Question

Differentiate from first principle:

$\left(ix\right)e^{\sqrt{2x}}$

Answer 1

Given:

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

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 above expression, we get:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{\sqrt{2x}}(e^{\sqrt{2(x+h)}-\sqrt{2x}}-1)}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{e^{\sqrt{2x}}(e^{\left(\frac{2(x+h)-2x}{\sqrt{2(x+h)}+\sqrt{2x}}\right)}-1)}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(e^{\left(\frac{2h}{\sqrt{2(x+h)}+\sqrt{2x}}\right)}-1)}{\frac{2h}{\sqrt{2(x+h)}+\sqrt{2x}}}\times \frac{2e^{\sqrt{2x}}}{\sqrt{2(x+h)}+\sqrt{2x}}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{2e^{\sqrt{2x}}}{\sqrt{2(x+0)}+\sqrt{2x}}[\because {lim}_{x\to 0}\frac{e^x-1}{x}=1]$

$\Rightarrow f^{\prime }\left(x\right)=\frac{e^{\sqrt{2x}}}{\sqrt{2x}}$

Therefore, the derivative of $e^{\sqrt{2x}}$ is $\frac{e^{\sqrt{2x}}}{\sqrt{2x}}$