#### Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 5 maths textbook solution

$\frac{e\sqrt{2x}}{\sqrt{2x}}$

Hint:

Use first principle formula to find the differentiation

Given:

$e\sqrt{2x}$

Solution:

Let,

$f\left ( x \right )=e^{\sqrt{2x}}$
$f\left ( x+h \right )=e^{\sqrt{2\left ( x+h \right )}}$

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

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

$\frac{d}{dx}\left ( e^{\sqrt{2x}} \right )=\lim_{h\rightarrow 0}\frac{e^{\sqrt{2\left ( x+h \right )}}-e^{\sqrt{2x}}}{h}$

Take 2$e^{\sqrt{2x}}$ common from numerator

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

$=\lim_{h\rightarrow 0}\cdot e^{\sqrt{2x}}\frac{\left [ \left ( \left [ e^{\sqrt{2\left ( x+h \right )}} \right ] \right ) -1\right ]}{h}$                        $\left [ \because \frac{a^{m}}{a^{n}} =a^{m-n}\right ]$

Multiply and divide by $\left ( \sqrt{2\left ( x+h \right )}-\sqrt{2x} \right )$

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

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

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

Now, $e^{\sqrt{2x}}$ is a function of x and unit applied on variable ‘h’. So,$e^{\sqrt{2x}}$  can be taken outside became w.r.t variable ‘h’, x will remain constant

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

Rationalising the numerator

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

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

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

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

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

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

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

$=\frac{ e^{\sqrt{2x}}}{\sqrt{2x}}$

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