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

Toolbox:
• $\big(\large\frac{u}{v}\big)'=\large\frac{u'v-uv'}{v^2}$
Step 1:
The given equation may be written as $x\sqrt{1+y}=-y\sqrt{1+x}$
Squaring both sides we get
$x^2(1+y)=y^2(1+x)$
$x^2+x^2y=y^2+y^2x$
$x^2-y^2=y^2x-x^2y$
$(x+y)(x-y)=-xy(x-y)$
$x+y=-xy$
$x=-xy-y$
$x=-y(x+1)$
$\large\frac{x}{x+1}$$=-y$
$y=-\large\frac{x}{x+1}$
Step 2:
Differentiating with respect to $x$
$\large\frac{dy}{dx}=-\frac{(1+x).1-x(0+1)}{(1+x)^2}$
$\large\frac{dy}{dx}=-\frac{1}{(1+x)^2}$