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Question
Let \[\cup = \sin^{- 1} \left( \frac{2 x}{1 + x^2} \right) \text { and }V = \tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text { then } \frac{d \cup}{dV} =\] ____________ .
Options
1/2
x
\[\frac{1 - x^2}{x^2 - 4}\]
1
Solution
1
\[\text { We have, u }= \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text { and }v = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{1 + x^2} \text { and } \frac{dv}{dx} = \frac{2}{1 + x^2} \]
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