Question

Differentiate from first principle:

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

Answer 1

Given:

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

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in the above expression, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{3(x+h)}-e^{3x}}{h}$


$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{3x}{e}^{3h}-e^{3x}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=3\underset{h\to 0}{lim}\frac{e^{3x}(e^{3h}-1)}{3h}$

$\Rightarrow f^{\prime }\left(x\right)=3e^{3x}\underset{h\to 0}{lim}\frac{e^{3h}-1}{3h}$


$\Rightarrow f^{\prime }\left(x\right)=3e^{3x}\left(1\right)(\because \underset{x\to 0}{lim}\frac{e^x-1}{x}=1)$


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

Therefore, the derivative of $e^{3x}\text{is}3e^{3x}.$