Question

Find the differential coefficient with respect to $x$ of $\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$.

Answer 1

Suppose that
$y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$
$\frac{dy}{dx}=\frac{d}{dx}\left[\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right]$
$=\frac{(e^{2x}-e^{-2x})\frac{d}{dx}(e^{2x}+e^{-2x})-(e^{2x}+e^{-2x})\frac{d}{dx}(e^{2x}-e^{-2x})}{(e^{2x}-e^{-2x}{)}^2}$
$=\frac{(e^{2x}-e^{-2x})(2e^{2x}-2e^{-2x})-(e^{2x}+e^{-2x})(2e^{2x}+2e^{-2x})}{(e^{2x}-e^{-2x}{)}^2}$
$=\frac{2\left[\right(e^{2x}-e^{-2x}{)}^2-(e^{2x}+e^{-2x}{)}^2]}{(e^{2x}-e^{-2x}{)}^2}$
$=\frac{2[e^{4x}+e^{-4x}-2-e^{4x}-e^{-4x}-2]}{(e^{2x}-e^{-2x}{)}^2}$
$=\frac{2\times (-4)}{(e^{2x}-e^{-2x}{)}^2}$
$=\frac{-8}{(e^{2x}-e^{-2x}{)}^2}$.