Question

Differentiate the following functions from first principles:

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

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(x\right)=e^{\sqrt{2x}} \\ \Rightarrow f\left(x+h\right)=e^{\sqrt{2\left(x+h\right)}} \\ \therefore \frac{d}{dx}\left\{f\left(x\right)\right\}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{{}^{\sqrt{2\left(x+h\right)}}}-e^{{}^{\sqrt{2x}}}}{h} \\ =\underset{h\to 0}{\lim }{e}^{\sqrt{2x}}\left[\frac{e^{\sqrt{2\left(x+h\right)}-\sqrt{2x}}-1}{h}\right] \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\left[\frac{e^{\sqrt{2\left(x+h\right)}-\sqrt{2x}}-1}{\sqrt{2\left(x+h\right)}-\sqrt{2x}}\right]\times \underset{h\to 0}{\lim }\frac{\sqrt{2\left(x+h\right)}-\sqrt{2x}}{h} \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{\sqrt{2\left(x+h\right)}-\sqrt{2x}}{h}\left[\because \underset{h\to 0}{\lim }\frac{e^h-1}{h}=1\right] \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{\sqrt{2\left(x+h\right)}-\sqrt{2x}}{h}\times \frac{\sqrt{2\left(x+h\right)}+\sqrt{2x}}{\sqrt{2\left(x+h\right)}+\sqrt{2x}}\left[\mathrm{Rationalising}\mathrm{the}\mathrm{numerator}\right] \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{2\left(x+h\right)-2x}{h\left(\sqrt{2\left(x+h\right)}+\sqrt{2x}\right)} \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{2x+2h-2x}{h\left(\sqrt{2\left(x+h\right)}+\sqrt{2x}\right)} \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{2h}{h\left(\sqrt{2\left(x+h\right)}+\sqrt{2x}\right)} \\ =e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{2}{\left(\sqrt{2\left(x+h\right)}+\sqrt{2x}\right)} \\ =\frac{e^{\sqrt{2x}}}{\sqrt{2x}} \\ \mathrm{Hence},\frac{d}{dx}\left(e^{\sqrt{2x}}\right)=\frac{e^{\sqrt{2x}}}{\sqrt{2x}} \end{gathered}$$