If $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$, where $|x| < 1, |y| < 1$ and $x≠y$, then

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

The correct answer is Option (2) → $(1+x)^2\frac{dy}{dx}+1=0$

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

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

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

$(x-y)(x+y+xy)=0$

$x\neq y\Rightarrow x+y+xy=0$

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

$u=\frac{x}{\sqrt{1+x}},\; v=\frac{y}{\sqrt{1+y}},\; u+v=0$

$\frac{du}{dx}+\frac{dv}{dy}\frac{dy}{dx}=0$

$\frac{du}{dx}=\frac{2+x}{2(1+x)^{3/2}},\; \frac{dv}{dy}=\frac{2+y}{2(1+y)^{3/2}}$

$\frac{dy}{dx} =-\frac{2+x}{2+y}\left(\frac{1+y}{1+x}\right)^{3/2}$

Using $(1+x)(1+y)=1$ gives $\frac{1+y}{1+x}=\frac{1}{(1+x)^2}$ and $\frac{2+x}{2+y}=1+x$

$\frac{dy}{dx}=-(1+x)\cdot\frac{1}{(1+x)^3}$

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

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