Question

Differentiate

$\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$

Answer 1

$\mathrm{Let}y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$
Differentiate with respect to x we get,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left[\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right] \\ =\left[\frac{\left(e^{2x}-e^{-2x}\right){\displaystyle \frac{d}{dx}}\left(e^{2x}+e^{-2x}\right)-\left(e^{2x}+e^{-2x}\right){\displaystyle \frac{d}{dx}}\left(e^{2x}-e^{-2x}\right)}{{\left(e^{2x}-e^{-2x}\right)}^2}\right]\left[\mathrm{Using}\mathrm{quotient}\mathrm{rule}\mathrm{and}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{\left(e^{2x}-e^{-2x}\right)\left[e^{2x}{\displaystyle \frac{d}{dx}}\left(2x\right)+e^{-2x}{\displaystyle \frac{d}{dx}}\left(-2x\right)\right]-\left(e^{2x}+e^{-2x}\right)\left[e^{2x}\frac{d}{dx}\left(2x\right)-e^{-2x}\frac{d}{dx}\left(-2x\right)\right]}{{\left(e^{2x}-e^{-2x}\right)}^2} \\ =\frac{\left(e^{2x}-e^{-2x}\right)\left(2e^{2x}-2e^{-2x}\right)-\left(e^{2x}+e^{-2x}\right)\left(2e^{2x}+2e^{-2x}\right)}{{\left(e^{2x}-e^{-2x}\right)}^2} \\ =\frac{2{\left(e^{2x}-e^{-2x}\right)}^2-2{\left(e^{2x}+e^{-2x}\right)}^2}{{\left(e^{2x}-e^{-2x}\right)}^2} \\ =\frac{2\left[e^{4x}+e^{-4x}-2e^{2x}{e}^{-2x}-e^{4x}-e^{-4x}-2e^{2x}{e}^{-2x}\right]}{{\left(e^{2x}-e^{-2x}\right)}^2} \\ =\frac{-8}{{\left(e^{2x}-e^{-2x}\right)}^2} \\ \mathrm{So},\frac{d}{dx}\left(\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right)=\frac{-8}{{\left(e^{2x}-e^{-2x}\right)}^2} \end{gathered}$$