Question

Let  \[\cup = \sin^{- 1} \left( \frac{2 x}{1 + x^2} \right) \text { and }V = \tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text { then } \frac{d \cup}{dV} =\] ____________ .

विकल्प

• 1/2

• x

• \[\frac{1 - x^2}{x^2 - 4}\]

• 1

Answer 1

1 

\[\text { We have, u }= \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text { and }v = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{1 + x^2} \text { and } \frac{dv}{dx} = \frac{2}{1 + x^2} \]

\[\therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2} = 1\]