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प्रश्न
If \[y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] write the value of \[\frac{dy}{dx}\text { for } x > 1\] ?
उत्तर
\[\text { We have, y } = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\]
\[\text { Putting x } = \tan\theta\]
\[ \Rightarrow 1 < \tan\theta < \infty \]
\[ \Rightarrow \frac{\pi}{4} < \theta < \frac{\pi}{2} \]
\[ \Rightarrow \frac{\pi}{2} < 2\theta < \pi\]
\[ \therefore y = \sin^{- 1} \left( \sin2\theta \right) \]
\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\left( \pi - 2\theta \right) \right\} \]
\[ \Rightarrow y = \pi - 2\theta \]
\[ \Rightarrow y = \pi - 2 \tan^{- 1} x\]
\[ \Rightarrow \frac{dy}{dx} = 0 - \frac{2}{1 + x^2} \]
\[ \Rightarrow \frac{dy}{dx} = - \frac{2}{1 + x^2} \]
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