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प्रश्न
Differentiate \[\cos^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
उत्तर
\[\text{ Let y } = \cos^{- 1} \left\{ \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right\}\]
\[\text{ Put, x } = \cos\theta\]
\[ y = \cos^{- 1} \left\{ \frac{\cos\theta + \sqrt{1 - \cos^2 \theta}}{\sqrt{2}} \right\}\]
\[ y = \cos^{- 1} \left\{ \frac{\cos\theta + \sin\theta}{\sqrt{2}} \right\}\]
\[ y = \cos^{- 1} \left\{ \cos\theta\left( \frac{1}{\sqrt{2}} \right) + \sin\theta\left( \frac{1}{\sqrt{2}} \right) \right\}\]
\[ y = \cos^{- 1} \left\{ \cos\theta\cos\frac{\pi}{4} + \sin\theta \sin\frac{\pi}{4} \right\}\]
\[ y = \cos^{- 1} \left\{ \cos\left( \theta - \frac{\pi}{4} \right) \right\} . . . \left( i \right)\]
\[\text{ Here }, - 1 < x < 1\]
\[ \Rightarrow - 1 < \cos\theta < 1 \]
\[ \Rightarrow \frac{3\pi}{4} < \theta < \frac{5\pi}{4} \]
\[ \Rightarrow \left( \frac{3\pi}{4} - \frac{\pi}{4} \right) < \left( \theta - \frac{\pi}{4} \right) < \frac{5\pi}{4} - \frac{\pi}{4}\]
\[ \Rightarrow \left( \frac{\pi}{2} \right) < \left( \theta - \frac{\pi}{4} \right) < \pi\]
\[\text{ So, from equation } \left( i \right), \]
\[ y = \left( \theta - \frac{\pi}{4} \right) ..........\left[ \text{ Since }, \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right] \]
\[ y = \cos^{- 1} x - \frac{\pi}{4} ............\left[ \text{ Since }, x = \sin\theta \right]\]
\[\text{ Differentiating it with respect to x }, \]
\[\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^2}} + 0\]
\[\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^2}}\]
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