ) - f ( 10 )}{h} = h \to 0 lim \frac{[ ( 10 + h ) ^{2} - 2 ] - ( 1 0 ^{2} - 2 )}{h} = h \to 0 lim \frac{1 0 ^{2} + 2 \cdot 10 h + h ^{2} - 2 - 1 0 ^{2} + 2}{h} = h \to 0 lim \frac{20 h + h ^{2}}{h} = h \to 0 lim (20 + h) = (20 + 0) = 20$

Thus, the derivative of $x^{2} - 2$ at $x = 10$ is 20.

Find the derivative of 99 x at $x = 100$ . Sol. Let $f (x) = 99 x$ . Accordingly,

$f^{'} (100) = h \to 0 lim \frac{f ( 100 + h ) - f ( 100 )}{h} = h \to 0 lim \frac{99 ( 100 + h ) - 99 ( 100 )}{h} = h \to 0 lim \frac{99 \times 100 + 99 h - 99 \times 100}{h} = h \to 0 lim \frac{99 h}{h} = h \to 0 lim (99) = 99$

Thus, the derivative of 99 x at $x = 100$ is 99 .

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

Sol. Let $f (x) = x$ . Accordingly,

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

Thus, the derivative of x at $x = 1$ is 1 .

Find the derivative of the following functions from first principle. (i) $x^{3} - 27$ (ii) $(x - 1) (x - 2)$ (iii) $\frac{1}{x ^{2}}$ (iv) $\frac{x + 1}{x - 1}$

Sol. (i) Let $f (x) = x^{3} - 27$ . Accordingly, from the first principle,

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

(ii) Let $f (x) = (x - 1) (x - 2)$ . Accordingly, from the first principle,

$= = = = = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} h \to 0 lim \frac{( x + h - 1 ) ( x + h - 2 ) - ( x - 1 ) ( x - 2 )}{h} h \to 0 lim \frac{( x ^{2} + h x - 2 x + h x + h ^{2} - 2 h - x - h + 2 ) - ( x ^{2} - 2 x - x + 2 )}{h} h \to 0 lim \frac{( h x + h x + h ^{2} - 2 h - h )}{h} h \to 0 lim \frac{2 h x + h ^{2} - 3 h}{h} h \to 0 lim (2 x + h - 3) = (2 x + 0 - 3) = 2 x - 3$

(iii) Let $f (x) = \frac{1}{x ^{2}}$ Accordingly, from the first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{\frac{1}{( x + h ) ^{2}} - \frac{1}{x ^{2}}}{h} = h \to 0 lim \frac{1}{h} [\frac{x ^{2} - ( x + h ) ^{2}}{x ^{2} ( x + h ) ^{2}}] = h \to 0 lim \frac{1}{h} [\frac{x ^{2} - x ^{2} - h ^{2} - 2 h x}{x ^{2} ( x + h ) ^{2}}] = h \to 0 lim \frac{1}{h} [\frac{- h ^{2} - 2 h x}{x ^{2} ( x + h ) ^{2}}] = h \to 0 lim [\frac{- h - 2 x}{x ^{2} ( x + h ) ^{2}}] = \frac{0 - 2 x}{x ^{2} ( x + 0 ) ^{2}} = \frac{- 2}{x ^{3}}$

(iv) Let $f (x) = \frac{x + 1}{x - 1}$ . Accordingly, from the first principle,

$= = = = = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} h \to 0 lim \frac{( \frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1} )}{h} h \to 0 lim \frac{1}{h} [\frac{( x - 1 ) ( x + h + 1 ) - ( x + 1 ) ( x + h - 1 )}{( x - 1 ) ( x + h - 1 )}] h \to 0 lim \frac{1}{h} [\frac{( x ^{2} + h x + x - x - h - 1 ) - ( x ^{2} + h x - x + x + h - 1 )}{( x - 1 ) ( x + h - 1 )}] h \to 0 lim \frac{1}{h} [\frac{- 2 h}{( x - 1 ) ( x + h - 1 )}] h \to 0 lim [\frac{- 2}{( x - 1 ) ( x + h - 1 )}] = \frac{- 2}{( x - 1 ) ( x - 1 )} = \frac{- 2}{( x - 1 ) ^{2}}$

For the function

$f (x) = \frac{x ^{100}}{100} + \frac{x ^{99}}{99} + \dots .. + \frac{x ^{2}}{2} + x + 1$

Prove that $f^{'} (1) = 100 f^{'} (0)$

Sol. The given function is

$f (x) = \frac{x ^{100}}{100} + \frac{x ^{99}}{99} + \dots .. + \frac{x ^{2}}{2} + x + 1 \frac{d}{d x} f (x) = \frac{d}{d x} [\frac{x ^{100}}{100} + \frac{x ^{99}}{99} + \dots .. + \frac{x ^{2}}{2} + x + 1] \frac{d}{d x} f (x) = \frac{d}{d x} (\frac{x ^{100}}{100}) + \frac{d}{d x} (\frac{x ^{99}}{99}) + \dots .. + \frac{d}{d x} (\frac{x ^{2}}{2}) + \frac{d}{d x} (x) + \frac{d}{d x} (1)$

On using theorem $\frac{d}{dx} (x^{n}) = nx^{n - 1}$ , we obtain

$\frac{d}{dx} f (x) = \frac{100 x ^{99}}{100} + \frac{99 x ^{98}}{99} + \dots .. + \frac{2 x}{2} + 1 + 0 = x^{99} + x^{98} + \dots .. + x + 1$

At $x = 0, f^{'} (0) = 1$ At $x = 1$

$f^{'} (1) = 1^{99} + 1^{98} + \dots .. + 1 + 1 = [1 + 1 + \dots .. + 1 + 1]_{100 terms} = 1 \times 100 = 100$

Thus, $f^{'} (1) = 100 \times f^{'} (0)$ . Hence, proved

Find the derivative of

$x^{n} + a x^{n - 1} + a^{2} x^{n - 2} + \dots .. + a^{n - 1} x + a^{n}$

for some fixed real number a. Sol. Let $f (x) = x^{n} + ax^{n - 1} + a^{2} x^{n - 2} + \dots .. + a^{n - 1} x + a^{n}$

$∴ f^{'} (x) = \frac{d}{d x} (x^{n} + a x^{n - 1} = \frac{d}{d x} (x^{n}) + a \frac{d}{d x} (x^{n - 1}) + \dots .. + a^{2} x^{n - 2} + \dots .. + a^{n - 1} x + a^{n}) + a^{2} \frac{d}{d x} (x^{n - 2}) a^{n - 1} \frac{d}{d x} (x) + a^{n} \frac{d}{d x} (1)$

On using $\frac{d}{dx} x^{n} = nx^{n - 1}$ , we obtain

$f^{'} (x) = = n x^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + \dots\dots + a^{n - 1} + a^{n} (0) n x^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + \dots\dots + a^{n - 1}$

For some constants a and b, find the derivative of (i) $(x - a) (x - b)$ (ii) $(ax^{2} + b)^{2}$ (iii) $\frac{x - a}{x - b}$

Sol.

(i) Let $f (x) = (x - a) (x - b)$ $\Rightarrow f (x) = x^{2} - (a + b) x + ab$ $∴ f^{'} (x) = \frac{d}{dx} (x^{2} - (a + b) x + ab)$

$= \frac{d}{dx} (x^{2}) - (a - b) \frac{d}{dx} (x) + \frac{d}{dx} (ab)$

On using theorem $\frac{d}{dx} (x^{n}) = nx^{n - 1}$ , we obtain

$f^{'} (x) = 2 x - (a + b) + 0 = 2 x - a - b$

(ii) Let $f (x) = (ax^{2} + b)^{2}$ $\Rightarrow f (x) = a^{2} x^{4} + 2 ab x^{2} + b^{2}$ $∴ f^{'} (x) = \frac{d}{dx} (a^{2} x^{4} + 2 abx^{2} + b^{2})$

$= a^{2} \frac{d}{d x} (x^{4}) + 2 ab \frac{d}{d x} (x^{2}) + \frac{d}{d x} (b^{2})$

On using theorem $\frac{d}{dx} x^{n} = nx^{n - 1}$ , we obtain

$f^{'} (x) = a^{2} (4 x^{3}) + 2 ab (2 x) + b^{2} (0) = 4 a^{2} x^{3} + 4 ab x = 4 a x (a x^{2} + b)$

(iii) Let $f (x) = \frac{x - a}{x - b} \Rightarrow f^{'} (x) = \frac{d}{dx} (\frac{x - a}{x - b})$ By quotient rule,

$f^{'} (x) = \frac{( x - b ) \frac{d}{d x} ( x - a ) - ( x - a ) \frac{d}{d x} ( x - b )}{( x - b ) ^{2}} = \frac{( x - b ) ( 1 ) - ( x - a ) ( 1 )}{( x - b ) ^{2}} = \frac{x - b - x + a}{( x - b ) ^{2}} = \frac{a - b}{( x - b ) ^{2}}$

Find the derivative of $\frac{x ^{n} - a ^{n}}{x - a}$ for some constant a. Sol. Let $f (x) = \frac{x ^{n} - a ^{n}}{x - a}$

$\Rightarrow f^{'} (x) = \frac{d}{d x} (\frac{x ^{n} - a ^{n}}{x - a})$

By quotient rule

$f^{'} (x) = \frac{( x - a ) \frac{d}{d x} ( x ^{n} - a ^{n} ) - ( x ^{n} - a ^{n} ) \frac{d}{d x} ( x - a )}{( x - a ) ^{2}} = \frac{( x - a ) ( n x ^{n - 1} - 0 ) - ( x ^{n} - a ^{n} )}{( x - a ) ^{2}} = \frac{n x ^{n} - an x ^{n - 1} - x ^{n} + a ^{n}}{( x - a ) ^{2}}$

Find the derivative of (i) $2 x - \frac{3}{4}$ (ii) $(5 x^{3} + 3 x - 1) (x - 1)$ (iii) $x^{- 3} (5 + 3 x)$ (iv) $x^{5} (3 - 6 x^{- 9})$ (v) $x^{- 4} (3 - 4 x^{- 5})$ (vi) $\frac{2}{x + 1} - \frac{x ^{2}}{3 x - 1}$

Sol.

(i) Let $f (x) = 2 x - \frac{3}{4}$

$f^{'} (x) = \frac{d}{dx} (2 x - \frac{3}{4}) = 2 \frac{d}{dx} (x) - \frac{d}{dx} (\frac{3}{4}) = 2 - 0 = 2$

(ii) Let $f (x) = (5 x^{3} + 3 x - 1) (x - 1)$ By Leibnitz product rule,

$f^{'} (x) = (5 x^{3} + 3 x - 1) \frac{d}{d x} (x - 1) + (x - 1) \frac{d}{d x} (5 x^{3} + 3 x - 1) = (5 x^{3} + 3 x - 1) (1) + (x - 1) (5.3 x^{2} + 3 - 0) = (5 x^{3} + 3 x - 1) + (x - 1) (15 x^{2} + 3) = 5 x^{3} + 3 x - 1 + 15 x^{3} + 3 x - 15 x^{2} - 3 = 20 x^{3} - 15 x^{2} + 6 x - 4$

(iii) Let $f (x) = x^{- 3} (5 + 3 x)$ By Leibnitz product rule,

$f^{'} (x) = x^{- 3} \frac{d}{d x} (5 + 3 x) + (5 + 3 x) \frac{d}{d x} (x^{- 3}) = x^{- 3} (0 + 3) + (5 + 3 x) (- 3 x^{- 3 - 1}) = x^{- 3} (3) + (5 + 3 x) (- 3 x^{- 4}) = 3 x^{- 3} - 15 x^{- 4} - 9 x^{- 3} = - 6 x^{- 3} - 15 x^{- 4} = - 3 x^{- 3} (2 + \frac{5}{x}) = \frac{- 3 x ^{- 3}}{x} (2 x + 5) = \frac{- 3}{x ^{4}} (5 + 2 x)$

(iv) Let $f (x) = x^{5} (3 - 6 x^{- 9})$ By Leibnitz product rule,

$f^{'} (x) = x^{5} \frac{d}{d x} (3 - 6 x^{- 9}) + (3 - 6 x^{- 9}) \frac{d}{d x} (x^{5}) = x^{5} {0 - 6 (- 9) x^{- 9 - 1}} + (3 - 6 x^{- 9}) (5 x^{4}) = x^{5} (54 x^{- 10}) + 15 x^{4} - 30 x^{- 5} = 54 x^{- 5} + 15 x^{4} - 30 x^{- 5} = 24 x^{- 5} + 15 x^{4} = 15 x^{4} + \frac{24}{x ^{5}}$

(v) Let $f (x) = x^{- 4} (3 - 4 x^{- 5})$ By Leibnitz product rule,

$f^{'} (x) = x^{- 4} \frac{d}{d x} (3 - 4 x^{- 5}) + (3 - 4 x^{- 5}) \frac{d}{d x} (x^{- 4}) = x^{- 4} {0 - 4 (- 5) x^{- 5 - 1}} + (3 - 4 x^{- 5}) (- 4) x^{- 4 - 1} = x^{- 4} (20 x^{- 6}) + (3 - 4 x^{- 5}) (- 4 x^{- 5}) = 20 x^{- 10} - 12 x^{- 5} + 16 x^{- 10} = 36 x^{- 10} - 12 x^{- 5} = - \frac{12}{x ^{5}} + \frac{36}{x ^{10}}$

(vi) Let $f (x) = \frac{2}{x + 1} - \frac{x ^{2}}{3 x - 1}$

$f^{'} (x) = \frac{d}{d x} (\frac{2}{x + 1}) - \frac{d}{d x} (\frac{x ^{2}}{3 x - 1})$

By quotient rule,

$f^{'} (x) = [\frac{( x + 1 ) \frac{d}{d x} ( 2 ) - 2 x \frac{d}{d x} ( x + 1 )}{( x + 1 ) ^{2}}] - [\frac{( 3 x - 1 ) \frac{d}{d x} ( x ^{2} ) - x ^{2} \frac{d}{d x} ( 3 x - 1 )}{( 3 x - 1 ) ^{2}}] = [\frac{( x + 1 ) ( 0 ) - 2 ( 1 )}{( x + 1 ) ^{2}}] - [\frac{( 3 x - 1 ) ( 2 x ) - ( x ^{2} ) ( 3 )}{( 3 x - 1 ) ^{2}}] = \frac{- 2}{( x + 1 ) ^{2}} - [\frac{6 x ^{2} - 2 x - 3 x ^{2}}{( 3 x - 1 ) ^{2}}] = \frac{- 2}{( x + 1 ) ^{2}} - [\frac{3 x ^{2} - 2 x}{( 3 x - 1 ) ^{2}}] = \frac{- 2}{( x + 1 ) ^{2}} - \frac{x ( 3 x - 2 )}{( 3 x - 1 ) ^{2}}$

Find the derivative of $cos x$ from first principle. Sol. Let $f (x) = cos x$ . Accordingly, from the first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{cos ( x + h ) - cos x}{h} = h \to 0 lim [\frac{cos x cos h - sin x sin h - cos x}{h}] = h \to 0 lim [\frac{- cos x ( 1 - cos h ) - sin x sin h}{h}] = h \to 0 lim [\frac{- cos ( 1 - cos h )}{h} - \frac{sin x sin h}{h}] = - cos x (h \to 0 lim \frac{1 - cos h}{h}) - sin x h \to 0 lim (\frac{sin h}{h}) = - cos x (0) - sin x (1) = - sin x$

$∴ f^{'} (x) = - sin x$

Find the derivative of the following functions: (i) $sin x cos x$ (ii) $sec x$ (iii) $5 sec x + 4 cos x$ (iv) $cosec x$ (v) $3 cot x + 5 cosec x$ (vi) $5 sin x - 6 cos x + 7$ (vii) $2 tan x - 7 sec x$

Sol.

(i) Let $f (x) = sin x cos x$ . Accordingly, from the first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{sin ( x + h ) cos ( x + h ) - sin x cos x}{h} = h \to 0 lim \frac{1}{2 h} [2 sin (x + h) cos (x + h) - 2 sin x cos x] = h \to 0 lim \frac{1}{2 h} [sin 2 (x + h) - sin 2 x] = h \to 0 lim \frac{1}{2 h} [2 cos \frac{2 x + 2 h + 2 x}{2} \cdot sin \frac{2 x + 2 h - 2 x}{2}] = h \to 0 lim \frac{1}{h} [cos \frac{4 x + 2 h}{2} \cdot sin \frac{2 h}{2}] = h \to 0 lim \frac{1}{h} [cos (2 x + h) sin h] = h \to 0 lim cos (2 x + h) \cdot h \to 0 lim \frac{sin h}{h} = cos (2 x + 0) \cdot 1 = cos 2 x$

(ii) Let $f (x) = sec x$ . Accordingly, from the first principle,

$= = = = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} h \to 0 lim \frac{sec ( x + h ) - sec ( x )}{h} h \to 0 lim \frac{1}{h} [\frac{1}{cos ( x + h )} - \frac{1}{cos x}] h \to 0 lim \frac{1}{h} [\frac{cos x - cos ( x + h )}{cos x cos ( x + h )}] \frac{1}{cos x} \cdot h \to 0 lim \frac{1}{h} [\frac{- 2 sin ( \frac{x + x + h}{2} ) sin ( \frac{x - x - h}{2} )}{cos ( x + h )}]$

$= \frac{1}{cos x} \cdot h \to 0 lim \frac{1}{h} [\frac{- 2 sin ( \frac{2 x + h}{2} ) sin ( - \frac{h}{2}}{cos ( x + h )} = \frac{1}{cos x} \cdot h \to 0 lim \frac{[ sin ( \frac{2 x + h}{2} ) \frac{s i n ( \frac{h}{2} )}{( \frac{h}{2} )} ]}{cos ( x + h )} = \frac{1}{cos x} \cdot h \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )} \cdot h \to 0 lim \frac{sin ( \frac{2 x + h}{2} )}{cos ( x + h )} = \frac{1}{cos x} \cdot 1 \cdot \frac{sin x}{cos x} = sec x tan x$

(iii) Let $f (x) = 5 sec x + 4 cos x$ . Accordingly, from the first principle,

$= = = = = = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} h \to 0 lim \frac{5 sec ( x + h ) + 4 cos ( x + h ) - [ 5 sec x + 4 cos x ]}{h} 5 h \to 0 lim \frac{[ sec ( x + h ) - sec x ]}{h} + 4 h \to 0 lim \frac{[ cos ( x + h ) - cos x ]}{h} 5 h \to 0 lim \frac{1}{h} [\frac{1}{cos ( x + h )} - \frac{1}{cos x}] + 4 h \to 0 lim \frac{1}{h} [cos (x + h) - cos x] 5 h \to 0 lim \frac{1}{h} [\frac{cos x - cos ( x + h )}{cos x cos ( x + h )}] + 4 h \to 0 lim \frac{1}{h} [cos x cos h - sin x sin h - cos x] \frac{5}{cos x} h \to 0 lim \frac{1}{h} [\frac{- 2 sin ( \frac{x + x + h}{2} ) sin ( \frac{x - x - h}{2} )}{cos ( x + h )}] + 4 h \to 0 lim \frac{1}{h} [- cos x (1 - cos h) - sin x sin h] \frac{5}{cos x} h \to 0 lim \frac{1}{h} [\frac{- 2 sin ( \frac{2 x + h}{2} ) sin ( - \frac{h}{2} )}{cos ( x + h )}] + 4 [- cos x h \to 0 lim \frac{( 1 - cos h )}{h} - sin x h \to 0 lim \frac{sin h}{h}]$

$= \frac{5}{cos x} h \to 0 lim \frac{sin ( \frac{2 x + h}{2} ) \cdot \frac{s i n ( \frac{h}{2} )}{\frac{h}{2}}}{cos ( x + h )} + 4 [(- cos x) \cdot (0) - (sin x) \cdot 1] = \frac{5}{cos x} [h \to 0 lim \frac{sin ( \frac{2 x + h}{2} )}{cos ( x + h )} \cdot h \to 0 lim \frac{sin ( \frac{h}{2} )}{\frac{h}{2}}] - 4 sin x = \frac{5}{cos x} \cdot \frac{sin x}{cos x} \cdot 1 - 4 sin x = 5 sec x tan x - 4 sin x$

(iv) Let $f (x) = cosec x$ . Accordingly, from the first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} f^{'} (x) = h \to 0 lim \frac{1}{h} [cosec (x + h) - cosec x] = h \to 0 lim \frac{1}{h} [\frac{1}{sin ( x + h )} - \frac{1}{sin x}] = h \to 0 lim \frac{1}{h} [\frac{sin x - sin ( x + h )}{sin ( x + h ) sin x}] = h \to 0 lim \frac{1}{h} [\frac{2 cos ( \frac{x + x + h}{2} ) sin ( \frac{x - x - h}{2} )}{sin ( x + h ) sin x}] = h \to 0 lim \frac{1}{h} [\frac{2 cos ( \frac{2 x + h}{2} ) sin ( - \frac{h}{2} )}{sin ( x + h ) sin x}] = h \to 0 lim \frac{- cos ( \frac{2 x + h}{2} ) \cdot \frac{s i n ( \frac{h}{2} )}{( \frac{h}{2} )}}{sin ( x + h ) sin x}$

$= h \to 0 lim (\frac{- cos ( \frac{2 x + h}{2} )}{sin ( x + h ) sin x}) \cdot \frac{h}{2} \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )} = (\frac{- cos x}{sin x sin x}) \cdot 1 = - cosec x cot x$

(v) Let f(x) = $3 cot x + 5 cosec x$ .

Accordingly, from the first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{3 cot ( x + h ) + 5 cosec ( x + h ) - 3 cot x - 5 cosec x}{h}$

$= 3 h \to 0 lim \frac{1}{h} [+ cot (x + h) - cot x] 5 h \to 0 lim \frac{1}{h} [cosec (x + h) - cosec x]$

Now, $lim_{h \to 0} \frac{1}{h} [cot (x + h) - cot x]$

$= lim_{h \to 0} \frac{1}{h} [\frac{c o s ( x + h )}{s i n ( x + h )} - \frac{c o s x}{s i n x}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{c o s ( x + h ) s i n x - c o s x s i n ( x + h )}{s i n x s i n ( x + h )}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{s i n ( x - x - h )}{s i n x s i n ( x + h )}] = lim_{h \to 0} \frac{1}{h} [\frac{s i n ( - h )}{s i n x s i n ( x + h )}]$

$= - (lim_{h \to 0} \frac{s i n h}{h}) \cdot (lim_{h \to 0} \frac{1}{s i n x \cdot s i n ( x + h )})$

$= - 1 \cdot \frac{1}{s i n x \cdot s i n ( x + 0 )} = \frac{- 1}{s i n ^{2} x} = - cosec^{2} x$

$= lim_{h \to 0} \frac{1}{h} [cosec (x + h) - cosec x]$

$= lim_{h \to 0} \frac{1}{h} [\frac{1}{s i n ( x + h )} - \frac{1}{s i n x}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{s i n x - s i n ( x + h )}{s i n ( x + h ) s i n x}]$

$= h \to 0 lim \frac{1}{h} [\frac{2 cos ( \frac{x + x + h}{2} ) \cdot sin ( \frac{x - x}{2}}{sin ( x + h ) sin x} = h \to 0 lim \frac{1}{h} [\frac{2 cos ( \frac{2 x + h}{2} ) \cdot sin ( - \frac{h}{2} )}{sin ( x + h ) sin x}] = h \to 0 lim \frac{- cos ( \frac{2 x + h}{2} ) \cdot \frac{s i n ( - \frac{h}{2} )}{( \frac{h}{2} )}}{sin ( x + h ) sin x} = h \to 0 lim (\frac{- cos ( \frac{2 x + h}{2} )}{sin ( x + h ) sin x}) \cdot \frac{h}{2} \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )} = (- \frac{cos x}{sin x sin x}) \cdot 1 = - cosec x cot x$

From (1), (2), and (3), we obtain

$f^{'} (x) = - 3 cosec^{2} x - 5 cosec x cot x$

(vi) Let $f (x) = 5 sin x - 6 cos x + 7$ . Accordingly, from the first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{1}{h} [5 sin (x + h) - 6 cos (x + h) + 7 - 5 sin x + 6 cos x - 7] = h \to 0 lim \frac{1}{h} [5 {sin (x + h) - sin x} - 6 {cos (x + h) - cos x}] = 5 h \to 0 lim \frac{1}{h} [sin (x + h) - sin x] - 6 h \to 0 lim \frac{1}{h} [cos (x + h) - cos x] = 5 h \to 0 lim \frac{1}{h} [2 cos (\frac{x + h + x}{2}) sin (\frac{x + h - x}{2})] - 6 h \to 0 lim \frac{cos x cos h - sin x sin h - cos x}{h}$

$= 5 h \to 0 lim \frac{1}{h} [2 cos (\frac{2 x + h}{2}) sin \frac{h}{2}] - 6 h \to 0 lim [\frac{- cos x ( 1 - cos h ) - sin x sin h}{h}] = 5 h \to 0 lim (cos (\frac{2 x + h}{2}) \frac{sin \frac{h}{2}}{\frac{h}{2}}) - 6 h \to 0 lim [\frac{- cos x ( 1 - cos h )}{h} - \frac{sin x sin h}{h}] = 5 [h \to 0 lim cos (\frac{2 x + h}{2})] [\frac{h}{2} \to 0 lim \frac{sin \frac{h}{2}}{\frac{h}{2}}] - 6 [(- cos x) (h \to 0 lim \frac{1 - cos h}{h}) - sin x h \to 0 lim (\frac{sin h}{h})] = 5 cos x \cdot 1 - 6 [(- cos x) (0) - sin x \cdot 1] = 5 cos x + 6 sin x$

(vii) Let $f (x) = 2 tan x - 7 sec x$ . Accordingly, from the first principle,

$= = = = = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} h \to 0 lim \frac{1}{h} [2 tan (x + h) - 7 sec (x + h) - 2 tan x + 7 sec x] h \to 0 lim \frac{1}{h} [2 {tan (x + h) - tan x} - 7 {sec (x + h) - sec x}] 2 h \to 0 lim \frac{1}{h} [tan (x + h) - tan x] - 7 h \to 0 lim \frac{1}{h} [sec (x + h) - sec x] 2 h \to 0 lim \frac{1}{h} [\frac{sin ( x + h )}{cos ( x + h )} - \frac{sin x}{cos x}] - 7 h \to 0 lim \frac{1}{h} [\frac{1}{cos ( x + h )} - \frac{1}{cos x}] 2 h \to 0 lim \frac{1}{h} [\frac{sin ( x + h ) cos x - sin x cos ( x + h )}{cos x cos ( x + h )}] - 7 h \to 0 lim \frac{1}{h} [\frac{cos x - cos ( x + h )}{cos x cos ( x + h )}]$

$= 2 h \to 0 lim \frac{1}{h} [\frac{sin ( x + h - x )}{cos x cos ( x + h )}] - 7 h \to 0 lim \frac{1}{h} [\frac{- 2 sin ( \frac{x + x + h}{2} ) sin ( \frac{x - x - h}{2} )}{cos x cos ( x + h )}] = 2 h \to 0 lim [(\frac{sinh}{h}) \frac{1}{cos x cos ( x + h )}] - 7 h \to 0 lim \frac{1}{h} [\frac{- 2 sin ( \frac{2 x + h}{2} ) sin ( \frac{- h}{2} )}{cos x cos ( x + h )}] = 2 (h \to 0 lim \frac{sin h}{h}) (h \to 0 lim \frac{1}{cos x cos ( x + h )}) - 7 (\frac{h}{2} \to 0 lim \frac{sin \frac{h}{2}}{\frac{h}{2}}) (h \to 0 lim \frac{sin ( \frac{2 x + h}{2} )}{cos x cos ( x + h )}) = 2 \cdot 1 \cdot \frac{1}{cos x cos x} - 7 \cdot 1 (\frac{sin x}{cos x cos x}) = 2 sec^{2} x - 7 sec x tan x$

MISCELLANEOUS EXERCISE

Find the derivative of the following functions from first principle: (i) -x (ii) $(- x)^{- 1}$ (iii) $sin (x + 1)$ (iv) $cos (x - \frac{π}{8})$

Sol.

(i) Let $f (x) = - x$ . Accordingly,

$f (x + h) = - (x + h)$

By first principle,

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

(ii) Let $f (x) = (- x)^{- 1} = \frac{1}{- x} = \frac{- 1}{x}$ . Accordingly,

$f (x + h) = \frac{- 1}{( x + h )}$

By first principle

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{1}{h} [\frac{- 1}{x + h} - (\frac{- 1}{x})] = h \to 0 lim \frac{1}{h} [\frac{- 1}{x + h} + \frac{1}{x}] = h \to 0 lim \frac{1}{h} [\frac{- x + ( x + h )}{x ( x + h )}] = h \to 0 lim \frac{1}{h} [\frac{- x + x + h}{x ( x + h )}] = h \to 0 lim \frac{1}{h} [\frac{h}{x ( x + h )}] = h \to 0 lim \frac{1}{x ( x + h )} = \frac{1}{x \cdot x} = \frac{1}{x ^{2}}$

(iii) Let $f (x) = sin (x + 1)$ . Accordingly $f (x + h) = sin (x + h + 1)$ By first principle

$f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$

$= lim_{h \to 0} \frac{1}{h} [sin (x + h + 1) - sin (x + 1)]$

$= lim_{h \to 0} \frac{1}{h} [2 cos (\frac{x + h + 1 + x + 1}{2}) sin (\frac{x + h + 1 - x - 1}{2})]$

$= lim_{h \to 0} \frac{1}{h} [2 cos (\frac{2 x + h + 2}{2}) sin (\frac{h}{2})]$

$= lim_{h \to 0} [cos (\frac{2 x + h + 2}{2}) \frac{s i n ( \frac{h}{2} )}{( \frac{h}{2} )}]$

$= h \to 0 lim cos (\frac{2 x + h + 2}{2}) \cdot \frac{h}{2} \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )} [As h \to 0 \Rightarrow \frac{h}{2} \to 0] = cos (\frac{2 x + 0 + 2}{2}) \cdot 1 [x \to 0 lim \frac{sin x}{x} = 1] = cos (x + 1)$

(iv) Let $f (x) = cos (x - \frac{π}{8})$ .

$Accordingly f (x + h) = cos (x + h - \frac{π}{8})$

By first principle,

$= = = = = = = f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} h \to 0 lim \frac{1}{h} [cos (x + h - \frac{π}{8}) - cos (x - \frac{π}{8})] h \to 0 lim \frac{1}{h} [- 2 sin \frac{( x + h - \frac{π}{8} + x - \frac{π}{8} )}{2} sin (\frac{x + h - \frac{π}{8} - x + \frac{π}{8}}{2})] h \to 0 lim \frac{1}{h} [- 2 sin (\frac{2 x + h - \frac{π}{4}}{2}) sin \frac{h}{2}] h \to 0 lim [- sin (\frac{2 x + h - \frac{π}{4}}{2}) \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )}] h \to 0 lim [- sin (\frac{2 x + h - \frac{π}{4}}{2})] \frac{h}{2} \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )} - sin (\frac{2 x + 0 - \frac{π}{4}}{2}) \cdot 1 As h \to 0 \Rightarrow \frac{h}{2} \to 0] - sin (x - \frac{π}{8})$

Find the derivative of the following functions Q. 2 to 16 (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):

(x+a) Sol. Let $f (x) = x + a$ . Accordingly, $f (x + h) = (x + h + a)$ By first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{x + h + a - x - a}{h} = h \to 0 lim (\frac{h}{h}) = h \to 0 lim (1) = 1$

$(px + q) (\frac{r}{x} + s)$ Sol. Let $f (x) = (px + q) (\frac{r}{x} + s)$

$f^{'} (x) = (px + q) (\frac{r}{x} + s)^{'} + (\frac{r}{x} + s) (px + q)^{'} = (px + q) (rx^{- 1} + s)^{'} + (\frac{r}{x} + s) (p) = (px + q) (- rx^{- 2}) + (\frac{r}{x} + s) p = (px + q) (\frac{- r}{x ^{2}}) + (\frac{r}{x} + s) p = \frac{- pr}{x} - \frac{qr}{x ^{2}} + \frac{pr}{x} + ps = ps - \frac{qr}{x ^{2}}$

$f (x) = (ax + b) (cx + d)^{2}$ Sol. Let $f (x) = (ax + b) (cx + d)^{2}$ By product rule,

$f^{'} (x) = (ax + b) \frac{d}{dx} (cx + d)^{2} + (cx + d)^{2} \frac{d}{dx} (ax + b) = (ax + b) \frac{d}{dx} (c^{2} x^{2} + 2 cdx + d^{2}) + (cx + d)^{2} \frac{d}{dx} (ax + b) = (ax + b) [\frac{d}{dx} (c^{2} x^{2}) + \frac{d}{dx} + (2 cdx) + \frac{d}{dx} + (d^{2})] + (cx + d)^{2} [\frac{d}{dx} ax + \frac{d}{dx} b] = (ax + b) (2 c^{2} x + 2 cd) + (cx + d)^{2} a = 2 c (ax + b) (cx + d) + a (cx + d)^{2}$

$\frac{ax + b}{cx + d}$ Sol. Let $f (x) = \frac{ax + b}{cx + d}$

$f^{'} (x) = \frac{( c x + d ) \frac{d}{d x} ( a x + b ) - ( a x + b ) \frac{d}{d x} ( c x + d )}{( c x + d ) ^{2}} = \frac{( c x + d ) ( a ) - ( a x + b ) ( c )}{( c x + d ) ^{2}} = \frac{a c x + a d - a c x - b c}{( c x + d ) ^{2}} = \frac{a d - b c}{( c x + d ) ^{2}}$

$\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$ Sol. Let $f (x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} = \frac{\frac{x + 1}{x}}{\frac{x - 1}{x}} = \frac{x + 1}{x - 1}$ where $x \neq = 0, 1$ By quotient rule

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

$\frac{1}{ax ^{2} + bx + c}$ Sol. Let $f (x) = \frac{1}{ax ^{2} + bx + c}$

By quotient rule,

$f^{'} (x) = \frac{( a x ^{2} + b x + c ) \frac{d}{d x} ( 1 ) - \frac{d}{d x} ( a x ^{2} + b x + c )}{( a x ^{2} + b x + c ) ^{2}} = \frac{( a x ^{2} + b x + c ) ( 0 ) - ( 2 a x + b )}{( a x ^{2} + b x + c ) ^{2}} = \frac{- ( 2 a x + b )}{( a x ^{2} + b x + c ) ^{2}}$

$\frac{ax + b}{px ^{2} + qx + r}$ Sol. Let $f (x) = \frac{ax + b}{px ^{2} + qx + r}$ By quotient rule,

$f^{'} (x) = \frac{( p x ^{2} + q x + r ) \frac{d}{d x} ( a x + b ) - ( a x + b ) \frac{d}{d x} ( p x ^{2} + q x + r )}{( p x ^{2} + q x + r ) ^{2}} = \frac{( p x ^{2} + q x + r ) ( a ) - ( a x + b ) ( 2 p x + q )}{( p x ^{2} + q x + r ) ^{2}} = \frac{a p x ^{2} + a q x + a r - 2 a p x ^{2} - a q x - 2 b p x - b q}{( p x ^{2} + q x + r ) ^{2}} = \frac{- a p x ^{2} - 2 b p x + a r - b q}{( p x ^{2} + q x + r ) ^{2}}$

$\frac{px ^{2} + qx + r}{ax + b}$ Sol. Let $f (x) = \frac{px ^{2} + qx + r}{ax + b}$ By quotient rule,

$f^{'} (x) = \frac{( a x + b ) \frac{d}{d x} ( p x ^{2} + q x + r ) - ( p x ^{2} + q x + r ) \frac{d}{d x} ( a x + b )}{( a x + b ) ^{2}} = \frac{( a x + b ) ( 2 p x + q ) - ( p x ^{2} + q x + r ) ( a )}{( a x + b ) ^{2}} = \frac{2 a p x ^{2} + a q x + 2 b p x + b q - a p x ^{2} - a q x - a r}{( a x + b ) ^{2}} = \frac{a p x ^{2} + 2 b p x + b q - a r}{( a x + b ) ^{2}}$

$\frac{a}{x ^{4}} - \frac{b}{x ^{2}} + cos x$ Sol. Let $f (x) = \frac{a}{x ^{4}} - \frac{b}{x ^{2}} + cos x$

$f^{'} (x) [\frac{d}{d x} (x^{n}) = \frac{d}{d x} (\frac{a}{x ^{4}}) - \frac{d}{d x} (\frac{b}{x ^{2}}) + \frac{d}{d x} (cos x) = a \frac{d}{d x} (x^{- 4}) - b \frac{d}{d x} (x^{- 2}) + \frac{d}{d x} (cos x) = a (- 4 x^{- 5}) - b (- 2 x^{- 3}) + (- sin x) = n x^{n - 1} and \frac{d}{d x} (cos x) = - sin x] = \frac{- 4 a}{x ^{5}} + \frac{2 b}{x ^{3}} - sin x$

$4 x - 2$ Sol. Let $f (x) = 4 x - 2$

$f^{'} (x) = \frac{d}{d x} (4 x - 2) = \frac{d}{d x} (4 x) - \frac{d}{d x} (2) = 4 \frac{d}{d x} (x^{\frac{1}{2}}) - 0 = 4 (\frac{1}{2} x^{\frac{1}{2} - 1}) = (2 x^{- \frac{1}{2}}) = \frac{2}{x}$

$(ax + b)^{n}$ Sol. Let $f (x) = (ax + b)^{n}$ . Accordingly,

$f (x + h) = {a (x + h) + b}^{n} = (ax + ah + b)^{n}$

By first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{( a x + ah + b ) ^{n} - ( a x + b ) ^{n}}{h} = h \to 0 lim \frac{( a x + b ) ^{n} ( 1 + \frac{ah}{a x + b} ) ^{n} - ( a x + b ) ^{n}}{h} = (a x + b)^{n} h \to 0 lim \frac{( ( 1 + \frac{ah}{a x + b} ) ^{n} - 1 )}{h}$

$= (a x + b)^{n} h \to 0 lim \frac{1}{h} [{1 + n (\frac{ah}{a x + b}) + \frac{n ( n - 1 )}{⌊ 2} (\frac{ah}{a x + b})^{2} + \dots ..} - 1]$

(Using Binomial Theorem)

$= (a x + b)^{n} lim_{h \to 0} \frac{1}{h} [n (\frac{ah}{a x + b}) + \frac{n ( n - 1 ) a ^{2} h ^{2}}{⌊ 2 ( a x + b ) ^{2}} + \dots ..$

$= (a x + b)^{n} h \to 0 lim [\frac{na}{( a x + b )} + \frac{n ( n - 1 ) a ^{2} h}{⌊ 2 ( a x + b ) ^{2}} + \dots ..] = (a x + b)^{n} [\frac{na}{( a x + b )} + 0] = na \frac{( a x + b ) ^{n}}{( a x + b )} = na (a x + b)^{n - 1}$

$(a x + b)^{n} (c x + d)^{m}$

Sol. Let $f (x) = (ax + b)^{n} (cx + d)^{m}$

$f^{'} (x) = (ax + b)^{n} \frac{d}{dx} (cx + d)^{m} + (cx + d)^{m} \frac{d}{dx} (ax + b)^{n}$

Now, let $f_{1} (x) = (cx + d)^{m}$

$f_{1} (x + h) f_{1}^{'} (x) = (c x + c h + d)^{m} = h \to 0 lim \frac{f _{1} ( x + h ) - f _{1} ( x )}{h} = h \to 0 lim \frac{( c x + c h + d ) ^{m} - ( c x + d ) ^{m}}{h} = (c x + d)^{m} h \to 0 lim \frac{1}{h} [(1 + \frac{c h}{c x + d})^{m} - 1] = (c x + d)^{m} h \to 0 lim \frac{1}{h} [(1 + \frac{m c h}{( c x + d )} + \frac{m ( m - 1 )}{2} \frac{( c ^{2} h ^{2} )}{( c x + d ) ^{2}} + \dots .) - 1]$

$= (cx + d)^{m} lim_{h \to 0} \frac{1}{h} [\frac{mch}{( cx + d )} + \frac{m ( m - 1 ) c ^{2} h ^{2}}{2 ( cx + d ) ^{2}}$

$+ \dots . ($ Terms containing higher degrees of h $)]$

$= (cx + d)^{m} h \to 0 lim [\frac{mc}{( cx + d )} + \frac{m ( m - 1 ) c ^{2} h}{2 ( cx + d ) ^{2}} + \dots ..] = (cx + d)^{m} [\frac{mc}{cx + d} + 0] = \frac{mc ( cx + d ) ^{m}}{( cx + d )} = mc (cx + d)^{m - 1} \frac{d}{dx} (cx + d)^{m} = mc (cx + d)^{m - 1}$

Similarly,

$\frac{d}{dx} (ax + b)^{n} = na (ax + b)^{n - 1}$

Therefore, from (1), (2), and (3), we obtain

$f^{'} (x) = (a x + b)^{n} {mc (c x + d)^{m - 1}} + (c x + d)^{m} {na (a x + b)^{n - 1}} = (a x + b)^{n - 1} (c x + d)^{m - 1} [mc (a x + b) + na (c x + d)]$

$sin (x + a)$

Sol. Let $f (x) = sin (x + a)$ Therefore, $f (x + h) = sin (x + h + a)$ By first principle,

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{sin ( x + h + a ) - sin ( x + a )}{h} = h \to 0 lim \frac{1}{h} [2 cos (\frac{x + h + a + x + a}{2}) sin (\frac{x + h + a - x - a}{2})]$

$= h \to 0 lim \frac{1}{h} [2 cos (\frac{2 x + 2 a + h}{2}) sin (\frac{h}{2})] = h \to 0 lim [cos (\frac{2 x + 2 a + h}{2}) {\frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )}}] = h \to 0 lim cos (\frac{2 x + 2 a + h}{2}) \frac{h}{2} \to 0 lim {\frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )}} [As h \to 0 \Rightarrow \frac{h}{2} \to 0] = cos (\frac{2 x + 2 a}{2}) \times 1 [x \to 0 lim \frac{sin x}{x} = 1] = cos (x + a)$

$cosec x cot x$

Sol. Let $f (x) = cosec x cot x$ By product rule,

$f^{'} (x) = cosec x (cot x)^{'} + cot x (cosec x)^{'}$

Let $f_{1} (x) = cot x$ Accordingly, $f_{1} (x + h) = cot (x + h)$ By first principle

$f_{1}^{'} (x) = h \to 0 lim \frac{f _{1} ( x + h ) - f _{1} ( x )}{h} = h \to 0 lim \frac{cot ( x + h ) - cot x}{h} = h \to 0 lim \frac{1}{h} (\frac{cos ( x + h )}{sin ( x + h )} - \frac{cos x}{sin x}) = h \to 0 lim \frac{1}{h} [\frac{sin x cos ( x + h ) - cos x sin ( x + h )}{sin x sin ( x + h )}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x - x - h )}{sin x sin ( x + h )}]$

$= \frac{1}{sin x} h \to 0 lim \frac{1}{h} [\frac{sin ( - h )}{sin ( x + h )}] = \frac{- 1}{sin x} \cdot (h \to 0 lim \frac{sin h}{h}) (h \to 0 lim \frac{1}{sin ( x + h )}) = \frac{- 1}{sin x} \cdot 1 \cdot (\frac{1}{sin ( x + 0 )}) = \frac{- 1}{sin ^{2} x} = - cosec^{2} x (cot x)^{'} = - cosec^{2} x$

Now, $f_{2} (x) = cosec x$ . Accordingly, $f_{2} (x + h) = cosec (x + h)$ By first principle,

$= = = = = = = = = f_{2}^{'} (x) = h \to 0 lim \frac{f _{2} ( x + h ) - f _{2} ( x )}{h} h \to 0 lim \frac{1}{h} [cosec (x + h) - cosec x] h \to 0 lim \frac{1}{h} [\frac{1}{sin ( x + h )} - \frac{1}{sin x}] h \to 0 lim \frac{1}{h} [\frac{sin x - sin ( x + h )}{sin x sin ( x + h )}] \frac{1}{sin x} h \to 0 lim \frac{1}{h} [\frac{2 cos ( \frac{x + x + h}{2} ) sin ( \frac{x - x - h}{2} )}{sin ( x + h )}] \frac{1}{sin x} h \to 0 lim \frac{1}{h} [\frac{2 cos ( \frac{2 x + h}{2} ) sin ( \frac{- h}{2} )}{sin ( x + h )}] \frac{1}{sin x} h \to 0 lim [\frac{- sin ( \frac{h}{2} )}{( \frac{h}{2} )} \cdot \frac{cos ( \frac{2 x + h}{2} )}{sin ( x + h )}] \frac{- 1}{sin x} \frac{h}{2} \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )} \cdot h \to 0 lim \frac{cos ( \frac{2 x + h}{2} )}{sin ( x + h )} \frac{- 1}{sin x} \cdot 1 \cdot \frac{cos ( \frac{2 x + 0}{2} )}{sin ( x + 0 )} = \frac{- 1}{sin x} \cdot \frac{cos x}{sin x} - cosec x \cdot cot x$

$∴ (cosec x)^{'} = - cosec x \cdot cot x$

From (1), (2), and (3), we obtain

$f^{'} (x) = cosec x (- cosec^{2} x) + cot x (- cosec x cot x) = - cosec^{3} x - cot^{2} x cosec x$

$\frac{c o s x}{1 + s i n x}$ Sol. Let $f (x) = \frac{c o s x}{1 + s i n x}$ By quotient rule,

$f^{'} (x) = \frac{( 1 + sin x ) \frac{d}{d x} ( cos x ) - ( cos x ) \frac{d}{d x} ( 1 + sin x )}{( 1 + sin x ) ^{2}} = \frac{( 1 + sin x ) ( - sin x ) - ( cos x ) ( cos x )}{( 1 + sin x ) ^{2}} = \frac{- sin x - sin ^{2} x - cos ^{2} x}{( 1 + sin ^{2} x ) ^{2}} = \frac{- sin x - ( sin ^{2} x + cos ^{2} x )}{( 1 + sin x ) ^{2}} = \frac{- sin x - 1}{( 1 + sin x ) ^{2}} = \frac{- ( 1 + sin x )}{( 1 + sin x ) ^{2}} = \frac{- 1}{( 1 + sin x )}$

$\frac{s i n x + c o s x}{s i n x - c o s x}$ Sol. Let $f (x) = \frac{s i n x + c o s x}{s i n x - c o s x}$ By quotient rule,

$f^{'} (x) = \frac{( sin x - cos x ) \frac{d}{d x} ( sin x + cos x ) - ( sin x + cos x ) \frac{d}{d x} ( sin x - cos x )}{( sin x - cos x ) ^{2}} = \frac{( sin x - cos x ) ( cos x - sin x ) - ( sin x + cos x ) ( cos x + sin x )}{( sin x - cos x ) ^{2}} = \frac{- ( sin x - cos x ) ^{2} - ( sin x + cos x ) ^{2}}{( sin x - cos x ) ^{2}} = \frac{- [ sin ^{2} x + cos ^{2} x - 2 sin x cos x + sin ^{2} x + cos ^{2} x + 2 sin x cos x ]}{( sin x - cos x ) ^{2}} = \frac{- [ 1 + 1 ]}{( sin x - cos x ) ^{2}} = \frac{- 2}{( sin x - cos x ) ^{2}}$

$\frac{s e c x - 1}{s e c x + 1}$ Sol. Let $f (x) = \frac{s e c x - 1}{s e c x + 1}$

$f (x) = \frac{\frac{1}{c o s x} - 1}{\frac{1}{c o s x} + 1} = \frac{1 - c o s x}{1 + c o s x}$

By quotient rule

$f^{'} (x) = \frac{( 1 + cos x ) \frac{d}{d x} ( 1 - cos x ) - ( 1 - cos x ) \frac{d}{d x} ( 1 + cos x )}{( 1 + cos x ) ^{2}} = \frac{( 1 + cos x ) ( sin x ) - ( 1 - cos x ) ( - sin x )}{( 1 + cos x ) ^{2}} = \frac{sin x + cos x sin x + sin x - sin x cos x}{( 1 + cos x ) ^{2}} = \frac{2 sin x}{( 1 + cos x ) ^{2}} = \frac{2 sin x}{( 1 + \frac{1}{s e c x} ) ^{2}} = \frac{2 sin x}{\frac{( s e c x + 1 ) ^{2}}{s e c ^{2} x}} = \frac{2 sin x sec ^{2} x}{( sec x + 1 ) ^{2}} = \frac{\frac{2 s i n x}{c o s x} sec x}{( sec x + 1 ) ^{2}} = \frac{2 sec x tan x}{( sec x + 1 ) ^{2}}$

$sin^{n} x$ Sol. Let $y = sin^{n} x$ Accordingly, for $n = 1, y = sin x$ $∴ \frac{dy}{dx} = cos x$ , i.e. $\frac{d}{dx} sin x = cos x$ For $n = 2, y = sin^{2} x$ $∴ \frac{dy}{dx} = \frac{d}{dx} (sin x sin x)$

$= (sin x)^{'} sin x + sin x (sin x)^{'}$

[By Leibnitz product rule]

$= cos x sin x + sin x cos x = 2 sin x cos x$

For $n = 3, y - sin^{3} x$ $∴ \frac{dy}{dx} = \frac{d}{dx} (sin x sin^{2} x)$ $= (sin x)^{'} sin^{2} x + sin x (sin^{2} x)^{'}$ [By Leibnitz product rule] $= cos x sin^{2} x + sin x (2 sin x cos x)$ [Using (1)]

$= cos x sin^{2} x + 2 sin^{2} x cos x = 3 sin^{2} x cos x$

We assert that $\frac{d}{dx} (sin^{n} x) = n^{2} n^{(n - 1)} x cos x$ Let our assertion be true for $n = k$ . i.e. $\frac{d}{dx} (sin^{k} x) = k sin^{(k - 1)} x cos x$ Consider

$\frac{d}{d x} (sin^{k + 1} x) = \frac{d}{d x} (sin x sin^{k} x) = (sin x)^{'} sin^{k} x + sin x (sin^{k} x)^{'}$

[By Leibnitz product rule] $= cos x sin^{k} x + sin x (k sin^{(k - 1)} x cos x)$ [Using (2)]

$= cos x sin^{k} x + k sin^{k} x cos x = (k + 1) sin^{k} x cos x$

Thus, our assertion is true for $n = k + 1$ . Hence, by mathematical induction

$\frac{d}{d x} (sin^{n} x) = n sin^{(n - 1)} x cos x$

$\frac{a + b s i n x}{c + d c o s x}$ Sol. Let $f (x) = \frac{a + b s i n x}{c + d c o s x}$

By quotient rule,

$f^{'} (x) = \frac{( c + d cos x ) \frac{d}{d x} ( a + b sin x ) - ( a + b sin x ) \frac{d}{d x} ( c + d cos x )}{( c + d cos x ) ^{2}} = \frac{( c + d cos x ) ( b cos x ) - ( a + b sin x ) ( - d sin x )}{( c + d cos x ) ^{2}} = \frac{c b cos x + b d cos ^{2} x + a d sin x + b d sin ^{2} x}{( c + d cos x ) ^{2}} = \frac{b c cos x + a d sin x + b d ( cos ^{2} x + sin ^{2} x )}{( c + d cos x ) ^{2}} = \frac{b c cos x + a d sin x + b d}{( c + d cos x ) ^{2}}$

$\frac{s i n ( x + a )}{c o s x}$

Sol. Let $f (x) = \frac{s i n ( x + a )}{c o s x}$ By quotient rule,

$f^{'} (x) f^{'} (x) = \frac{cos x \frac{d}{d x} [ sin ( x + a )] - sin ( x + a ) \frac{d}{d x} cos x}{cos ^{2} x} = \frac{cos x \frac{d}{d x} [ sin ( x + a )] - sin ( x + a ) ( - sin x )}{cos ^{2} x}$

Let $g (x) = sin (x + a)$ . Accordingly, $g (x + h) = sin (x + h + a)$ By first principle,

$g^{'} (x) = h \to 0 lim \frac{g ( x + h ) - g ( x )}{h} = h \to 0 lim \frac{1}{h} [sin (x + h + a) - sin (x + a)] = h \to 0 lim \frac{1}{h} [2 cos (\frac{x + h + a + x + a}{2}) sin (\frac{x + h + a - x - a}{2})] = h \to 0 lim \frac{1}{h} [2 cos (\frac{2 x + 2 a + h}{2}) sin (\frac{h}{2})] = h \to 0 lim [cos (\frac{2 x + 2 a + h}{2}) {\frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )}}]$

$= h \to 0 lim cos (\frac{2 x + 2 a + h}{2}) \cdot \frac{h}{2} \to 0 lim {\frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )}} = (cos \frac{2 x + 2 a}{2}) \times 1 [h \to 0 lim \frac{sinh}{h} = 1] = cos (x + a)$

From (1) and (2), we obtain

$f^{'} (x) = \frac{cos x \cdot cos ( x + a ) + sin x sin ( x + a )}{cos ^{2} x} = \frac{cos ( x + a - x )}{cos ^{2} x} = \frac{cos a}{cos ^{2} x}$

$x^{4} (5 sin x - 3 cos x)$

Sol. Let $f (x) = x^{4} (5 sin x - 3 cos x)$ By product rule,

$f^{'} (x) = x^{4} \frac{d}{d x} (5 sin x - 3 cos x) + (5 sin x - 3 cos x) \frac{d}{d x} (x^{4}) = x^{4} [5 \frac{d}{d x} (sin x) - 3 \frac{d}{d x} (cos x)] + (5 sin x - 3 cos x) \frac{d}{d x} (x^{4}) = x^{4} [5 cos x - 3 (- sin x)] + (5 sin x - 3 cos x) (4 x^{3}) = x^{3} [5 x cos x + 3 sin x + 20 sin x - 12 cos x]$

$(x^{2} + 1) cos x$

Sol. Let $f (x) = (x^{2} + 1) cos x$ By product rule

$f^{'} (x) = (x^{2} + 1) \frac{d}{d x} (cos x) + cos x \frac{d}{d x} (x^{2} + 1) = (x^{2} + 1) (- sin x) + cos x (2 x) = - x^{2} sin x - sin x + 2 x cos x$

$(a x^{2} + sin x) (p + q cos x)$

Sol. Let $f (x) = (ax^{2} + sin x) (p + q cos x)$ By product rule,

$f^{'} (x) = (a x^{2} + sin x) \frac{d}{d x} (p + q cos x) + (p + q cos x) \frac{d}{d x} (a x^{2} + sin x) = (a x^{2} + sin x) (- q sin x) + (p + q cos x) (2 a x + cos x) = - q sin x (a x^{2} + sin x) + (p + q cos x) (2 a x + cos x)$

$(x + cos x) (x - tan x)$ Sol. Let $f (x) = (x + cos x) (x - tan x)$ By product rule,

$f^{'} (x) = (x + cos x) \frac{d}{d x} (x - tan x) + (x - tan x) \frac{d}{d x} (x + cos x) = (x + cos x) [\frac{d}{d x} (x) - \frac{d}{d x} (tan x)] + (x - tan x) (1 - sin x) = (x + cos x) [1 - \frac{d}{d x} tan x] + (x - tan x) (1 - sin x)$

Let $g (x) = tan x$ . Accordingly, $g (x + h) = tan (x + h)$ By first principle

$g^{'} (x) = h \to 0 lim \frac{g ( x + h ) - g ( x )}{h} = h \to 0 lim (\frac{tan ( x + h ) - tan x}{h}) = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h )}{cos ( x + h )} - \frac{sin x}{cos x}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h ) cos x - sin x cos ( x + h )}{cos ( x + h ) cos x}] = \frac{1}{cos x} h \to 0 lim \frac{1}{h} [\frac{sin ( x + h - x )}{cos ( x + h )}] = \frac{1}{cos x} h \to 0 lim \frac{1}{h} [\frac{sin h}{cos ( x + h )}] = \frac{1}{cos x} (h \to 0 lim \frac{sin h}{h}) (h \to 0 lim \frac{1}{cos ( x + h )}) = \frac{1}{cos x} \cdot 1 \cdot \frac{1}{cos ( x + 0 )} = \frac{1}{cos ^{2} x} = sec^{2} x$

Therefore, from (1) and (2) we obtain

$f^{'} (x) = (x + cos x) (1 - sec^{2} x) + (x - tan x) (1 - sin x) = (x + cos x) (- tan^{2} x) + (x - tan x) (1 - sin x) = - tan^{2} x (x + cos x) + (x - tan x) (1 - sin x)$

$\frac{4 x + 5 s i n x}{3 x + 7 c o s x}$ Sol. Let $f (x) = \frac{4 x + 5 s i n x}{3 x + 7 c o s x}$ By quotient rule,

$f^{'} (x) = \frac{( 3 x + 7 cos x ) \frac{d}{d x} ( 4 x + 5 sin x ) - ( 4 x + 5 sin x ) \frac{d}{d x} ( 3 x + 7 cos x )}{( 3 x + 7 cos x ) ^{2}} = \frac{( 3 x + 7 cos x ) [ 4 \frac{d}{d x} ( x ) + 5 \frac{d}{d x} ( sin x ) ] - ( 4 x + 5 sin x ) [ 3 \frac{d}{d x} x + 7 \frac{d}{d x} cos x ]}{( 3 x + 7 cos x ) ^{2}} = \frac{( 3 x + 7 cos x ) ( 4 + 5 cos x ) - ( 4 x + 5 sin x ) ( 3 - 7 sin x )}{( 3 x + 7 cos x ) ^{2}} = \frac{12 x + 15 x cos x + 28 cos x + 35 cos ^{2} x - 12 x + 28 x sin x - 15 sin x + 35 sin ^{2} x}{( 3 x + 7 cos x ) ^{2}} = \frac{15 x cos x + 28 cos x + 28 x sin x - 15 sin x + 35 ( cos ^{2} x + sin ^{2} x )}{( 3 x + 7 cos x ) ^{2}} = \frac{35 + 15 x cos x + 28 cos x + 28 x sin x - 15 sin x}{( 3 x + 7 cos x ) ^{2}}$

$\frac{x ^{2} c o s ( \frac{π}{4} )}{s i n x}$ Sol. Let $f (x) = \frac{x ^{2} c o s ( \frac{π}{4} )}{s i n x}$ By quotient rule,

$f^{'} (x) = cos \frac{π}{4} \cdot [\frac{sin x \frac{d}{d x} ( x ^{2} ) - x ^{2} \frac{d}{d x} ( sin x )}{sin ^{2} x}] = cos \frac{π}{4} \cdot [\frac{sin x \cdot 2 x - x ^{2} cos x}{sin ^{2} x}] = \frac{x cos \frac{π}{4} [ 2 sin x - x cos x ]}{sin ^{2} x}$

$\frac{x}{1 + t a n x}$ Sol. Let $f (x) = \frac{x}{1 + t a n x}$

$f^{'} (x) = \frac{( 1 + tan x ) \frac{d}{d x} ( x ) - x \frac{d}{d x} ( 1 + t}{( 1 + tan x ) ^{2}} f^{'} (x) = \frac{( 1 + tan x ) - x \cdot \frac{d}{d x} ( 1 + tan x )}{( 1 + tan x ) ^{2}}$

Let $g (x) = 1 + tan x$ . Accordingly, $g (x + h) = 1 + tan (x + h)$ By first principle,

$g^{'} (x) = h \to 0 lim \frac{g ( x + h ) - g ( x )}{h} = h \to 0 lim [\frac{1 + tan ( x + h ) - 1 - tan x}{h}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h )}{cos ( x + h )} - \frac{sin x}{cos x}] = h \to 0 lim \frac{1}{h} \frac{sin ( x + h ) cos x - sin x cos ( x + h )}{cos ( x + h ) cos x}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h - x )}{cos ( x + h ) cos x}] = h \to 0 lim \frac{1}{h} [\frac{sin h}{cos ( x + h ) cos x}] = (h \to 0 lim \frac{sin h}{h}) \cdot (h \to 0 lim \frac{1}{cos ( x + h ) cos x}) = 1 \times \frac{1}{cos ^{2} x} = sec^{2} x$

From (1) and (2), we obtain

$f^{'} (x) = \frac{1 + t a n x - x s e c ^{2} x}{( 1 + t a n x ) ^{2}}$

$(x + sec x) (x - tan x)$

Sol. Let $f (x) = (x + sec x) (x - tan x)$ By product rule

$f^{'} (x) = (x + sec x) \frac{d}{d x} (x - tan x) + (x - tan x) \frac{d}{d x} (x + sec x) = (x + sec x) [\frac{d}{d x} (x) - \frac{d}{d x} tan x] + (x - tan x) [\frac{d}{d x} (x) + \frac{d}{d x} sec x] = (x + sec x) [1 - \frac{d}{d x} tan x] + (x - tan x) [1 + \frac{d}{d x} sec x]$

Let $f_{1} (x) = tan x, f_{2} (x) = sec x$ Accordingly, $f_{1} (x + h) = tan (x + h)$

$and f_{2} (x + h) = sec (x + h) f_{1}^{'} (x) = h \to 0 lim (\frac{f _{1} ( x + h ) - f _{1} ( x )}{h}) = h \to 0 lim [\frac{tan ( x + h ) - tan x}{h}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h )}{cos ( x + h )} - \frac{sin x}{cos x}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h ) cos x - sin x cos ( x + h )}{cos ( x + h ) cos x}] = h \to 0 lim \frac{1}{h} [\frac{sin ( x + h - x )}{cos ( x + h ) cos x}] = h \to 0 lim \frac{1}{h} [\frac{sin h}{cos ( x + h ) cos x}] = (h \to 0 lim \frac{sin h}{h}) \cdot (h \to 0 lim \frac{1}{cos ( x + h ) cos x}) = 1 \times \frac{1}{cos ^{2} x} = sec^{2} x$

$\Rightarrow \frac{d}{dx} tan x = sec^{2} x$

$f_{2}^{'} (x) = lim_{h \to 0} (\frac{f _{2} ( x + h ) - f _{2} ( x )}{h})$

$= lim_{h \to 0} \frac{1}{h} [\frac{c o s x - c o s ( x + h )}{c o s ( x + h ) c o s x}]$

$= \frac{1}{cos x} \cdot h \to 0 lim \frac{sin ( \frac{2 x + h}{2} ) { \frac{s i n ( \frac{h}{2} )}{( \frac{h}{2} )} }}{cos ( x + h )} = sec x - {h \to 0 lim sin (\frac{2 x + h}{2})} {\frac{h}{2} \to 0 lim \frac{sin ( \frac{h}{2} )}{( \frac{h}{2} )}} = sec x \frac{sin x \cdot 1}{cos x} \Rightarrow \frac{d}{d x} sec x = sec x tan x$

From (1), (2) and (3) we obtain

$f^{'} (x) = (x + sec x) (1 - sec^{2} x) + (x - tan x) (1 + sec x tan x)$

$\frac{x}{s i n ^{n} x}$

Sol. Let $f (x) = \frac{x}{s i n ^{n} x}$

By quotient rule

$f^{'} (x) = \frac{s i n ^{n} x \frac{d}{d x} x - x \frac{d}{d x} s i n ^{n} x}{s i n ^{2 n} x}$

It can be easily shown that

$\frac{d}{d x} sin^{n} x = n sin^{n - 1} x cos x$

Therefore,

$f^{'} (x) = \frac{sin ^{n} x \frac{d}{d x} x - x \frac{d}{d x} sin ^{n} x}{sin ^{2 n} x} = \frac{sin ^{n} x \cdot 1 - x ( n sin ^{n - 1} x cos x )}{sin ^{2 n} x} = \frac{sin ^{n - 1} x ( sin x - n x cos x )}{sin ^{2 n} x} = \frac{sin x - n x cos x}{sin ^{n + 1} x}$

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Get chapter-wise NCERT Class 11 Maths solutions with accurate answers, important formulas, and clear explanations. Each solution is presented in simple language to help students understand Maths concepts, solve NCERT questions step by step, strengthen problem-solving skills, and prepare confidently for exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 13	Statistics
Chapter 14	Probability

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 12

Exercise	Number of Questions	Important Concepts Covered
Exercise 12.1	32 Questions and Solutions	Limits of functions, properties of limits, left-hand and right-hand limits, algebraic methods, Squeeze Theorem
Exercise 12.2	11 Questions and Solutions	Derivatives, geometrical interpretation of derivatives, differentiation rules, derivatives of functions
Miscellaneous Exercise	30 Questions and Solutions	Mixed practice on limits, algebraic methods, Squeeze Theorem, derivatives, and differentiation rules

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 12

Left-Hand and Right-Hand Limits: Learn how to find the left-hand limit (LHL) and right-hand limit (RHL) and determine whether a limit exists by comparing their values.
Step-by-Step "First Principle" Derivations:
Every standard derivative (like \sin x, x^n, or \cos x) is derived using the first principle, helping students understand the core logic of calculus.
Important Limit Methods: Learn how to simplify algebraic and trigonometric expressions and apply standard limit results to solve limit problems step by step.
Product and Quotient Rule Clarity:
The solutions provide multiple examples of using the product and quotient rules to ensure students can handle complex algebraic and trigonometric functions.
Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Calculus Visualization:The solutions explain how derivatives represent the slope of a tangent and the rate of change, helping Physics students understand how calculus is used to describe motion, velocity, and other changing quantities.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises are covered with clear, step-by-step solutions to help students understand and apply the concepts of Limits and Derivatives.

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Why are NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives important for students?

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What concepts are covered in NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives?

Are NCERT Solutions for Class 11 Maths Chapter 12 by ALLEN useful for competitive exams?

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NCERT Solutions Class 11 Maths Chapter 12 – Limits and Derivatives

1.0Class 11 Maths Chapter 12 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 12

EXERCISE - 12.2

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 12

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 12

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