#### Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 1 maths textbook solution

$ae^{\left ( ax+b \right )}$

Hint:

Use first principle to find the differentiation

Given:

$\left ( e^{ax+b} \right )$

Solution:

Let

$f\left ( x \right )=e^{ax+b}$

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

Now, we will use formula of differentiation by first principle

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

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

$=\lim_{h\rightarrow 0}\frac{e^{ax+ah+b}-e^{ax+b}}{h}$

$=\lim_{h\rightarrow 0}\frac{\left ( e^{ax+b} \right )\left ( e^{ah} \right )-e^{ax+b}}{h}$

$=\lim_{h\rightarrow 0}\frac{e^{ax+b}\left ( e^{ah}-1 \right )}{h}$

Multiply and divide by a

$=\lim_{h\rightarrow 0}e^{ax+b}\frac{\left ( e^{ah}-1 \right )}{ah}\times a$

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

$=ae^{ax+b}$

Hence, the differentiation of $e^{ax+b}$ is $ae^{ax+b}$