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प्रश्न
Differentiate \[\cos^{- 1} \left\{ 2x\sqrt{1 - x^2} \right\}, \frac{1}{\sqrt{2}} < x < 1\] ?
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उत्तर
\[\text{ Let, y } = \cos^{- 1} \left\{ 2x\sqrt{1 - x^2} \right\}\]
\[\text {Put x } = \cos\theta\]
\[ y = \cos^{- 1} \left\{ 2\cos\theta\sqrt{1 - \cos^2 \theta} \right\}\]
\[ y = \cos^{- 1} \left\{ 2\cos\theta \sin\theta \right\}\]
\[ y = \cos^{- 1} \left\{ \sin2\theta \right\} \left[ Since, \sin2\theta = 2\sin\theta\cos\theta \right]\]
\[ y = \cos^{- 1} \left[ \cos\left( \frac{\pi}{2} - 2\theta \right) \right] . . . \left( i \right)\]
\[\text{Now,} \]
\[ \frac{1}{\sqrt{2}} < x < 1\]
\[ \Rightarrow \frac{1}{\sqrt{2}} < \cos\theta < 1\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]
\[ \Rightarrow 0 < 2\theta < \frac{\pi}{2}\]
\[ \Rightarrow 0 > - 2\theta > - \frac{\pi}{2}\]
\[ \Rightarrow \frac{\pi}{2} > \left( \frac{\pi}{2} - 2\theta \right) > 0\]
\[\text{ Hence, from equation } \left( i \right)\]
\[y = \frac{\pi}{2} - 2\theta \left[ Since, \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right]\]
\[y = \frac{\pi}{2} - 2 \cos^{- 1} x \left[ Since, x = \cos\theta \right] \]
\[\text{ differentiating it with respect to x }, \]
\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{\pi}{2} \right) - 2\frac{d}{dx}\left( \cos^{- 1} x \right)\]
\[\frac{d y}{d x} = 0 - 2\left( \frac{- 1}{\sqrt{1 - x^2}} \right)\]
\[\frac{d y}{d x} = \frac{2}{\sqrt{1 - x^2}}\]
