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

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प्रश्न

If \[y = \sqrt{x} + \frac{1}{\sqrt{x}}\], prove that  \[2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}\] ?

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उत्तर

\[\text{ We have, y } = \sqrt{x} + \frac{1}{\sqrt{x}}\]

Differentiate with respect to x,

\[\Rightarrow \frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x} \right) + \frac{d}{dx}\left( \frac{1}{\sqrt{x}} \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2\sqrt{x}} + \left( \frac{- \frac{1}{2\sqrt{x}}}{x} \right)\]

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

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

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

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

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

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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 63 | पृष्ठ ३८

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