If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$

Question

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

My Attempt $$ x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\ 2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\ \frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\ \frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy} $$ How do I proceed further and find the derivative ?

Answer 1

$$ x\sqrt{1+y}+y\sqrt{1+x}=0\implies x\sqrt{1+y}=-y\sqrt{1+x}\\ \implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+y^2x\\ \implies \boxed{(1+x)\color{red}{y^2}-x^2\color{red}{y}-x^2=0} $$

$B^2-4AC=x^4+4x^2(1+x)=x^4+4x^2+4x^3=x^2(x^2+4x+4)=x^2(x+2)^2$ So, $$ \begin{align} y&=\frac{x^2\pm\sqrt{x^2(x+2)^2}}{2(x+1)}=\frac{x^2\pm|{x(x+2)}|}{2(x+1)}\\ &=\frac{x^2+x(x+2)}{2(x+1)}\text{ or }\frac{x^2-x(x+2)}{2(x+1)}\\ &=\frac{x^2+x^2+2x}{2(x+1)}\text{ or }\frac{x^2-x^2-2x}{2(x+1)}\\ &=\frac{2x(x+1)}{2(x+1)}\text{ or }\frac{-2x}{2(x+1)}\\ y&=x \text{ or } \frac{-x}{x+1} \end{align} $$ As $y\neq x$ we have the derivative, $$\color{blue}{ y'=\frac{-1}{(1+x)^2} } $$

Note: $x+\sqrt{1+y}=−y\sqrt{1+x}\implies(x−y)(x+y+xy)=0\implies x−y=0$ or $(x+y+xy)=0$. Note that $x−y=0\implies x=y$ is equivalent to the original function only when $x=0$, and $x\sqrt{1+y}=−y\sqrt{1+x}\Leftrightarrow(x+y+xy)=0$ for all $x∈D_\text{original function}$

Answer 2

Hint:

Squaring both sides of $$x\sqrt{1+y}=-y\sqrt{1+x}$$

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

$$\iff0=(x-y)(x+y+xy)$$

But $x\sqrt{1+y}=-y\sqrt{1+x}\implies x,y$ are of opposite sign, hence $x\ne y$ unless $x=y=0$

Otherwise, $$y(1+x)=-x\implies y=\dfrac{1-(x+1)}{1+x}=\dfrac1{1+x}-1$$

Answer 3

$x^2+x^2y = y^2 + xy^2 $

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

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

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

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

$x \neq y$ since $x\sqrt{1+y} = -y\sqrt{1+x}$ unless (0,0)

$x + xy + y = 0 $

$x + y(x + 1) = 0 $

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

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