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

$y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}$

$\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{a}{x}}\times \frac{1}{a}-\frac{1}{2}\sqrt{\frac{x}{a}}\times \frac{a}{x^2}\Rightarrow \frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{ax}}-\frac{1}{2}\sqrt{\frac{a}{x}}\times \frac{1}{x}$

$2x\left(\frac{dy}{dx}\right)=(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}})$

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