Question

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

Answer 1

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

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

Squaring both sides we get

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

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

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

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

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

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

$\Rightarrow x=-xy-y$

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

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

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

Differentiating with respect to x, we get

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

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

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