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Question
If \[u = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ and v} = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] where \[- 1 < x < 1\], then write the value of \[\frac{du}{dv}\] ?
Solution
\[\text{ We have, u } = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ and }v = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{1 + x^2} \text{ and} \frac{dv}{dx} = \frac{2}{1 + x^2} \]
\[ \therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2} = 1\]
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