Question

If $x\sqrt{1+y}+y\sqrt{1+x}=0$ and $x\ne y$, show that $\frac{dy}{dx}=\frac{-1}{(1+x{)}^2}$

Answer 1

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

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

Squaring both sides, we get

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

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

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

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

$\Rightarrow x+y=-xy$

$\Rightarrow x+y+xy=0$

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

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

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

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

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

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

Hence proved.