We have already seen that the derivative of a function $f$ at $x=a$ is $$f'(a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a} = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}.$$ We can apply the same idea to get a formula for the derivative of $f$ at every point:$$f'(\Box) = \lim_{h \to 0} \frac{f(\Box+h)-f(\Box)}{h},$$ where $\Box$ stands for "anything you like".

In particular, if we take $x$ as input and associate to it $f'(x)$ as output, then we have a whole new function, which we call the derivative of the original function $f$. The value of $f'$ at a point $x$ is the same as the instantaneous rate of change of $f$ at that point. Wherever $f$ is increasing, $f'$ is positive. Wherever $f$ is decreasing, $f'$ is negative. Wherever the graph of $f$ is flat, $f'$ is (close to) zero.

The following video develops this idea, both in general and for the particular function $f(x)=x^2$.

Returning to our running example, $s(t)=4t^2+19$, we can compute: \begin{eqnarray*} s'(t) & = & \lim_{h \to 0} \frac{s(t+h)-s(t)}{h} \cr \cr &=& \lim_{h \to 0} \frac{4(t+h)^2+3 - (4t^2+3)}{h} \cr \cr &=& \lim_{h\to 0} \frac{4t^2 + 8th + 4h^2 + 3 - (4t^2+3)}{h} \cr \cr &=& \lim_{h \to 0} \frac{8th + 4h^2}{h} \cr \cr &=& \lim_{h \to 0} 8t+4h = 8t. \end{eqnarray*} So the derivative of the function $s(t)=4t^2+3$ is the function $s'(t)=8t$.