NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives Exercise 12.2 - 2025-26

Exercise 12.2

1. Find the derivative of \[{x^2} - 2\]at $x = 10$

Ans: Let $f\left( x \right) = {x^2} - 2$

Accordingly,

$f'\left( 10 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 10+h \right)-f\left( 10 \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left( {10 + h} \right)}^2} - 2} \right] - \left( {{{10}^2} - 2} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{{10}^2} + \left( {2 \times 10 \times h} \right) + {h^2} - 2 - {{10}^2} + 2}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{20h + {h^2}}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {20 + h} \right) = 20 + 0 = 20$

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

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

Ans: Let $f\left( x \right) = x$

Accordingly,

$f'\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 + h} \right) - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( 1 \right) = 1$

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

3. Find the derivative of $99x$at $x = 100$

Ans: Let $f\left( x \right) = 99x$

Accordingly,

$f'\left( 100 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 100+h \right)-f\left( 100 \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{99\left( {100 + h} \right) - \left( {99 \times 100} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {99 \times 100} \right) + 99h - \left( {99 \times 100} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{99h}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {99} \right) = 99$

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

4. Find the derivative of the following functions using the first principle.

(i). ${x^3} - 27$

Ans: Let $f\left( x \right) = {x^3} - 27$

Accordingly, from the first principle,

     $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left( {x + h} \right)}^3} - 27} \right] - \left( {{x^3} - 27} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + {h^3} + 3{x^2}h + 3x{h^2} - {x^3}}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {{h^2} + 3{x^2} + 3xh} \right)$

$ \Rightarrow 0 + 3{x^2} + 0 = 3{x^2}$

(ii). $\left( {x - 1} \right)\left( {x - 2} \right)$

Ans: Let $f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)$

Accordingly, from the first principle,

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h - 1} \right)\left( {x + h - 2} \right) - \left( {x - 1} \right)\left( {x - 2} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} + hx - 2x + hx + {h^2} - 2h - x - h + 2} \right) - \left( {{x^2} - 2x - x + 2} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {hx + hx + {h^2} - 2h - h} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{2hx + {h^2} - 3h}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {2x + h - 3} \right) = 2x - 3$

(iii). $\frac{1}{{{x^2}}}$

Ans: Let $f\left( x \right) = \frac{1}{{{x^2}}}$

Accordingly, from the first principle,

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {x + h} \right)}^2}}} - \frac{1}{{{x^2}}}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{{x^2} - {{\left( {x + h} \right)}^2}}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{{x^2} - {x^2} - 2hx - {h^2}}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - {h^2} - 2hx}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - {h^2} - 2x}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right] = \frac{{0 - 2x}}{{{x^2}{{\left( {x + 0} \right)}^2}}}$

$ \Rightarrow \frac{{ - 2}}{{{x^3}}}$

(iv). $\frac{{x + 1}}{{x - 1}}$

Ans: Let $f\left( x \right) = \frac{{x + 1}}{{x - 1}}$

Accordingly, from the first principle,

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{x + h + 1}}{{x + h - 1}} - \frac{{x + 1}}{{x - 1}}} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\left( {x - 1} \right)\left( {x + h + 1} \right) - \left( {x + 1} \right)\left( {x + h - 1} \right)}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\left( {{x^2} + hx + x - x - h - 1} \right) - \left( {{x^2} + hx - x + x + h - 1} \right)}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2h}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - 2}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right] = \frac{{ - 2}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}$

$ \Rightarrow \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}$

5. For the function $f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + .... + \frac{{{x^2}}}{2} + x + 1$prove that      $f'\left( 1 \right)=100f'\left( 0 \right)$

Ans: The given function is,

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

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

$\frac{d}{{dx}}f\left( x \right) = \frac{d}{{dx}}\left( {\frac{{{x^{100}}}}{{100}}} \right) + \frac{d}{{dx}}\left( {\frac{{{x^{99}}}}{{99}}} \right) + ... + \frac{d}{{dx}}\left( {\frac{{{x^2}}}{2}} \right) + \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( 1 \right)$

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

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

$\Rightarrow {x^{99}} + {x^{98}} + .... + x + 1$

$\therefore \text{ }f'\left( x \right)={{x}^{99}}+{{x}^{98}}+.....x+1$

At $x = 0,$

$f'\left( 0 \right)=1$

At $x = 1,$

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

$\Rightarrow 1 \times 100 = 100$

Thus, $f'\left( 1 \right)=100f'\left( 0 \right)$

6. Find the derivative of ${x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ..... + {a^{n - 1}}x + {a^n}$ , where$a$ is some fixed real number.

Ans: Let$f\left( x \right) = {x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ..... + {a^{n - 1}}x + {a^n}$

$\frac{d}{{dx}}f\left( x \right) = \frac{d}{{dx}}\left( {{x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ..... + {a^{n - 1}}x + {a^n}} \right)$

$\Rightarrow \frac{{d\left( {{x^n}} \right)}}{{dx}} + a\frac{{d\left( {{x^{n - 1}}} \right)}}{{dx}} + {a^2}\frac{{d\left( {{x^{n - 2}}} \right)}}{{dx}} + .... + {a^{n - 1}}\frac{{d\left( x \right)}}{{dx}} + {a^n}\frac{{d\left( 1 \right)}}{{dx}}$

On using theorem $\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

So, we obtain

\[f'\left( x \right)=n{{x}^{n-1}}+a\left( n-1 \right){{x}^{n-2}}+{{a}^{2}}\left( n-2 \right){{x}^{n-3}}+....+{{a}^{n-1}}+{{a}^{n}}\left( 0 \right)\]

$\therefore \text{ }f'\left( x \right)=n{{x}^{n-1}}+a\left( n-1 \right){{x}^{n-2}}+{{a}^{2}}\left( n-2 \right){{x}^{n-3}}+...+{{a}^{n-1}}$

7. For some constants $a$and $b$, find the derivative of the following functions.

(a). $\left( {x - a} \right)\left( {x - b} \right)$

Ans: Let $f\left( x \right) = \left( {x - a} \right)\left( {x - b} \right)$

$\Rightarrow f\left( x \right) = {x^2} - \left( {a - b} \right)x + ab$

$\therefore \text{ }f'\left( x \right)=\frac{d}{dx}\left( {{x}^{2}}-\left( a+b \right)x+ab \right)$

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

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

$f'\left( x \right)=2x-\left( a+b \right)+0=2x-a-b$

(b). ${\left( {a{x^2} + b} \right)^2}$

Ans: Let, $f\left( x \right) = {\left( {a{x^2} + b} \right)^2}$

$\Rightarrow f\left( x \right) = {a^2}{x^4} + 2ab{x^2} + {b^2}$

$\therefore \text{ }f'\left( x \right)=\frac{d}{dx}\left( {{a}^{2}}{{x}^{4}}+2ab{{x}^{2}}+{{b}^{2}} \right)$

$\Rightarrow {a^2}\frac{d}{{dx}}\left( {{x^4}} \right) + 2ab\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}{b^2}$

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

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

$\Rightarrow 4{a^2}{x^3} + 4abx = 4ax\left( {a{x^2} + b} \right)$

(c). $\frac{{x - a}}{{x - b}}$

Ans: Let, $f\left( x \right) = \frac{{x - a}}{{x - b}}$

$\Rightarrow \text{ }f'\left( x \right)=\frac{d}{dx}\left( \frac{x-a}{x-b} \right)$

By quotient rule,

$f'\left( x \right)=\frac{\left( x-b \right)\frac{d}{dx}\left( x-a \right)-\left( x-a \right)\frac{d}{dx}\left( x-b \right)}{{{\left( x-b \right)}^{2}}}$

$\Rightarrow \frac{{\left( {x - b} \right) \times 1 - \left( {x - a} \right) \times 1}}{{{{\left( {x - b} \right)}^2}}}$

$\Rightarrow \frac{{x - b - x + a}}{{{{\left( {x - b} \right)}^2}}} = \frac{{a - b}}{{{{\left( {x - b} \right)}^2}}}$

8. Find the derivative of $\frac{{{x^n} - {a^n}}}{{x - a}}$for any constant $a$

Ans: Let $f\left( x \right) = \frac{{{x^n} - {a^n}}}{{x - a}}$

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

By quotient rule,

$f'\left( x \right)=\frac{\left( x-a \right)\frac{d}{dx}\left( {{x}^{n}}-{{a}^{n}} \right)-\left( {{x}^{n}}-{{a}^{n}} \right)\frac{d}{dx}\left( x-a \right)}{{{\left( x-a \right)}^{2}}}$

$\Rightarrow \frac{{\left( {x - a} \right)\left( {n{x^{n - 1}} - 0} \right) - \left( {{x^n} - {a^n}} \right)}}{{{{\left( {x - a} \right)}^2}}}$

$\Rightarrow \frac{{n{x^n} - an{x^{n - 1}} - {x^n} + {a^n}}}{{{{\left( {x - a} \right)}^2}}}$

9. Find the derivative of the following functions

(a). $2x - \frac{3}{4}$

Ans: Let $f\left( x \right) = 2x - \frac{3}{4}$

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

$\Rightarrow 2\frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {\frac{3}{4}} \right)$

$\Rightarrow 2 - 0 = 2$

(b). $\left( {5{x^3} + 3x - 1} \right)\left( {x - 1} \right)$

Ans: Let $f\left( x \right) = \left( {5{x^3} + 3x - 1} \right)\left( {x - 1} \right)$

By Leibnitz product rule,

\[f'\left( x \right)=\left( 5{{x}^{3}}+3x-1 \right)\frac{d}{dx}\left( x-1 \right)+\left( x-1 \right)\frac{d}{dx}\left( 5{{x}^{3}}+3x-1 \right)\]

$\Rightarrow \left( {\left( {5{x^3} + 3x - 1} \right) \times 1} \right) + \left( {x - 1} \right)\left( {5 \times 3{x^2} + 3 - 0} \right)$

$\Rightarrow \left( {5{x^3} + 3x - 1} \right) + \left( {x - 1} \right)\left( {15{x^2} + 3} \right)$

$\Rightarrow 5{x^3} + 3x - 1 + 15{x^3} + 3x - 15{x^2} - 3$

$\Rightarrow 20{x^3} - 15{x^2} + 6x - 4$

(c). ${x^{ - 3}}\left( {5 + 3x} \right)$

Ans: Let, $f\left( x \right) = {x^{ - 3}}\left( {5 + 3x} \right)$

By Leibnitz product rule,

$f'\left( x \right) = {x^{ - 3}}\frac{d}{{dx}}\left( {5 + 3x} \right) + \left( {5 + 3x} \right)\frac{d}{{dx}}\left( {{x^{ - 3}}} \right)$

$\Rightarrow {x^{ - 3}}\left( {0 + 3} \right) + \left( {5 + 3x} \right)\left( {3{x^{ - 3 - 1}}} \right)$

$\Rightarrow 3{x^{ - 3}} + \left( {5 + 3x} \right)\left( { - 3{x^{ - 4}}} \right) = 3{x^{ - 3}} - 15{x^{ - 4}} - 9{x^{ - 3}}$

$\Rightarrow  - 6{x^{ - 3}} - 15{x^{ - 4}} =  - 3{x^{ - 3}}\left( {2 + \frac{5}{x}} \right) = \frac{{ - 3{x^{ - 3}}}}{x}\left( {2x + 5} \right)$

$\Rightarrow \frac{{ - 3}}{{{x^4}}}\left( {5 + 2x} \right)$

(d). ${x^5}\left( {3 - 6{x^{ - 9}}} \right)$

Ans: Let, $f\left( x \right) = {x^5}\left( {3 - 6{x^{ - 9}}} \right)$

By Leibnitz product rule,

$f'\left( x \right) = {x^5}\frac{d}{{dx}}\left( {3 - 6{x^{ - 9}}} \right) + \left( {3 - 6{x^{ - 9}}} \right)\frac{d}{{dx}}{x^5}$

$\Rightarrow {x^5}\left\{ {0 - 6\left( { - 9} \right){x^{ - 9 - 1}}} \right\} + \left( {3 - 6{x^{ - 9}}} \right)\left( {5{x^4}} \right)$

$\Rightarrow {x^5}\left( {54{x^{ - 10}}} \right) + 15{x^4} - 30{x^{ - 5}} = 54{x^{ - 5}} + 15{x^4} - 30{x^{ - 5}}$

$\Rightarrow 24{x^{ - 5}} + 15{x^4} = 15{x^4} + \frac{{24}}{{{x^5}}}$

(e). ${x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)$

Ans: Let $f\left( x \right) = {x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)$

By Leibnitz product rule,

$f'\left( x \right) = {x^{ - 4}}\frac{d}{{dx}}\left( {3 - 4{x^{ - 5}}} \right) + \left( {3 - 4{x^{ - 5}}} \right)\frac{d}{{dx}}\left( {{x^{ - 4}}} \right)$

$\Rightarrow {x^{ - 4}}\left\{ {0 - 4\left( { - 5} \right){x^{ - 5 - 1}}} \right\} + \left( {3 - 4{x^{ - 5}}} \right)\left( { - 4} \right){x^{ - 4 - 1}}$

$\Rightarrow {x^{ - 4}}\left( {20{x^{ - 6}}} \right) + \left( {3 - 4{x^{ - 5}}} \right)\left( { - 4{x^{ - 5}}} \right)$

$\Rightarrow 20{x^{ - 10}} - 12{x^{ - 5}} + 16{x^{ - 10}} = 36{x^{ - 10}} - 12{x^{ - 5}}$

$\Rightarrow \frac{{36}}{{{x^{10}}}} - \frac{{12}}{{{x^5}}}$

(f). $\frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$

Ans: $f\left( x \right) = \frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$

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

By quotient rule,

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

$\Rightarrow \left[ {\frac{{\left( {x + 1} \right)\left( 0 \right) - 2\left( 0 \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right] - \left[ {\frac{{\left( {3x - 1} \right)\left( {2x} \right) - {x^2}\left( 3 \right)}}{{{{\left( {3x - 1} \right)}^2}}}} \right]$

$\Rightarrow \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \left[ {\frac{{6{x^2} - 2x - 3{x^2}}}{{{{\left( {3x - 1} \right)}^2}}}} \right] = \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \left[ {\frac{{3{x^2} - 2x}}{{{{\left( {3x - 1} \right)}^2}}}} \right]$

$\Rightarrow \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \frac{{x\left( {3x - 2} \right)}}{{{{\left( {3x - 1} \right)}^2}}}$

10. Find the derivative of $\cos x$from the first principle

Ans: Let $f\left( x \right) = \cos x$

Accordingly from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos \left( {x + h} \right) - \cos \left( x \right)}}{h}} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos x\cosh  - \sin x\sinh  - \cos x}}{h}} \right] = \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right)}}{h} - \frac{{\sin x\sinh }}{h}} \right]$

$\Rightarrow  - \cos x\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{1 - \cosh }}{h}} \right)} \right] - \sin x\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sinh }}{h}} \right)} \right]$

We know, $\mathop {\lim }\limits_{h \to 0} \frac{{1 - \cosh }}{h} = 0$ and $\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h} = 1$

$\Rightarrow f'\left( x \right) =  - \cos x \times 0 - \sin x \times 1$

$\therefore f'\left( x \right) =  - \sin x$

11. Find the derivative of the functions below:

(i). $\sin x\cos x$

Ans: Let $f\left( x \right) = \sin x\cos x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right)\cos \left( {x + h} \right) - \sin x\cos x}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\sin \left( {x + h} \right)\cos \left( {x + h} \right) - 2\sin x\cos x} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\sin 2\left( {x + h} \right) - \sin 2x} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{2x + 2h + 2x}}{2}.\sin \frac{{2x + 2h - 2x}}{2}} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{4x + 2h}}{2}.\sin \frac{{2h}}{2}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\cos \left( {2x + h} \right)\sinh } \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \cos \left( {2x + h} \right).\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}$

$\Rightarrow \cos \left( {2x + h} \right) \times 1 = \cos 2x$

(ii). $\sec x$

Ans: Let $f\left( x \right) = \sec x$

Accordingly, from the first principle

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\sec \left( {x + h} \right) - \sec x}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\frac{{\left[ { - 2\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\cos \left( {x + h} \right)}}$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}$

$\Rightarrow \frac{1}{{\cos x}} \times 1 \times \frac{{\sin x}}{{\cos x}} = \sec x\tan x$

(iii). $5\sec x + 4\cos x$

Ans: Let, 

$f\left( x \right) = 5\sec x + 4\cos x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \frac{{\lim }}{{h \to 0}}\frac{{5\sec \left( {x + h} \right) + 4\cos \left( {x + h} \right) - \left[ {5\sec x + 4\cos x} \right]}}{h}$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sec \left( {x + h} \right) - \sec x} \right]}}{h} + 4\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \left( {x + h} \right) - \cos x} \right]}}{h}$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right] + 4\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {x + h} \right) - \cos x} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right] + 4\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos x\cosh  - \sin x\sinh  - \cos x} \right]$

$ \Rightarrow \frac{5}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right] + 4\left[ { - \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 - \cos x} \right)}}{h} - \sin x\mathop {\lim \frac{{\sinh }}{h}}\limits_{h \to 0} } \right]$

$ \Rightarrow \frac{5}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\cos \left( {x + h} \right)}} + 4\left[ { - \cos x\left( 0 \right) - \sin x\left( 1 \right)} \right]$

$ \Rightarrow \frac{5}{{\cos x}}\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right] - 4\sin x$

$ \Rightarrow \frac{5}{{\cos x}} \times \frac{{\sin x}}{{\cos x}} \times 1 - 4\sin x = 5\sec x\tan x - 4\sin x$

(iv). $\cos ecx$

Ans: Let $f\left( x \right) = \cos ecx$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x - \sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ { - \cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\sin x\sin \left( {x + h} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right)\mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}$

$ \Rightarrow \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right) \times 1 =  - \cos ecx\cot x$

(v). $3\cot x + 5\cos ecx$

Ans: Let $f\left( x \right) = 3\cot x + 5\cos ecx$

Accordingly, from the first principle

 $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {3\cot \left( {x + h} \right) + 5\cos ec\left( {x + h} \right) - 3\cot x - 5\cos ecx} \right]$

\[ \Rightarrow 3\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cot \left( {x + h} \right) - \cot x} \right] + 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right]\,\,\,\,\,\,\,\,\,\,....\left( 1 \right)\]

Now, $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cot \left( {x + h} \right) - \cot x} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos \left( {x + h} \right)}}{{\sin \left( {x + h} \right)}} - \frac{{\cos x}}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos \left( {x + h} \right)\sin x - \cos x\sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( {x - x - h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( { - h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}.\mathop {\lim }\limits_{h \to 0} \left[ {\frac{1}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow  - 1 \times \frac{1}{{\sin x\sin \left( {x + h} \right)}} = \frac{{ - 1}}{{{{\sin }^2}x}} =  - \cos e{c^2}x\,\,\,\,\,\,......(2)$

$\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x - \sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}}}{{\sin x\sin \left( {x + h} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right)\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}} = \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right).1$

$ \Rightarrow  - \cos ecx\cot x\,\,\,\,\,\,......(3)$

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

$f'\left( x \right) =  - 3\cos e{c^2}x - 5\cos ecx\cot x$

(vi). $5\sin x - 6\cos x + 7$

Ans: Let $f\left( x \right) = 5\sin x - 6\cos x + 7$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {5\sin \left( {x + h} \right) - 6\cos \left( {x + h} \right) + 7 - 5\sin x + 6\cos x - 7} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\sin \left( {x + h} \right) - \sin x} \right] - 6\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {x + h} \right) - \cos x} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{x + h + x}}{2}} \right).\sin \left( {\frac{{x + h - x}}{2}} \right)} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos x\cosh  - \sin x\sinh  - \cos x}}{h}} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{2x + h}}{2}} \right).\sin \left( {\frac{h}{2}} \right)} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right) - \sin x\sinh }}{h}} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right)}}{h} - \frac{{\sin x\sinh }}{h}} \right]$

$ \Rightarrow 5\left[ {\mathop {\lim }\limits_{h \to 0} \cos \left( {\frac{{2x + h}}{2}} \right)} \right]\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right] - 6\left[ { - \cos x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{1 - \cosh }}{h}} \right) - \sin x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}} \right)} \right]$

$ \Rightarrow 5\cos x.1 - 6\left[ {\left( { - \cos x} \right).\left( 0 \right) - \sin x.1} \right]$

$ \Rightarrow 5\cos x + 6\sin x$

(vii). $2\tan x - 7\sec x$

Ans: Let $f\left( x \right) = 2\tan x - 7\sec x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\tan \left( {x + h} \right) - 7\sec \left( {x + h} \right) - 2\tan x + 7\sec x}\right]$

  $ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\tan \left( {x + h} \right) - \tan x} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\sec \left( {x + h} \right) - \sec x} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( {x + h} \right)}}{{\cos \left( {x + h} \right)}} - \frac{{\sin x}}{{\cos x}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos ec\left( {x + h} \right)}} - \frac{1}{{\cos ecx}}} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x\sin \left( {x + h} \right) - \sin x\cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x + h - x}}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sinh }}{h}} \right)\frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}} \right)\left[ {\mathop {\lim }\limits_{h \to 0} \frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right)$

$ \Rightarrow 2 \times 1 \times 1 \times \frac{1}{{\cos x\cos x}} - 7\left( {1 \times \frac{{\sin x}}{{\cos x\cos x}}} \right) = 2{\sec ^2}x - 7\sec x\tan x$

Conclusion

Class 11 Maths Exercise 12.2 Solutions has helped Students understand the concept of limits and how to find them for different functions. You already have practiced using techniques like factoring, rationalizing, and standard limit forms to determine the limit as the input gets close to a specific value. By learning these methods, students now have a good foundation in the basics of calculus. Class 11 Maths Limits and Derivatives Exercise 12.2 will help students understand complex topics as they move on to more advanced topics like derivatives and integrals.

Class 11 Maths Chapter 12: Exercises Breakdown

CBSE Class 11 Maths Chapter 12 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

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