Question

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

Answer 1

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

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

Square on both sides

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

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

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

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

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

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

$x+y=-xy$

$x=-xy-y$

$x=-y(x+1)$

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

Differentiate $y$ with respect to $x$ using Chain rule, we get ;

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

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