Losing the $\pm$ when differentiating $x^2+y^2=1$ implicitly

Question

When differentiating the explicit function: $y=\pm \sqrt{1-x^2}$ there are two branches and the $\pm$ is conserved by considering both branches: $\frac{dy}{dx}=\pm \frac{x}{\sqrt{1-x^2}}$

But if we differentiate the function implicitly we lose the positive branch: $$x^2+y^2=1$$ $$2x+2y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^2}}$$

How do I differentiate this function implicitly without losing the $\pm$ ?

Answer 1

From the second line, we just get that $\frac{dy}{dx}=\frac{-x}{y}$... then if $y=\pm \sqrt{1-x^2}$, you don't lose anything as

$\frac{dy}{dx}=\frac{-x}{\pm \sqrt{1-x^2}}=\mp \frac{x}{ \sqrt{1-x^2}}$