Why is $\frac{d}{dx}(x^2+y^2)=2x+2y(y')$ but $\frac{d}{dy}(x^2+y^2)$ is only $2y$? I was reviewing implicit differentiation for DVQ and spent some time understanding why:
$\frac{d}{dx}(x^2+y^2)$=$2x+2y(y')$
and I felt like I got it. Then today $\frac{d}{dy}$$(x^2+y^2)$ came up in a problem and I thought, "Great, that must be $2y+2x(x')$. But it is somehow just $2y$. What is the difference?
 A: There were some very helpful responses that I am going to summarize as an official answer. Thanks to mm8511 and Georg Lehner for the clear answers and GoodDeeds and Clement C. for the helpful comments.
$\frac{d}{dx}(x^2+y^2)=2x+2y(y')$ is solved with implicit differentiation. The problem is asking for differentiation in terms of $x$. For $x^2$ this is simply $2x$ but for $y^2$ the chain rule must be used since it is not already in terms of x. With the chain rule you will find $\frac{dy}{dx}\frac{d}{dy}(y^2)$ which is $2y(y')$.
Now if your repeated the last problem with the independent variable as $y$, $\frac{d}{dy}$, you would repeat the above to arrive at  $\frac{d}{dy}(x^2+y^2)=2x(x')+2y$.
If you tried to compute  $\frac{d}{dy}(x^2+y^2)$ but $x$ was not a function of $y$ in any way, then you would end up with just $2y$, but this is a bit of a contradiction in notation and was not the source of my error.
My error was that partial derivatives were used. A partial derivative calls for, in summary, only one variable to be differentiated in an equation with more than one variable. The other variables are held constant. So the true is question why:
$\frac{\partial}{\partial y}(x^2+y^2)=2y$
and the answer is because we are only differentiating the $y$ variable while the $x$ is being held constant.
A: I need somebody to explain me why $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$
This explains your question. It is basically an application of the chain rule, since f=y(x).
(note, I do not have enough rep to comment, I apologize for posting this as an answer). 
