We summarize the results of the last two slides with the following
| Definition:
The derivative of a function $f$ at a number $a$, denoted by $f'(a)$, is the value
$$f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a},$$
if this limit exists.
Equivalently, $$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.$$ The derivative can also be computed with centered differences: $$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a-h)}{2h}.$$ |
| Besides being a rate of change, $f'(a)$ is the slope of the tangent line to $y=f(x)$ at $(a, f(a))$, and is the conversion factor from small changes in $x$ to small changes in $f(x)$. |
In our running example, $s(t)=4t^2+3$, we previously computed $s'(2)$ by working with $\displaystyle{\lim_{t \to 2} \frac{s(t)-s(2)}{t-2}}$. Let's repeat the calculation in terms of $h$: \begin{eqnarray*} s(2+h) &=& 4(2+h)^2 +3 \cr & = & 4(4 + 4h + h^2) + 3 \cr &=& 19 + 16h + 4h^2 \cr s(2) &=& 19 \cr s(2+h)-s(2) & = & 16 h + 4h^2 \cr \cr \frac{s(2+h)-s(2)}{h} & = & 16 + 4h \cr \cr \lim_{h \to 0} \frac{s(2+h)-s(2)}{h} & = & 16 \cr \cr s'(2) & = & 16, \end{eqnarray*} which (naturally) is the answer we obtained before. When computing derivatives using $h$, the calculations look a little bit different, but they're essentially the same.
If we had worked instead with centered differences, the calculation would work like this: \begin{eqnarray*} s(2+h) & = & 19 + 16 h + 4h^2 \cr s(2-h) & = & 19 - 16h + 4h^2 \hbox{ by replacing $h$ with $-h$ in the last line}\cr s(2+h)-s(2-h) & = & 32 h \cr \cr \frac{s(2+h)-s(2-h)}{2h} & = & 16 \cr \cr s'(2) = \lim_{h \to 0} \frac{s(2+h)-s(2-h)}{2h} & = & 16. \end{eqnarray*} Different computation, but the same answer.