The derivative of $y=x^2$ and other values of x that equal zero? Put simply, if the derivative of $y=x^2$ is $0$ when $x$ is $0$, how does that explain all the other values being greater than zero? From what I know, a derivative is the slope of a line tangent to a function. It can also be explained as the slope from a point on a graph to another point that is very, very close to it. So if the slope is $0$ when $x$ is $0$, then wouldn't there be two values (one right before and one right after) that are equal to zero as well? And it could be said that this value approaches zero, so it pretty much is zero. In that case, wouldn't the next very, very small point also be zero, and so on forever? How would the function ever increase in value?
 A: 
a derivative is the slope of a line tangent to a function

Actually, it's the slope of a line tangent to the graph of a function.

It can also be explained as the slope from a point on a graph to another point that is very, very close to it.

Not really. The derivative of $f$ at $x_0$ is the limit of the slopes of the lines joining $\bigl(x_0,f(x_0)\bigr)$ to points of the form $\bigl(x,f(x)\bigr)$, as $x$ gets closer to $x_0$. And a limit of things which are all different from $0$ may well be $0$.
A: For every $x\neq 0$ you have $x^2 > 0$. Values with small distances to $0$ yield small squares, but they are non-zero. The secant through $(0,0)$ and $(x,x^2)$ with $x\neq 0$ has slope $\frac{\Delta y}{\Delta x}=\frac{x^2}{x}=x$ which is non-zero but approaches $0$ when $x$ approaches $0$. Only the tangent through $(0,0)$ has slope equal to $0$, and it doesn't intersect the graph at any other point.
If this doesn't answer your question, please try to be more specific.
A: There is no number "right after $0$" to connect to $0$ to form a slope of $0$, or at which the slope will still be $0$.
You need the idea of a limit to build a logically sound way to deal with the fact that there is no smallest positive number. That's what calculus is all about. When you start learning it you rely on developing intuition about limits - look carefully at the algebra that tells you  the derivative of $f(x) = x^2$ is $f'(x) = 2x$.
In later studies you will see a more formal definition of limits.
A: The derivative of $y=x^2$ with respect to $x$ is
$$
\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x} \, .
$$
Informally, people regard the derivative as the change in $y$ divided by a small change in $x$. In other words, they regard $(x^2)'$ as
$$
\frac{(x+\Delta x)^2-x^2}{\Delta x}
$$
for some very small value of $\Delta x$. While it is fine to use this notion as an informal conception of what the derivative is, you must remember that this is not quite correct. Instead, the derivative is the limit of the above quotient as $\Delta x \to 0$. When $x=0$, the quotient simplifies to
$$
\frac{(\Delta x)^2}{\Delta x}=\Delta x
$$
Therefore,
$$
(x^2)'|_{x=0}= \lim_{\Delta x \to 0}\Delta x=0 \, .
$$
Even though the slope of the secant line is never zero, the limit of the slope of the secant lines is zero.
