Question

If $x\sqrt{1+y}+y\sqrt{1+x}=0$, for $-1<x<1$, prove that $\frac{dy}{dx}=\frac{1}{{(1+x)}^2}$

Answer 1

Given, $x\sqrt{1+y}+y\sqrt{1+x}=0$

If we differentiate this function with respect to $x$, then we get

$\sqrt{1+y}+\frac{x}{2\sqrt{1+y}}\frac{dy}{dx}+\frac{y}{2\sqrt{1+x}}+\frac{dy}{dx}\sqrt{1+x}=0$
Taking out $\frac{dy}{dx}$ one side, we get $\frac{dy}{dx}=\frac{1}{{(1+x)}^2}$