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Question
If \[y = x \sin^{- 1} x + \sqrt{1 - x^2}\] ,prove that \[\frac{dy}{dx} = \sin^{- 1} x\] ?
Solution
\[\text{ We have, y } = x \sin^{- 1} x + \sqrt{1 - x^2}\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right]\]
\[ = \frac{d}{dx}\left( x \sin^{- 1} x \right) + \frac{d}{dx}\left( \sqrt{1 - x^2} \right)\]
\[ = \left[ x \frac{d}{dx} \sin^{- 1} x + \sin^{- 1} x\frac{d}{dx}\left( x \right) \right] + \frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right) \]
\[ = \left[ \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x \right] - \frac{2x}{2\sqrt{1 - x^2}}\]
\[ = \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x - \frac{x}{\sqrt{1 - x^2}}\]
\[ = \sin^{- 1} x\]
\[\]
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