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