Question

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

Answer 1

$x\sqrt{1+y}$ given

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

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

Squaring on both sides we obtain

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

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

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

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

$x+y=-xy$

$xy+y=-x$

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

differentiating both sides w. r. t. x

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

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

Hence proved