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Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsDerivatives2 Marks
Question
If $\text{y}=\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}},$ prove that $\text{2xy}\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)$
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Answer

We have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{a}}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}+\sqrt{\text{a}}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{a}}}\frac{1}{2\sqrt{\text{x}}}+\sqrt{\text{a}}\Big(\frac{-1}{2}\Big)\times\frac{1}{\text{x}\sqrt{\text{x}}}$
$=\frac{1}{\text{2x}}\Bigg(\sqrt{\frac{\text{x}}{\text{a}}}+\Big(-\sqrt{\frac{\text{a}}{\text{x}}}\Big)\Bigg)$
$\Rightarrow\text{2x}\frac{\text{dy}}{\text{dx}}=\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\frac{\text{a}}{\text{x}}}$
Mulitiplying both the side by $\text{y}=\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}$
$\Rightarrow\text{2xy}\frac{\text{dy}}{\text{dx}}=\Big(\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\frac{\text{a}}{\text{x}}}\Big)\Big(\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}\Big)$
$=\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)$
Hence, proved.

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