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प्रश्न
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
उत्तर
\[\text{We have, y } = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\frac{d}{dx}\left( x \sin^{- 1} x \right) - \left( x \sin^{- 1} x \right)\frac{d}{dx}\left( \sqrt{1 - x^2} \right)}{\left( \sqrt{1 - x^2} \right)^2} \right] \]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\left\{ x\frac{d}{dx}\left( \sin^{- 1} x \right) + \sin^{- 1} x\frac{d}{dx}\left( x \right) \right\} - \left( x \sin^{- 1} x \right)\frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)} \right] \]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\left\{ \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x \right\} - \frac{x \sin^{- 1} x\left( - 2x \right)}{2\sqrt{1 - x^2}}}{\left( 1 - x^2 \right)} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{x + \sqrt{1 - x^2} \sin^{- 1} x + \frac{x^2 \sin^{- 1} x}{\sqrt{1 - x^2}}}{\left( 1 - x^2 \right)} \right]\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d y}{d x} = x + \frac{\sqrt{1 - x^2} \sin^{- 1} x}{1} + \frac{x^2 \sin^{- 1} x}{\sqrt{1 - x^2}}\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\left( 1 - x^2 \right) \sin^{- 1} x + x^2 \sin^{- 1} x}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\sin^{- 1} x - x^2 \sin^{- 1} x + x^2 \sin^{- 1} x}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\sin^{- 1} x}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d y}{d x} = x + \frac{y}{x} \left[ \because y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} \right]\]
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