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प्रश्न
If \[y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), 0 < x < 1,\] prove that \[\frac{dy}{dx} = \frac{4}{1 + x^2}\] ?
उत्तर
\[\text{ Let, y } = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) + se c^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right)\]
\[ \Rightarrow y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) + \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)\]
\[\text{ Put, x } = \tan\theta\]
\[ \therefore y = \sin^{- 1} \left( \frac{2 \tan\theta}{1 + \tan^2 \theta} \right) + \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \]
\[ \Rightarrow y = \sin^{- 1} \left( \sin 2\theta \right) + \cos^{- 1} \left( \cos 2\theta \right) . . . \left( i \right)\]
\[\text{ Here}, 0 < x < 1\]
\[ \Rightarrow 0 < \tan\theta < 1\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]
\[ \Rightarrow 0 < 2\theta < \frac{\pi}{2}\]
\[\text{ So, from equation} \left( i \right), \]
\[ y = 2\theta + 2\theta ........[\text{ Since}, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right], \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right]\]
\[ \Rightarrow y = 4\theta\]
\[ \Rightarrow y = 4 \tan^{- 1} x ...........\left[ \text{ Since}, x = \tan\theta \right]\]
Differentiate it with respect to x,
\[\therefore \frac{d y}{d x} = \frac{4}{1 + x^2}\]
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