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This question already has an answer here:

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 $2y$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The problem is that $y$ is a function of $x$, that is $y=y(x)$.

Using the chain rule you can compute \begin{equation} \frac{d(y^2(x))}{dx} = 2y \frac{dy}{dx} \end{equation}

You should read the chain rule.

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