#### Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 7 maths textbook solution

Ex_10.1_Q_7

$\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$

Hint:

Use first principle formula to find the differentiation

Given:

$e^{\sqrt{\cot x}}$

Solution:

Let,

\begin{aligned} &f(x)=e^{\sqrt{\cot x}} \\ &f(x+h)=e^{\sqrt{\cot (x+h)}} \end{aligned}

Now, we will use the formula of first principle to find the differentiation

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot (x+h)}}-e^{\sqrt{\cot x}}}{h} \end{aligned}

Take $e^{\sqrt{\cot x}}$ common from numerator

$=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(\frac{e^{\sqrt{\cot (x+h)}}}{e^{\sqrt{\cot x}}}-1\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right)}{h}$                                   $\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right]$

$\because e^{\sqrt{\cot x}}$ Is a function of variable x and limit is applied on variable ‘h’

So,

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{e^{(\sqrt{\cot (x+h)}-\sqrt{\cot x})}-1}{h}$

Multiply and divide by  $(\sqrt{\cot (x+h)}-\sqrt{\cot x})$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\left[e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right]}{(\sqrt{\cot (x+h)}-\sqrt{\cot x})} \times\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}\right]$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0}(1) \times\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}\right]$                                           $\left[\because \lim _{h \rightarrow 0} \frac{e^{x}-1}{x}=1\right]$

Rationalising the numerator

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0}\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h} \times \frac{\sqrt{\cot (x+h)}+\sqrt{\cot x}}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}\right]$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$                                          $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$$\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\frac{\cot (x+h) \times \cot x+1}{\cot (x+h-x)}}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$                            $\left[\cot (A+B)=\frac{\cot A \cot B+1}{\cot A-\cot B}\right]$

$=\left(e^{\sqrt{\cot x}}\right) \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{h(\operatorname{coth})(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{\left(\frac{h}{\tanh }\right)(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$                         $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\tanh }{h} \times\left[\frac{\cot (x+h) \cot x+1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}\right]$

$=e^{\sqrt{\cot x}} \times 1 \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$                              $\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]$

$=e^{\sqrt{\cot x}} \times \frac{\cot ^{2} x+1}{2 \sqrt{\cot x}}$

$=e^{\sqrt{\cot x}} \times \frac{\cos e c^{2} x}{2 \sqrt{\cot x}}$                                                    $\left[\because 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\right]$

$\therefore \frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$

Hence, the differentiation of $e^{\sqrt{\cot x}}$ is $\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$