Question

Differentiate the following function w.r.t $x$
$\sqrt{x+\frac{1}{x}}$

Answer 1

$=\frac{d}{dx}\left(\sqrt{x+\frac{1}{x}}\right)$

$=\frac{1}{2}{(x+\frac{1}{x})}^{\frac{1}{2}-1}\cdot \frac{d}{dx}[x+\frac{1}{x}]$

$=\frac{\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[\frac{1}{x}\right]}{2\sqrt{x+\frac{1}{x}}}$

$=\frac{1-\frac{\frac{d}{dx}\left[x\right]}{x^2}}{2\sqrt{x+\frac{1}{x}}}$

$=\frac{1-\frac{1}{x^2}}{2\sqrt{x+\frac{1}{x}}}$