Definition of the derivative

Average rate of change

If we want to know how fast something is changing, we look at the rate of change. Think about your phone’s battery percentage throughout the day. If you check it at two different points in time, time $a$ and time $b$, then the average rate of change is the difference in battery percentages, divided by the time that has passed:

This tells you how quickly the battery drained, on average, between time $a$ and time $b$, and with it you might be able to estimate how it’ll last before needing a recharge (this is a simplified version of how your phone already estimates that time left).

From algebra, this is just the slope - the change in $y$-values divided by the change in $x$-values - of the line that connects two points on $f(x)$. You may have seen the slope of a line written as

$$m=\frac{\Delta y}{\Delta x}$$

where the delta symbol $\Delta$ means “change.”

Instantaneous rate of change

In reality, the battery doesn’t deplete at a steady rate. Watching videos and keeping it on uses more power, while leaving it off and idle slows the drain. How does the phone “know” how quickly it’s draining at any specific moment? The average rate of change gives a general idea, but to figure out the instantaneous rate of change at a specific moment in time, we have to zoom in closer to see how the battery changed over a tiny interval in time.

This is where limits come in - instead of two static points $a$ and $b$, fix point $a$ and bring the variable $x$ closer and closer to $a$, then observe what happens to the battery function as $x$ approaches $a$. Because the difference between one moment $a$ and the next moment $x$ becomes infinitesimally small, the limit is the instantaneous rate of change, or how quickly the function is changing right at point $a$. The instantaneous rate of change at $a$ is also called the derivative of $f(x)$ at $x=a$, defined as:

The notation $f^{\prime }(a)$ is read as “$f$ prime of a” and the prime symbol (${}^{\prime }$) is often used to indicate a derivative. Later on you’ll also work with second derivatives, represented by $f^{\prime \prime }(a)$ (“$f$ double prime of a”)

Other limit definition of the derivative

We can also define a general derivative function that gives the instantaneous rate of change - how fast $f(x)$ is changing - if you plug in any value $x$ into $f^{\prime }(x)$.

This time, fix point $x$, and look at how the function changes when we move some amount $h$ away from it. In this form, the average rate of change is called the difference quotient:

In some problems, $h$ is replaced with the symbol $\Delta x$, but the idea is the same. This formula might look intimidating, but it’s really just the slope formula applied over a small interval $h$ around $x$ - a change in the function values divided by a change in the input values.

To get the instantaneous rate of change, or the derivative, at the specific point $x$, the difference $h$ needs to be as small as possible. This is where limits are used again - to see what happens to a function when $h$ approaches 0.

Then another limit definition of the derivative of $f(x)$ is:

This limit process creates a new function $f^{\prime }(x)$. Plugging in any value of $x$ into the derivative function will give the instantaneous rate of change at that point.

Examples

1. Find the derivative of $f(x)=x^2$ using the limit definition.

Let’s use both forms to show the difference:

• 2nd form:

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(x+h{)}^2-x^2}{h}\\ & =\underset{h\to 0}{\lim }\frac{\cancel{x^2}+2xh+h^2\cancel{-x^2}}{h}\\ & =\underset{h\to 0}{\lim }\frac{2xh+h^2}{h}\\ & =\underset{h\to 0}{\lim }\frac{\cancel{h}(2x+h)}{\cancel{h}}\\ & =\underset{h\to 0}{\lim }(2x+h)\\ f^{\prime }(x)& =\boxed{2x}\end{array}$$

• 1st form:

$$\begin{array}{rl}{f}^{\prime }(a)& =\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}\\ & =\underset{x\to a}{\lim }\frac{x^2-a^2}{x-a}\\ & =\underset{x\to a}{\lim }\frac{(x+a)\cancel{(x-a)}}{\cancel{(x-a)}}\\ & =\underset{x\to a}{\lim }(x+a)\\ & =a+a\\ f^{\prime }(a)& =2a\end{array}$$

The answers are the same other than different letters used. The 2nd form is just the general derivative that uses the same variable $x$ as the original function does, but plugging in any value of $x$ into $2x$ or $a$ into $2a$ will result in the same instantaneous rate of change if $x=a$.

2. If $f(x)=\sqrt{x}$, find $f^{\prime }(9)$.

Solution

As a bonus exercise, try to figure out the derivative of $f(x)=\sqrt{x}$ using the second form of the limit definition on your own. The answer should be $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ and plugging in $x=9$ will result in the same instantaneous rate of change found using the 1st limit definition.

Working backwards

Most derivative problems involve finding $f^{\prime }(x)$ given a function $f(x)$. However, some AP problems present a derivative $f^{\prime }(x)$ as a limit expression and ask you to determine the original function $f(x)$ and the point $a$ it’s evaluated at.

The key to these problems is to recognize which form of the limit definition is being used:

$$f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$$

$$\text{or}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Examples

1. Find a function $f(x)$ and a number $a$ such that the limit represents $f^{\prime }(a)$:

$$\underset{h\to 0}{\lim }\frac{(3+h{)}^2-9}{h}$$

Solution

2. Find $f(x)$ and a number $a$ such that the limit represents $f^{\prime }(a)$:

$$\underset{x\to 5}{\lim }\frac{-1+e^{(3x-15)}}{x-5}$$

Solution

Using tables

Some questions will present a table and ask you to find the average rate of change on the interval.

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Estimate the average rate of change over the time interval $4\le t\le 7$ based on the table.

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

Most if not all of the questions in AP Calculus assign time as the independent variable, or $t$, so the water depth, $D(t)$, is a function of time.

Because the water level is increasing over time, we should expect the rate of change to be positive.

The average rate of change is

$$\frac{D(7)-D(4)}{7-4}=\frac{16-8}{3}=\boxed{\frac{8}{3}}$$

Because depth, on top, is in inches, and time, on the bottom, is in minutes, the units for this answer are in inches per minute. This means that, on average, the water depth increased by $\frac{8}{3}$ inches per minute between minute $4$ and $7$.

You may also be required to “estimate the instantaneous rate of change” given a table. In these cases, you’re actually expected to calculate the average rate of change to use as an estimate for the instantaneous rate of change, only because of the limited information presented.

2. Using the same table, estimate the instantaneous rate of change of the water depth at $3$ minutes and $9$ minutes.

Solution