Why is $a$ the derivative of $f(x)=ax$? I thought there was some kind of process to calculate a derivative. Can this be graphed? I know about the power rule, the chain rule, etc. but I don't know what is happening here.
 A: You could look at this a few ways...
For instance, consider what the derivative "means." $f'(x)$, for a function $f(x)$, encodes the rate of change of the function at the point $x$. It is a generalization of the notion of slope from familiar algebra: it is now just the rate of change at a given point. Of course, this makes linear functions such as $f(x)=ax$ special. The slope of this line is $a$, so it makes sense intuitively that $f'(x)=a$ for all $x$.
Or we could calculate it rigorously. Recall:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
(provided the limit exists of course). Let $f(x)=ax$. Then the above becomes
$$f'(x) = \lim_{h \to 0} \frac{a(x+h)-ax}{h} = \lim_{h \to 0} \frac{ax+ah-ax}{h} = \lim_{h \to 0} \frac{ah}{h} = \lim_{h \to 0} a = a$$


I thought there was some kind of process to calculate a derivative.

I mean, there is, sort of. You can apply the definition as above, or you can use known formulas. For instance, you have, for constants $c$ and differentiable $f$,
$$\frac{d}{dx} (cf(x)) = c \cdot \frac{d}{dx} f(x)$$
or, in other notation, $(cf(x))' = cf'(x)$. Moreover, we know that
$$\frac{d}{dx} x^n = nx^{n-1}$$
from one of the familiar derivative laws, on the premise $n \ne 0$. You can apply both to your case, since $f(x) = ax = ax^1$:
$$f'(x) = (ax^1)' = a(x^1)' = a(1x^{1-1}) = a(x^0) = a(1) = a$$
A: You can use the slope to prove it. We know that the derivative of a function $f(x)$ at a point $(x_1, f(x_1))$ is the same as the slope $m$ of the tangent line to the graph of $y=f(x)$ at that point. That is,
$$f'(x_1) = m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$
where $(x_2,f(x_2))$ is some other point on the tangent line.
In your case, $f(x) = ax$, so
$$
\begin{align}
f'(x_1) = \frac{ax_2 - ax_1}{x_2 - x_1} = \frac{a(x_2 - x_1)}{x_2 - x_1} = a.
\end{align}
$$
A: By definition, the derivative of a real function with a real variable is:
$$ f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
(this has a geometric meaning). Thus, in your case we have:
$$ (ax)' = \lim_{h \rightarrow 0} \frac{a(x+h)-ax}{h} = \lim_{h \rightarrow 0} \frac{ah}{h} = a $$
