# Question about implicit differentiation and y derivative [duplicate]

Suppose I have: $$x^2 + y^2 = 1$$ Then, $$d/dx(x^2) + d/dx(y^2) = d/dx(1)$$ $$2x + 2y*y' = 0$$ $$y' = -2x/2y = -x/y$$

What I don't get is when we do $d/dx(y^2)$, where do I get the $y'$ from? In $d/dx(x^2)$, its simply $2x$ ... for $d/dx(y^2)$, why isn't it just $2y$

## marked as duplicate by Git Gud, zhoraster, Adam Hughes, Ferra, Daniel W. FarlowDec 7 '16 at 18:38

The problem is that $y$ is a function of $x$, that is $y=y(x)$.
Using the chain rule you can compute $$\frac{d(y^2(x))}{dx} = 2y \frac{dy}{dx}$$