Computing an instantaneous rate of change of any function

We can get the instantaneous rate of
change of any function, not just of position.
If $f$ is a function of $x$, then the instantaneous rate of change
at $x=a$ is theaverage rate of
change over a short interval, as we make that
interval smaller and smaller. In other words, we want to look
at $$\lim_{x \to a} \frac{\Delta f}{\Delta x} = \lim_{x \to a}
\frac{f(x)-f(a)}{x-a}.$$ This is the slope of the line tangent to
$y=f(x)$ at the point $(a,f(a))$. It can also be written as a limit
$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h},$$ where $h$ is substituted
for $x-a$.

If this limit exists, we call it the derivative
of $f$ at $x=a$.

Example We use this definition
to compute the derivative at $x=3$ of the function $f(x)=\sqrt
x$. We will need to compute either
$$(1)\quad \lim_{x\to 3}\frac{f(x)-f(3)}{x-3},\qquad\text{ or
}\qquad (2)\quad \lim_{h\to 0}\frac{f(3+h)-f(3)}{h}.$$

DO:Work
through these examples carefully. Try to compute both
limits above before looking at the solutions below.

We see that the derivative of $f$ at
$x=3$ is $\frac{1}{2\sqrt 3}$, i.e. $f'(3)=\frac{1}{2\sqrt 3}$.

Just as in the prior slide computing instantaneous velocity,
notice that in both computations (until the denominator was
cancelled) if we plugged $x=3$ in, we would get the
indeterminate form $\frac{0}{0}$ (try
it) so we had to do
more work.

Also, again notice that the work required in either computation
is similar. Be sure you practice
method (2)!