Question

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

Answer 1

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

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

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

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

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

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

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

$\Rightarrow x+y=-xy$

$\Rightarrow x+xy+y=0$

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

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

$\Rightarrow \frac{dy}{dx}=-\left[\frac{1.\left(x+1\right)-x\times 1}{{\left(x+1\right)}^2}\right]$

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

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