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