Question

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

Answer 1

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

$\Rightarrow x\sqrt{1+y}=-y\sqrt{1+x}$

Squaring both sides, we get

$x^2(1+y)=y^2(1+x)$

$\Rightarrow x^2+x^{2}y=y^2+x{y}^2$

$\Rightarrow x^2-y^2=xy(y-x)$

$\Rightarrow (x-y)(x+y)=xy(y-x)$

$\therefore x+y=-xy$

$\Rightarrow (1+x)y=-x$

$\Rightarrow y=\frac{-x}{1+x}$

Differentiating both sides w.r.t. $x$ we get

$\frac{dy}{dx}=-\frac{(1+x)\frac{d}{dx}\left(x\right)-x\frac{d}{dx}(1+x)}{{(1+x)}^2}$

$=-\frac{1+x-x}{{(1+x)}^2}$

$\therefore \frac{dy}{dx}=-\frac{1}{{(1+x)}^2}$

Hence proved.