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प्रश्न
Differentiate \[\sin^{- 1} \left\{ \sqrt{\frac{1 - x}{2}} \right\}, 0 < x < 1\] ?
उत्तर
\[\text{ Let, y }= \sin^{- 1} \left\{ \sqrt{\frac{1 - x}{2}} \right\}\]
\[\text{ Put x }= \cos 2\theta\]
\[ y = \sin^{- 1} \left\{ \sqrt{\frac{1 - \cos2\theta}{2}} \right\}\]
\[ y = \sin^{- 1} \left\{ \sqrt{\frac{2 \sin^2 \theta}{2}} \right\}\]
\[ y = \sin^{- 1} \left( \sin\theta \right) . ... . \left( 1 \right)\]
\[\text{ Here,} 0 < x < 1\]
\[ \Rightarrow 0 < \cos 2\theta < 1\]
\[ \Rightarrow 0 < 2\theta < \frac{\pi}{2}\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]
\[\text{ So, from equation } \left( 1 \right), \]
\[ y = \theta \left[ \text{ Since }, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ y = \frac{1}{2} \cos^{- 1} x \left[ \text{ Since, x } = \cos 2\theta \right]\]
\[\text{ Differentiating it with respect to x }, \]
\[\frac{d y}{d x} = - \frac{1}{2\sqrt{1 - x^2}}\]
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