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Answer:

$\left ( 3e^{3x} \right )$

Hint:

Use first principle to find the differentiation of $\left ( e^{3x} \right )$

Given:

$\left ( e^{3x} \right )$

Solution:

Let

$f\left ( x \right )= e^{3x}$

$f\left ( x+h \right )= e^{3\left ( x+h \right )}$

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

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

$\frac{d}{dx}\left ( e^{3x} \right )=\lim_{h\rightarrow 0}\frac{e^{3\left ( x+h \right )}-e^{3x}}{h}$

$=\lim_{h\rightarrow 0}\frac{e^{3x}\times e^{3h}-e^{3x}}{h}$

$=\lim_{h\rightarrow 0}\frac{e^{3x}\left ( e^{3h}-1 \right )}{h}$

Multiply and divide by 3

$=\lim_{h\rightarrow 0}e^{3x}\frac{\left ( e^{3h}-1 \right )}{3h}\times 3$

$=\lim_{h\rightarrow 0}e^{3x}\times 1\times 3$                                          $\left [ \because \lim_{h\rightarrow 0} \left ( \frac{e^{x}-1}{x} \right )=1\right ]$

$=3e^{3x}$

Hence, the differentiation of $e^{3x}$ is $3e^{3x}$

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