# If $x\sqrt{1+y}+y\sqrt{1+x}=0,\normalsize\; Prove\; that\; \Large\frac{dy}{dx}=\frac{-1}{(1+x)^2}$

Toolbox:
• Shift one term to R.H.S square and then differentiate
Step 1
$x \sqrt {1+y}=-y\sqrt{1+x}$ squaring both sides
Step 2
$\Rightarrow x^2(1+y)=y^2(1+x)$
$\Rightarrow x^2 + x^2y-y^2-yx^2=0$
Step 3
$(x-y)(x+y+xy)=0$
$\Rightarrow x+y+xy = 0$
$y = \frac{-x}{1+x}$ differentiate both sides to get the answer.

edited Apr 4, 2013