#### Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 4 maths textbook solution

$-\left ( \sin x \right )\cdot e^{\cos x}$

Hint:

Use first principle to find the differentiation

Given:

$\left ( e^{\cos x} \right )$

Solution:

Let

$f\left ( x \right )= e^{\cos x}$

$f\left ( x+h \right )= e^{\cos \left ( x+h \right )}$

Now, we will use formula of differentiation by first principle

$\frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$

$\frac{d}{dx}\left ( e^{\cos x}\right )=\lim_{h\rightarrow 0}\frac{e^{\cos \left ( x+h \right )}-e^{\cos x}}{h}$

Take $e^{\cos x}$ common from numerator

$=\lim_{h\rightarrow 0}\left ( e^{\cos x} \right )\frac{\left [ \frac{e^{cos\left ( x+h \right )}}{e^{\cos x}}-1 \right ]}{h}$

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

Multiply and divide $\left [ \cos \left ( x+h \right )-\cos x \right ]$

$=\lim_{h\rightarrow 0}\frac{e^{\cos x}\left [ e^{\cos \left ( x+h \right )-\cos x} -1\right ]}{\left ( \cos \left ( x+h \right )-\cos x \right )\times h}\times\left [ \cos \left ( x+h \right )-\cos x \right ]$

$=\lim_{h\rightarrow 0}e^{\cos x}\times \left ( \frac{\cos \left ( x+h \right )-\cos x}{h} \right )$                                     $\left [ \because \lim_{h\rightarrow 0}\frac{e^{x}-1}{x}=1 \right ]$

$=\lim_{h\rightarrow 0}e^{\cos x}\times \left ( \frac{\cos \left ( x+h \right )-\cos x}{h} \right )$                $\left [ \because \cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2} \right ]$

$=\lim_{h\rightarrow 0}e^{\cos x}\times \left [ \frac{-2\sin \frac{x+h+x}{2}\times \sin \frac{x+h-x}{2}}{h} \right ]$

$e^{\cos x}$ Is a function of $f\left ( x \right )$ and limit is applied on variable ‘h’

So,$e^{\cos x}$ can be taken outside

$= e^{\cos x}\times \lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2x+h}{2}\ \right )\times \sin \left ( \frac{h}{2} \right )}{h}$

Multiply the denominator by 2 and divide the denominator by 2

$= e^{\cos x}\times \lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2x+h}{2} \right )}{2}\times \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}$

$= e^{\cos x}\times \lim_{h\rightarrow 0}\left [ -\sin \left ( \frac{2x+h}{2} \right ) \right ]\times 1$                                        $\left [ \because \lim_{h\rightarrow 0}\frac{\sin x}{x}=1 \right ]$

$=e^{\cos x}\times\left ( -\sin x \right )=-\sin xe^{\cos x}$

Hence, the differentiation of $e^{\cos x}$ is $\left [ -\sin xe^{\cos x} \right ]$