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