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