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#### Provirde solution RD Sharma maths class 12 chapter 10 differentiationability  exercise 10 .1 question 1 maths textbook solution

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

Hint:

Use first principle to find the differentiation of

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

Given:

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

Solution:

Let

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

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

So, we will use formula of differentiation by first principle,

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

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

$=\lim_{h\rightarrow 0}\frac{e^{-x}\times e^{-h}-e^{-x}}{h}$

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

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

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

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

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

Hence, the differentiation of $e^{-x}$ is $\left ( -e^{-x} \right )$