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Question
Differentiate log (1 + x2) with respect to tan−1 x ?
Solution
\[\text { Let u } = \log\left( 1 + x^2 \right) \text { and v } = \tan^{- 1} x \]
\[ \Rightarrow \frac{du}{dx} = \frac{1}{\left( 1 + x^2 \right)}\frac{d}{dx}\left( 1 + x^2 \right) = \frac{2x}{\left( 1 + x^2 \right)} \text { and } \frac{dv}{dx} = \frac{1}{1 + x^2} \]
\[ \therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2x}{1 + x^2} \times \frac{1 + x^2}{1} = 2x\]
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