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