Question

Differentiate

${\tan }^{-1}\left(e^x\right)$

Answer 1

$\mathrm{Let}y={\tan }^{-1}\left(e^x\right)$
Differentiate it with respect to x we get,
$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\left({\tan }^{-1}{e}^x\right) \\ =\frac{1}{1+{\left(e^x\right)}^2}\frac{d}{dx}\left(e^x\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{1}{1+e^{2x}}\times e^x \\ =\frac{e^x}{1+e^{2x}} \\ \mathrm{So},\frac{d}{dx}\left({\tan }^{-1}{e}^x\right)=\frac{e^x}{1+e^{2x}} \end{gathered}$$