\[If y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, \text{ prove that } 2xy\frac{dy}{dx} = \left( \frac{x}{a} - \frac{a}{x} \right)\]

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

I F Y = √ X a + √ a X , Prove that 2 X Y D Y D X = ( X a − a X ) - Mathematics

प्रश्न

$I f y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, prove that 2 x y \frac{d y}{d x} = (\frac{x}{a} - \frac{a}{x})$

उत्तर

$y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} = \frac{1}{\sqrt{a}} x^{\frac{1}{2}} + \sqrt{a} x^{\frac{- 1}{2}}$
$\frac{d y}{d x} = \frac{1}{\sqrt{a}} \frac{1}{2} x^{\frac{- 1}{2}} + \sqrt{a} (\frac{- 1}{2}) x^{\frac{- 3}{2}}$
$L H S = 2 x y \frac{d y}{d x}$
$= 2 x (\frac{1}{\sqrt{a}} x^{\frac{1}{2}} + \sqrt{a} x^{\frac{- 1}{2}}) (\frac{1}{\sqrt{a}} \frac{1}{2} x^{\frac{- 1}{2}} + \sqrt{a} (\frac{- 1}{2}) x^{\frac{- 3}{2}})$
$= 2 x (\frac{1}{2 a} - \frac{1}{2 x} + \frac{1}{2 x} - \frac{a}{2 x^{2}})$
$= 2 x (\frac{1}{2 a} - \frac{a}{2 x^{2}})$
$= (\frac{x}{a} - \frac{a}{x})$
$= R H S$
$Hence, proved .$

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The Concept of Derivative - Algebra of Derivative of Functions

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Q 22 Q 21 Q 23

अध्याय 30: Derivatives - Exercise 30.3 [पृष्ठ ३४]

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अध्याय 30 Derivatives
Exercise 30.3 | Q 22 | पृष्ठ ३४

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The Concept of Derivative - Algebra of Derivative of Functions

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I F Y = √ X a + √ a X , Prove that 2 X Y D Y D X = ( X a − a X ) - Mathematics

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I F Y = √ X a + √ a X , Prove that 2 X Y D Y D X = ( X a − a X ) - Mathematics

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