Question

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

Answer 1

Given,

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

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

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

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

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

$x+y=-xy$

$x=-y-xy$

$y(1+x)=-x$

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

Therefore,

$\frac{dy}{dx}=-\frac{(1+x).1-x(0+1)}{(1+x{)}^2}$

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