Question

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

Answer 1

It is given that
$x\sqrt{1+y}+y\sqrt{1+x}=0$
⇒ $x\sqrt{1+y}=-y\sqrt{1+x}$
squaring both sides, we obtain x2(1+y)=y2(1+x)
⇒ x2+x2y=y2+xy2
⇒ x2-y2=xy2-x2y
⇒ x2-y2=xy(y-x)
⇒ (x+y)(x-y)=xy(y-x)
∴x+y=-xy
⇒ (1+x)y=-x
⇒ y=-x/(1+x)
Differentiating both sides with respect to x, we obtain
$\frac{dy}{dx}=(1+x)\frac{d}{dx}(x)-x\frac{d}{dx}\frac{(1+x)}{(1+x)^2}=-(1+x)-\frac{x}{(1+x)^2}$
=$-\frac{1}{(1+x)^2}$
Hence, it proved.