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प्रश्न
\[If y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, \text{ prove that } 2xy\frac{dy}{dx} = \left( \frac{x}{a} - \frac{a}{x} \right)\]
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उत्तर
\[y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} = \frac{1}{\sqrt{a}} x^\frac{1}{2} + \sqrt{a} x^\frac{- 1}{2} \]
\[\frac{dy}{dx} = \frac{1}{\sqrt{a}}\frac{1}{2} x^\frac{- 1}{2} + \sqrt{a}\left( \frac{- 1}{2} \right) x^\frac{- 3}{2} \]
\[LHS = 2xy \frac{dy}{dx}\]
\[ = 2x \left( \frac{1}{\sqrt{a}} x^\frac{1}{2} + \sqrt{a} x^\frac{- 1}{2} \right)\left( \frac{1}{\sqrt{a}}\frac{1}{2} x^\frac{- 1}{2} + \sqrt{a}\left( \frac{- 1}{2} \right) x^\frac{- 3}{2} \right)\]
\[ = 2x\left( \frac{1}{2a} - \frac{1}{2x} + \frac{1}{2x} - \frac{a}{2 x^2} \right)\]
\[ = 2x\left( \frac{1}{2a} - \frac{a}{2 x^2} \right)\]
\[ = \left( \frac{x}{a} - \frac{a}{x} \right)\]
\[ = RHS \]
\[\text{ Hence, proved } . \]
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