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Question
Differentiate \[\tan^{- 1} \left( e^x \right)\] ?
Solution
\[\text{Let} y = \tan^{- 1} \left( e^x \right)\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \tan^{- 1} e^x \right)\]
\[ = \frac{1}{1 + \left( e^x \right)^2}\frac{d}{dx}\left( e^x \right) \left[ \text{Using chain rule} \right]\]
\[ = \frac{1}{1 + e^{2x}} \times e^x \]
\[ = \frac{e^x}{1 + e^{2x}}\]
\[So, \frac{d}{dx}\left( \tan^{- 1} e^x \right) = \frac{e^x}{1 + e^{2x}}\]
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