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The derivative of

$$x^2 + y^2$$ is $$2x$$

I figured it out by using the calculator.

Why does $y$ become $0$? Do I always think $y$ as $0$ in that situation?

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    If it's a partial derivative w.r.t. $x$ then $y$ is considered constant. – 2017-02-24
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    When you derive with respect to a variable, say $x$, all other variable become "dummy numbers". In this case, simply imagine that $y^2 = 42$ and you therefore take the derivative of $x^2 + 42$ which is indeed $2x$. – 2017-02-24
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    You need to say what you are differentiating with respect to. Finding the partial derivative with respect to $y $ of $f (x,y)=x^2+y^2$ yields $2y $ – 2017-02-24

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If you have a function like:

$$f(x,y)=x^2+y^2$$

then the derivative is a vector called Gradient and it is given by

$$\nabla f(x,y)=\begin{pmatrix} 2x\\ 2y \end{pmatrix}$$

on the other hand if you take a partial derivative w.r.t. $x$ then

$$\partial _{x}f(x,y) =2x$$

on that case you consider $y$ as a constant.

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    @tom: you are very welcome! – 2017-02-25
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$2x$ is the partial derivative of $f(x,y)=x^2+y^2$ when you look into infinitesimal variations of $x$; in fact in this case you are not concerned about the value of $y$ as you are just focused into the $x$ variable.

On the other hand, the partial derivative of $f(x,y)$ with respect to $y$ is $2y$