Before we start talking about instantaneous
rate of change, let's talk about average
rate of change. A simple example is average
velocity. If you drive 180 miles in 3 hours, then
your average speed is 60 mph. We get this by dividing the distance
traveled by the time. Using the familiar $d=rt$, where rate
means the average velocity, we see that $r=\frac{d}{t}$. We
will use: $$ v_{avg} = \frac{\Delta s}{\Delta t},$$ where $\Delta s$
is the distance traveled and $\Delta t$ is the time elapsed. We use
the Greek letter $\Delta$ to mean "change in". If we start at
position $s(t_0)$ at time $t_0$ and end up at position $s(t_1)$ at
time $t_1$, then \begin{eqnarray*} \Delta s & = & s(t_1) -
s(t_0), \cr \Delta t & = & t_1 - t_0. \end{eqnarray*} If you
plot position against time on a graph, $v_{avg}$ is the slope of a secant line.

To get the instantaneous velocity
at a particular time $t=a$, we average over shorter and shorter
time intervals. That is, we compute the average
velocity between time $a$ and $a + \Delta t$, and
then take a limit as $\Delta t \to 0$. On a graph, this is taking
the slope of secant lines between points that are getting closer
and closer. In the limit, we get the slope of a tangent line.

Whether we think in terms of velocity or slope, we get a limit:
$$\lim_{\Delta t \to 0} \frac{s(a+\Delta t)-s(a)}{\Delta t}.$$
This can also be written as: $$\lim_{t \to a}
\frac{s(t)-s(a)}{t-a},$$ where $t=a+\Delta t$. This quantity (if
the limit exists) is called the derivative
of $s(t)$ at time $t=a$.

In the animations below, \(a=3\), the \(x\) coordinate of point
\(P\); and \(a+\Delta x\) is the \(x\)-coordinate of point
\(Q\). The movement is the point $Q$ approaching the point
$P$, which means $3+\Delta x$ is getting closer to $3$, which
means $\Delta x$ is getting closer to $0$.
The red tangent line (at $P$) is the limit of the blue secant
lines through $P$ and $Q$. Note that $\Delta x$ can be positive
(first animation) or negative (second animation). For the
derivative to exist, the limits as $\Delta x \to 0^+$ and $\Delta
x \to 0^-$ must give the same answer.