# NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2

## Chapter 13 Ex.13.2 Question 1

Find the derivative of $${x^2} - 2$$ at  $$x = 0.$$

### Solution

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

Accordingly,

\begin{align}f'\left( {10} \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left(10 + h\right) - f\left(10 \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left(10 + h\right)}^2} - 2} \right] - \left( {{10}^2 - 2} \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{{10^2} + 2.10.h + h^2 - 2 - {10^2} + 2}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{20h + {h^2}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \left(20 + h \right)\\&= 20 + 0\\&= 20\end{align}

Thus, the derivative of $${x^2} - 2$$ at $$x = {\rm{1}}0$$ is $$20.$$

## Chapter 13 Ex.13.2 Question 2

Find the derivative of $$\,99x$$ at $$x = 100$$.

### Solution

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

Accordingly,

\begin{align}f'\left( {100} \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left(100 + h\right) - f\left(100 \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{99\left(100 + h\right) - 99\left(100 \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{99 \times 100 + 99h - 99 \times 100}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{99h}{h}\\&= \mathop {\lim }\limits_{h \to 0} \left(99\right)\\&= 99\end{align}

Thus, the derivative of $$\,99x$$ at $$x = {\rm{100}}$$ is $$99.$$

## Chapter 13 Ex.13.2 Question 3

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

### Solution

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

Accordingly,

\begin{align}f{\rm{ '}}\left( 1 \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 + h} \right) - 1}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{h}{h}\\&= \mathop {\lim }\limits_{h \to 0} \left( 1 \right)\\&= 1\end{align}

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

## Chapter 13 Ex.13.2 Question 4

Find the derivative of the following functions from first principle.

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

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

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

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

### Solution

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

Accordingly, from the first principle

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left(x + h\right) - f\left( x \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left(x + h\right)}^3} - 27} \right] - \left( {x^3 - 27} \right)}}{h}\\&= \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} \frac{{h^3 + 3{x^2}h + 3x{h^2}}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \left( {h^3 + 3{x^2}h + 3x{h^2}} \right)\\&= 0 + 3{x^2} + 0\\&= 3{x^2}\end{align}

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

Accordingly, from the first principle

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\&= \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}\\&= \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}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{2hx + h^2 - 3h}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \left(2x + h - 3 \right)\\&= 2x - 3\end{align}

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

Accordingly, from the first principle

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left(x + h \right) - f\left( x \right)}}{h}\\&= \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]\\&= \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(x + h)}^2}} \right]\\&= \mathop {\lim }\limits_{h \to 0} \left[ {\frac{ - h - 2x}{{{x^2}{(x + h)}^2}}} \right]\\&= \frac{{0 - 2x}}{x^2{(x + 0)^2}}\\&= \frac{{ - 2}}{{{x^3}}}\end{align}

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

Accordingly, from the first principle

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}} \right)}}{h}\\&= \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]\\&= \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]\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{- 2h}{{\left(x - 1 \right)\left(x + h - 1 \right)}}} \right]\\&= \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)}\\&= \frac{ - 2}{{\left(x - 1 \right)^2}}\end{align}

## Chapter 13 Ex.13.2 Question 5

For the function

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

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

### Solution

The given function is $$f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + \ldots + \frac{{{x^2}}}{2} + x + 1$$

\begin{align}\frac{d}{dx}f\left( x \right) &= \frac{d}{dx}\left[ {\frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \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) + \dots\\&\quad+ \frac{d}{dx}\left( {\frac{{x^2}}2} \right) + \frac{d}{dx}\left( x \right) + \frac{d}{dx} \qquad \qquad \left( 1 \right)\end{align}

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

\begin{align}\frac{d}{dx}f\left( x \right)& = \frac{100{x^99}}{100} + \frac{99{x^{98}}}{99} + \ldots + \frac{2x}{2} + 1 + 0\\&= {x^99} + {x^98} + \ldots + x + 1\end{align}

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

At $$x = 0,$$

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

At $$x = 1,$$

\begin{align}f'\left( 1 \right) &= {1^99} + {1^98} + \ldots + 1 + 1\\&= {\left[ 1 + 1 + ..... + 1 + 1 \right]_{100{\rm{\,terms }}}}\\&= 1 \times 100\\&= 100f'\left( 0 \right)\end{align}

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

## Chapter 13 Ex.13.2 Question 6

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

### Solution

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

\begin{align}\frac{d}{dx}f\left( x \right) &= \frac{d}{dx}\left( {x^n + a{x^n - 1} + a^2{x^{n - 2}} + \dots + {a^{n - 1}}x + a^n} \right)\\&= \frac{d}{dx}\left(x^n\right) + a\frac{d}{dx}\left( {x^{n - 1}} \right) + {a^2}v\left( {x^{n - 2}} \right) \\&\quad+ ... + a^{n - 1}\frac{d}{dx}\left( x \right) + a^n\frac{d}{dx} \qquad \qquad\left( 1 \right)\end{align}

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

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

Thus, $$f'\left( x \right) = n{x^{n - 1}} + a\left(n - 1\right){x^{n - 2}} + a^2\left(n - 2\right)x^{n - 3} + \ldots + a^{n - 1}$$

## Chapter 13 Ex.13.2 Question 7

For some constants $$a$$ and $$b$$, find the derivative of

(i) $$\left( x - a \right)\left( x - b \right)$$

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

(iii) $$\frac{x - a}{x - b}$$

### Solution

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

Therefore,

\begin{align}f\left( x \right) &= x^2 - \left(a + b\right)x + ab\\&= \frac{d}{dx}\left[ {x^2 - \left(a + b\right)x + ab} \right]\\&= \frac{d}{dx}\left( {x^2} \right) - \left(a + b\right)\frac{dy}{dx}\left( x \right) + \frac{d}{dx}\left( {ab} \right)\end{align}

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

\begin{align}f\left( x \right) &= {\rm{2}}x - \left( {a + b} \right) + 0\\&= 2x - a - b\end{align}

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

Therefore,

\begin{align}f\left( x \right) &= {a^2}{x^4} + 2ab{x^2} + {b^2}\\ &= \frac{d}{{dx}}\left( {{a^2}{x^4} + 2ab{x^2} + {b^2}} \right)\\ &= {a^2}\frac{d}{{dx}}\left( {{x^4}} \right) + 2ab\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}{b^2}\end{align}

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

\begin{align}f\left( x \right) &= {a^2}\left( {4{x^3}} \right) + 2ab\left( {2x} \right) + {b^2}\left( 0 \right)\\ &= 4{a^2}{x^3} + 4abx\\ &= 4ax\left( {a{x^2} + b} \right)\end{align}

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

Therefore,

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

By quotient rule

\begin{align}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}}\\ &= \frac{{\left(x - b\right)(1) - \left( {x - a} \right)(1)}}{{{\left(x - b\right)}^2}}\\ &= \frac{x - b - x + a}{{{\left(x - b\right)^2}}}\\ &= \frac{a - b}{{\left(x - b\right)^2}}\end{align}

## Chapter 13 Ex.13.2 Question 8

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

### Solution

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

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

By quotient rule,

\begin{align}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}}}\\ &= \frac{{\left(x - a\right)\left( {n{x^n - 1} - 0} \right) - \left( {x^n - a^n} \right)}}{{{{\left(x - a \right)}^2}}}\\ &= \frac{{n{x^n} - an{x^{n - 1}} - x^n + a^n}}{{{\left(x - a\right)}^2}}\end{align}

## Chapter 13 Ex.13.2 Question 9

Find the derivative of

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

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

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

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

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

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

### Solution

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

\begin{align}f'\left( x \right) &= \frac{d}{dx}\left( 2x - \frac{3}{4} \right)\\&= 2\frac{d}{dx}\left( x \right) - \frac{d}{dx}\left( \frac{3}{4} \right)\\&= 2 - 0\\&= 2\end{align}

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

By Leibnitz product rule,

\begin{align}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)\\&= \left( {5{x^3} + 3x - 1} \right)\left( 1 \right) + \left( {x - 1} \right)\left( {5.3{x^2} + 3 - 0} \right)\\&= \left( {5{x^3} + 3x - 1} \right) + \left( {x - 1} \right)\left( {15{x^2} + 3} \right)\\&= 5{x^3} + 3x - 1 + 15{x^3} + 3x - 15{x^2} - 3\\&= 20{x^3} - 15{x^2} + 6x - 4\end{align}

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

By Leibnitz product rule,

\begin{align}f'\left( x \right) &= x^{ - 3}\frac{d}{{dx}}\left(5 + 3x \right) + \left(5 + 3x\right)\frac{d}{{dx}}\left( {x^{ - 3}} \right)\\&= x^{ - 3}\left( {0 + 3} \right) + \left(5 + 3x\right)\left( { - 3{x^{ - 3 - 1}}} \right)\\&= {x^{ - 3}}\left( 3 \right) + \left( 5 + 3x \right)\left( - 3{x^{ - 4}} \right)\\&= 3{x^{ - 3}} - 15{x^{ - 4}} - 9{x^{ - 3}}\\&= - 6{x^lig{ - 3}} - 15{x^{ - 4}}\\&= - 3{x^{ - 3}}\left( {2 + \frac{5}{x}} \right)\\&= \frac{{ - 3{x^{ - 3}}}}{x}\left( {2x + 5} \right)\\&= \frac{{ - 3}}{{{x^4}}}\left( {5 + 2x} \right)\end{align}

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

By Leibnitz product rule,

\begin{align}f'\left( x \right) &= x^5\frac{d}{dx}\left( 3 - 6{x^{ - 9}} \right) + \left(3 - 6{x^{ - 9}} \right)\frac{d}{{dx}}\left(x^5 \right)\\&= x^5\left\{ {0 - 6\left(- 9 \right){x^{ - 9 - 1}}} \right\} + \left( {3 - 6{x^{ - 9}}} \right)\left( {5{x^4}} \right)\\&= {x^5}\left( 54{x^{ - 10}} \right) + 15{x^4} - 30{x^{ - 5}}\\&= 24{x^{ - 5}} + 15{x^4}\\&= 15{x^4} + \frac{{24}}{{{x^5}}}\end{align}

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

By Leibnitz product rule,

\begin{align}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)\\&= x^{ - 4}\left\{ {0 - 4\left( - 5 \right){x^{ - 5 - 1}}} \right\} + \left( {3 - 4{x^{ - 5}}} \right)\left( - 4\right){x^{ - 4 - 1}}\\&= {x^{ - 4}}\left( {20{x^{ - 6}}} \right) + \left( {3 - 4{x^{ - 5}}} \right)\left( { - 4{x^{ - 5}}} \right)\\&= 20{x^{ - 10}} - 12{x^{ - 5}} + 16{x^{ - 10}}\\&= 36{x^{ - 10}} - 12{x^{ - 5}}\\&= \frac{12}{{x^5}} + \frac{36}{{x^{10}}}\end{align}

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

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

By quotient rule,

\begin{align}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]\\&= \left[ {\frac{{\left( {x + 1} \right)\left( 0 \right) - 2\left( 1 \right)}}{(x + 1)^2}} \right] - \left[ {\frac{{\left( {3x - 1} \right)\left( {2x} \right) - {x^2}\left( 3 \right)}}{\left( {3x - 1} \right)}^2} \right]\\&= \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} - \frac{{x\left( {3x - 2} \right)}}{{\left(3x - 1\right)}^2}\end{align}

## Chapter 13 Ex.13.2 Question 10

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

### Solution

Let $$f\left( x \right) = \cos x$$.

Accordingly, from the first principle,

\begin{align} {f}'\left( x \right)&=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h} \\& =\underset{h\to 0}{\mathop{\lim }}\,\left[ \frac{\cos \left( x+h \right)-\cos \left( x \right)}{h} \right] \\& =\underset{h\to 0}{\mathop{\lim }}\,\left[ \frac{\cos x\cos h-\sin x\sin h-\cos x}{h} \right] \\& =\underset{h\to 0}{\mathop{\lim }}\,\left[ \frac{-\cos x\left( 1-\cos h \right)}{h}\frac{-\sin x\sin h}{h} \right] \\& =-\cos x\left[ \underset{h\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos h}{h} \right) \right]-\sin x\left[ \underset{h\to 0}{\mathop{\lim }}\,\left( \frac{\sin h}{h} \right) \right] \\& =-\cos x\left( 0 \right)-\sin x\left( 1 \right) \qquad \quad \left[ \because \underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cosh }{h}=0\text{ }and\text{ }\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h}=1 \right] \\& =-\sin x\end{align}

## Chapter 13 Ex.13.2 Question 11

Find the derivative of following functions:

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

(ii) $$\sec x$$

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

(iv) $${\rm{cosec}}\;x$$

(v) $$3\cot x + 5\;{\rm{cosec}}\;x$$

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

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

### Solution

(i) Let$$f\left( x \right) = \sin x\cos x$$.

Accordingly, from the first principle,

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right)\cos \left( {x + h} \right) - \sin x\cos x}}{h}\\&= \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]\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\sin 2\left( {x + h} \right) - \sin 2x} \right]\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{2x + 2h + 2x}}{2}.\sin \frac{{2x + 2h - 2x}}{h}} \right]\\&= \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}{h}\left[ {\cos \left( {2x + h} \right)\sin h} \right]\end{align}

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \cos \left( {2x + h} \right).\mathop {\lim }\limits_{h \to 0} \frac{{\sin h}}{h}\\&= \cos \left( {2x + h} \right).1\\&= \cos 2x\end{align}

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

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{\sec \left( {x + h} \right) - \sec x}{h}\\&= \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]\\&= \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]\\&= \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]\\&= \frac{1}{\cos x}.\mathop {\lim }\limits_{h \to 0} \frac{1}{2h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}}}\cos \left( {x + h} \right)} \right]\\&= \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)}}\\&= \frac{1}{\cos x}.1.\frac{\sin x}{\cos x}\\&= \sec x\tan x\end{align}

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

Accordingly, from the first principle,

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\ &= \mathop {\lim }\limits_{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}\\ &= 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}\\ &= 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 x\cos h - \sin x\sin h - \cos x} \right]\\ &= \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 \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 - \cos x} \right)}}{h} - \sin x\mathop {\lim }\limits_{h \to 0} \frac{{\sin h}}{h}} \right]\\ &= \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]\\ &= \frac{5}{{\cos x}}.\mathop {\lim }\limits_{h \to 0} \left[ {\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\\ &= \frac{5}{{\cos x}}.\frac{{\sin x}}{{\cos x}}.1 - 4\sin x\\ &= 5\sec x\tan x - 4\sin x\end{align}

(iv) Let $$f\left( x \right) = {\rm{cosec}}\;x.$$

Accordingly, from the first principle,

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{f\left(x + h \right) - f\left( x \right)}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {{\rm{cosec}}\left( {x + h} \right) - {\rm{cosec}}\;x} \right]\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]\\&= \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]\end{align}

\begin{align}f'\left( x \right) &= \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{{\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)}}\\&\mathop {\lim }\limits_{h \to 0} \left( {\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)}}} \right).\mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\\&= \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right).1\\&= - {\rm{cosec}}\;x\cot x\end{align}

(v) Let $$f\left( x \right) = 3\cot x + 5\;{\rm{cosec}}\;x.$$

Accordingly, from the first principle

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {3\cot \left( {x + h} \right) + 5\;{\rm{cosec}}\left( {x + h} \right) - 3\cot x - 5\;{\rm{cosec}}\;x} \right]\\&= 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[ {{\rm{cosec}}\left( {x + h} \right) - {\rm{cosec}}\;x} \right] \qquad \ldots \left(1\right)\end{align}

Now,

\begin{align}\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]\\&= \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]\\&= \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]\\&= - 1.\frac{1}{{\sin x\sin \left( {x + 0} \right)}} = \frac{{ - 1}}{{\sin }^2}x = - {\rm{cose}}{{\rm{c}}^2}x \qquad \ldots \left( {\rm{2}} \right)\end{align}

\begin{align}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {{\rm{cosec}}\left( {x + h} \right) - {\rm{cosec}}\;x} \right] &= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]\\&= \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{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]\\&= \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)}}\\&= \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\\&= - {\rm{cosec}}\;x\cot x \qquad \ldots \left( 3 \right)\end{align}

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

$$f'\left( x \right) = - 3\;{\rm{cose}}{{\rm{c}}^2}x - 5\;{\rm{cosec}}\;x\cot x$$

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

Accordingly, from the first principle,

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\&= \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]\\&= 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]\\&= 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]\\&\quad - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos x\cos h - \sin x\sin h - \cos x}}{h}} \right]\\&= 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] \\&\quad- 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cos h} \right) - \sin x\sin h}}{h}} \right]\end{align}

\begin{align}f'\left( x \right) &= 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]\\&\quad - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cos h} \right)}}{h} - \frac{{\sin x\sin h}}{h}} \right]\\&= 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]\\&\quad - 6\left[ { - \cos x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos h}}{h}} \right) - \sin x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin h}}{h}} \right)} \right]\\&= 5\cos x.1 - 6\left[ {\left( { - \cos x} \right).\left( 0 \right) - \sin x.1} \right]\\&= 5\cos x + 6\sin x\end{align}

(vii) Let $$f\left( x \right) = 2\tan x - 7\sec x.$$

Accordingly, from the first principle,

\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\&= \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]\\&= 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]\\&= 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] \\& \quad- 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right]\\&= 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] \\& \quad- 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]\\&= 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x + h - x}}{{\cos x\cos \left( {x + h} \right)}}} \right] \\& \quad- 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]\\&= 2\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sinh }}{h}} \right)\frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] \\& \quad - 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]\\&= 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] \\& \quad - 7\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right)\\&= 2 \times 1 \times 1\frac{1}{{\cos x\cos x}} - 7 \times 1\left( {\frac{{\sin x}}{{\cos x\cos x}}} \right)\\&= 2{\sec ^2}x - 7\sec x\tan x\end{align}

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