Question

If $\sqrt{x}+\sqrt{y}=\sqrt{a}$ prove that $\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$.

Answer 1

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

On differentiating , we get

$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$