Differentiate the following functions from first principles :

We have to find the derivative of $\mathrm{e}^{\sqrt{2} x}$ with the first principle method, so,

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

by using the first principle formula, we get,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2 x}}}{h}$

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

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

[By using $\left.\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{2 x}}}{h} \times(\sqrt{2(x+h)}-\sqrt{2 x}) \times \frac{\sqrt{2(x+h)}+\sqrt{2 x}}{\sqrt{2(x+h)}+\sqrt{2 x}}$

[By rationalising]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{2 x}}}{h} \times \frac{(2(x+h)-2 x)}{\sqrt{2(x+h)}+\sqrt{2 x}}$

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