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.