Question

If $y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}},$ prove that $2xy\frac{dy}{dx}=(\frac{x}{a}-\frac{a}{x})$

Answer 1

$\frac{dy}{dx}=\frac{d}{dx}(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}})$

$=\frac{1}{\sqrt{a}}\frac{d}{dx}\sqrt{x}+\sqrt{a}\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)$

$=\frac{1}{\sqrt{a}}\frac{1}{2\sqrt{x}}+\sqrt{a}\left(\frac{-1}{2}\right)\times \frac{1}{x\sqrt{x}}$

$=\frac{1}{2x}(\sqrt{\frac{x}{a}}+(-\sqrt{\frac{a}{x}}))$

$\Rightarrow 2x\frac{dy}{dx}=\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}$

Multiplying both sides by $y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}$

$=2xy\frac{dy}{dx}=(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}})(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}})$

$=(\frac{x}{a}-\frac{a}{x})$

Hence, proved.