the basic idea

Instantaneous rate of change is essentially the rate of change of a function at a single point, or the "slope" of a function at a single point.

It may be easy to determine the instantaneous rate of change for a function like $y = mx + b$, since the slope is given to you as $m$, however, it may be harder to determine for more complex functions.

Approximating instantaneous rate of change

One way to calculate the instanuous rate of change at some point $(a, f(a))$ is to use the average rate of change formula such that $x_1 = a$ and $y_1 = f(a)$, but then $x_2 = a + h$ and $y_2 = f(a + h)$ for some small value of $h$.

From this information, we can come up with a formula for the approximate instantaneous rate of change:

$\text { instantaneous rate of change } \approx \frac { f(a + h) \ - \ f(a) } { (x + h) \ - \ (x) } = \frac { f(a + h) \ - \ f(a) } { h }$

Where $h$ is a small value. The smaller the value of $h$, the more accurate the value obtained from the equation.

visualization

The following animation illustrates how the instantaneous rate of change, as well as its formula works:

Calculating instantaneous rate of change

Although our current method of finding the instantaneous rate of change is somewhat accurate at small values of $h$, there is a much better method that removes the inaccuracies.

If we combine the previous formula with a limit for $h$ as it gets closer to $0$, we can get the exact value of the instantaneous rate of change at any point!

Here's the formula:

$\text { instantaneous rate of change } = \ ^{\lim}_{h \to 0} \frac {f(a+h) - f(a)} {h}$

example

Find the instantaneous rate of change of the function $f(x) = x^2 + 4x$ at $x = 2$

answer

Use the formula from above:

$\ ^{\lim}_{h \to 0} \frac {f(a + h) - f(a)} {h}$

$= \ \ ^{\lim}_{h \to 0} \frac { [(a + h)^2 + 4(a + h)] - [a^2 + 4a] } {h}$

$= \ \ ^{\lim}_{h \to 0} \frac { a^2+2ah+h^2+4a+4h - a^2+4a} {h}$

$= \ \ ^{\lim}_{h \to 0} \frac { \cancel{a^2} + 2ah + h^2 + \cancel{4a} + 4h - \cancel{a^2} + \cancel{4a} } {h}$

$= \ \ ^{\lim}_{h \to 0} \frac { 2ah + h^2 + 4h } {h}$

$= \ \ ^{\lim}_{h \to 0} \frac { h(2a + h + 4) } {h}$

$= \ \ ^{\lim}_{h \to 0} \frac { \cancel{h} (2a + h + 4) } { \cancel{h} }$

$= \ \ ^{\lim}_{h \to 0} (2a + h + 4)$

$= 2a + 4$

$= 2(2) + 4 \quad$ (since $x = a = 2$)

$= 8$

$\therefore$ The instantaneous rate of change of $f(x)$ at $x = 2$ is $8$