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