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If Y = X Sin − 1 X √ 1 − X 2 , Prove that ( 1 − X 2 ) D Y D X = X + Y X ? - Mathematics

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प्रश्न

If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] ,  prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?

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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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पाठ 11: Differentiation - Exercise 11.02 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.02 | Q 64 | पृष्ठ ३८

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