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