Question

Differentiate \[\tan^{- 1} \left( e^x \right)\] ?

Answer 1

\[\text{Let} y = \tan^{- 1} \left( e^x \right)\]

Differentiate it with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left( \tan^{- 1} e^x \right)\]

\[ = \frac{1}{1 + \left( e^x \right)^2}\frac{d}{dx}\left( e^x \right) \left[ \text{Using chain rule} \right]\]

\[ = \frac{1}{1 + e^{2x}} \times e^x \]

\[ = \frac{e^x}{1 + e^{2x}}\]

\[So, \frac{d}{dx}\left( \tan^{- 1} e^x \right) = \frac{e^x}{1 + e^{2x}}\]