Find $\frac{dy}{dx},$ if $x\sqrt{1-y}+y\sqrt{1-x}=0$

Question

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

Attempt:

On differentiating both sides w.r.t x, I got the following result, which doesn't match with the expected expression.

$$\frac{d}{dx} \{ x\sqrt{1-y} + y\sqrt{1-x} \} = 0$$ $$\frac{dx}{dx}\sqrt{1-y}+x\frac{d}{dx}\{\sqrt{1-y}\}+\frac{dy}{dx}\{\sqrt{1-x}\}+y\frac{d}{dx}\{\sqrt{1-x}\}=0$$ $$\sqrt{1-y}+x\frac{1}{2\sqrt{1-y}}\frac{dy}{dx}+\sqrt{1-x}\frac{dy}{dx}+y\frac{1}{2\sqrt{1-x}}=0$$ $$\frac{dy}{dx}\{x\frac{1}{2\sqrt{1-y}}+\sqrt{1-x}\}=-1\times\{\sqrt{1-y}+y\frac{1}{2\sqrt{x-1}}\}$$ $$\frac{dy}{dx}=-\frac{\sqrt{1-y}+y\frac{1}{2\sqrt{x-1}}}{x\frac{1}{2\sqrt{y-1}}+\sqrt{x-1}}$$

Any help will be appreciated.

Answer 1

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

For real $x,y$ both must have opposite signs.

Squaring both sides $$x^2(1-y)=y^2(1-x)$$

$$0=x^2-y^2+xy(x-y)=(x-y)(x+y+xy)$$

Unless $x=y\implies x=y=0$ $$x+y+xy=0\iff y=?$$

Answer 2

$$x\sqrt{1-y}+y\sqrt{1-x}=0$$ Squaring on both sides gives $$x^2(1-y)=y^2(1-x)$$ Now differentiate and $(uv)^{\prime}=uv^{\prime}+vu^{\prime}$

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