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प्रश्न
Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
उत्तर
\[\text{ Let, y } = \sin^{- 1} \left\{ \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right\}\]
\[\text{ putting } x = \sin\theta\]
\[ \therefore y = \sin^{- 1} \left( \frac{\sin\theta + \sqrt{1 - \sin^2 \theta}}{\sqrt{2}} \right)\]
\[ \Rightarrow y = \sin^{- 1} \left( \frac{\sin\theta + \cos\theta}{\sqrt{2}} \right)\]
\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\theta\left( \frac{1}{\sqrt{2}} \right) + \cos\theta\left( \frac{1}{\sqrt{2}} \right) \right\}\]
\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\theta \cos\frac{\pi}{4} + \cos\theta \sin\frac{\pi}{4} \right\}\]
\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\left( \theta + \frac{\pi}{4} \right) \right\} . . . . . \left( 1 \right)\]
\[\text{ Here }, - 1 < x < 1\]
\[ \Rightarrow - 1 < \sin\theta < 1 \]
\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2} \]
\[ \Rightarrow \left( - \frac{\pi}{2} + \frac{\pi}{4} \right) < \left( \frac{\pi}{4} + \theta \right) < \frac{3\pi}{4}\]
\[ \Rightarrow - \frac{\pi}{4} < \left( \frac{\pi}{4} + \theta \right) < \frac{3\pi}{4}\]
\[\text{ So, from } \left( 1 \right), \]
\[ y = \theta + \frac{\pi}{4} ..........\left[ \text{ Since }, \sin^{- 1} \left( \sin\alpha \right) = \alpha, \text{ if }\alpha \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right] \]
\[ \Rightarrow y = \sin^{- 1} x + \frac{\pi}{4} \]
\[\text{ Differentiating it with respect to x }, \]
\[ \frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}} + 0\]
\[ \therefore \frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}}\]
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