Question

Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?

Answer 1

\[\text{ Let, y } = \sin^{- 1} \left\{ \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right\}\]

\[\text{ putting } x = \sin\theta\]

\[ \therefore y = \sin^{- 1} \left( \frac{\sin\theta + \sqrt{1 - \sin^2 \theta}}{\sqrt{2}} \right)\]

\[ \Rightarrow y = \sin^{- 1} \left( \frac{\sin\theta + \cos\theta}{\sqrt{2}} \right)\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\theta\left( \frac{1}{\sqrt{2}} \right) + \cos\theta\left( \frac{1}{\sqrt{2}} \right) \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\theta \cos\frac{\pi}{4} + \cos\theta \sin\frac{\pi}{4} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin\left( \theta + \frac{\pi}{4} \right) \right\} . . . . . \left( 1 \right)\]

\[\text{ Here }, - 1 < x < 1\]

\[ \Rightarrow - 1 < \sin\theta < 1 \]

\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2} \]

\[ \Rightarrow \left( - \frac{\pi}{2} + \frac{\pi}{4} \right) < \left( \frac{\pi}{4} + \theta \right) < \frac{3\pi}{4}\]

\[ \Rightarrow - \frac{\pi}{4} < \left( \frac{\pi}{4} + \theta \right) < \frac{3\pi}{4}\]

\[\text{ So, from } \left( 1 \right), \]

\[ y = \theta + \frac{\pi}{4} ..........\left[ \text{ Since }, \sin^{- 1} \left( \sin\alpha \right) = \alpha, \text{ if }\alpha \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right] \]

\[ \Rightarrow y = \sin^{- 1} x + \frac{\pi}{4} \]

\[\text{ Differentiating it with respect to x }, \]

\[ \frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}} + 0\]

\[ \therefore \frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}}\]