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:
$$ v_{avg} = \frac{\Delta s}{\Delta t},$$
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 veloctity 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:
$$\lim_{x \to a} \frac{s(x)-s(a)}{x-a},$$
where $x=a+\Delta t$.
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 \(a+\Delta t\) is the \(x\)-coordinate of point \(Q\). The red tangent line
is the limit of the blue secant lines. Note that $\Delta t$
can be positive (first animation) or negative (second animation). For the
derivative to exist, the limits as $\Delta t \to 0^+$ and $\Delta t \to 0^-$
must give the same answer.