Implicit function differentiation methods

Question

Q: If $$x\sqrt{1+y}+y\sqrt{1+x}=0$$ where $x,y\in\mathbb{R}$ then prove that: $$\dfrac{dy}{dx}=-\dfrac{1}{(1+x)^2}$$ My approach: $$x\sqrt{1+y}+y\sqrt{1+x}=0$$ $$\implies x\sqrt{1+y}=-y\sqrt{1+x}$$ $$\implies x^2(1+y)=y^2(1+x)$$ $$\implies x^2-y^2=xy(y-x)$$ $$\implies x+y=-xy$$ $$\implies y=-\frac{x}{1+x}$$ Now having converted to explicit form $y=f(x)$ we can differentiate and get the answer. I want to know if there is any other approach which avoids squaring or cancellation of terms?

Answer 1

let $u = \sqrt{(1+x)(1+y)}$, multiply your condition by $\sqrt{1+x}$ and insert $u$, we get $(u-1)(u+(1+x))=0$ so $u= 1.$ Then differentiate $u^2 = (1+x)(1+y)=1$ to reach the conclusion. This is just an alternative way but not really simpler than your solution though.