Unit 2 · Topic 2.2 · about 30 minutes · 7 parts · 12 scored checks, 2 of them multiple choice, plus 2 from earlier lessons

AP Calculus AB 2.2: Defining the Derivative of a Function and Using Derivative Notation

Q: What is the difference between f'(a) and f'(x)?

f'(x) is the derivative as a function, a rule that gives the rate of change at any input. f'(a) is one value of it, a number: the rate at the input a, and the slope of the tangent line there.

Q: Is dydx a fraction?

No. It is one symbol for the derivative, the limit of change in ychange in x. The notation is a reminder of that quotient. Treat it as a single symbol, not as one number divided by another. Later topics sometimes write dy and dx apart in a step, but dydx itself is still the limit.

Q: What units does a derivative have?

The output's units per unit of input. If V(t) is in liters and t in minutes, V'(t) is in liters per minute, and a negative value means the volume is going down.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 2: 10 to 15% of the multiple-choice section

What you'll be able to do

Write the derivative of a function as the limit of a difference quotient, in each of its notations, and work out that limit.
Write an equation for the line tangent to a curve at a given point.

Key vocabulary

Term	What it means	The exam trap
Derivative of a function	The function $f^{'}$ whose value at $x$ is $\lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ , provided this limit exists.	Keep writing $\lim_{h \to 0}$ on every line until you actually take the limit; a line without it is a difference quotient, not the derivative.
Derivative notation: $f^{'} (x)$ , $y^{'}$ and $\frac{d y}{d x}$	For $y = f (x)$ , three names for the same derivative, read "f prime of x," "y prime" and "dee y, dee x."	The letters follow the function and its input, so a function $Q$ of $r$ has derivative $Q^{'} (r)$ or $\frac{d Q}{d r}$ .
Line tangent to a curve	The line through a point on a graph whose slope is the derivative at that point, the limit of the slopes of secant lines through that point.	A tangent line can cross the graph somewhere else; it is tangent at the one point named.
Representations of the derivative	A derivative can be given graphically, numerically, analytically or verbally: as slopes or a graph, a table, a formula, or words with units.	A table can list $f$ and $f^{'}$ side by side, so read the column heading before you use a number.

Two stacked graphs. The top graph is f(x) = x cubed over 4 minus 3x for x from -4 to 4, with short tangent lines at x = -3, 0 and 2 whose slopes are 3.75, -3 and 0. The bottom graph is the graph of f prime, with dots at (-3, 3.75), (0, -3) and (2, 0), one for each tangent slope above it.

Part 1 Think first One limit for every input?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

Maya used the definition of the derivative three times for $f (x) = x^{2}$ , one input at a time, and got $f^{'} (1) = 2$ , $f^{'} (2) = 4$ and $f^{'} (3) = 6$ . Her homework now asks for $f^{'} (10)$ . What is it?

In short. The derivative of $f$ is the function $f^{'}$ whose value at each input $x$ is $\lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ , wherever that limit exists, and its value $f^{'} (a)$ is the slope of the line tangent to the graph of $f$ at $x = a$ .

For $y = f (x)$ , the derivative can be written $f^{'} (x)$ , $y^{'}$ or $\frac{d y}{d x}$ , and its units are output units per unit of input.
The same derivative can be shown as a formula, a table of values, a graph or a sentence with units.
The line tangent to the graph of $f$ at $x = a$ passes through $(a, f (a))$ with slope $f^{'} (a)$ : $y - f (a) = f^{'} (a) (x - a)$ .

Part 2 Learn it Four short ideas, each checked as you go

2a. The derivative is a function

In 2.1 Defining Average and Instantaneous Rates of Change at a Point you found $f^{'} (a)$ at one number $a$ , and each new input meant a new limit. Do the same algebra with the letter $x$ in place of the number, and the limit comes out as a formula that works at every input.

The derivative of $f$ is the function $f^{'}$ whose value at $x$ is

f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h},

provided this limit exists. While the limit is taken, $x$ stays fixed and only $h$ moves.

Take $f (x) = 2 x^{2} + x$ . Then $f (x + h) = 2 x^{2} + 4 x h + 2 h^{2} + x + h$ , and subtracting all of $f (x)$ leaves $4 x h + 2 h^{2} + h$ . For $h \neq 0$ ,

\frac{4 x h + 2 h^{2} + h}{h} = 4 x + 2 h + 1,

which approaches $4 x + 1$ as $h \to 0$ . So $f^{'} (x) = 4 x + 1$ , and one limit now answers every input: $f^{'} (1) = 5$ and $f^{'} (- 1) = - 3$ .

At an input where the limit does not exist, as for $| x |$ at 0, $f^{'}$ has no value. The derivative is defined only where its limit is.

Quick check · Calculate

No calculator. Let $g (x) = 3 x - x^{2}$ . Use the definition of the derivative to find $g^{'} (x)$ . What is $g^{'} (5)$ ?

Your answer

2b. Three ways to write it

For $y = f (x)$ , the derivative has three names you will see: $f^{'} (x)$ , read "f prime of x"; $y^{'}$ , read "y prime"; and $\frac{d y}{d x}$ , read "dee y, dee x." They name the same function. For $y = 2 x^{2} + x$ from 2a, $f^{'} (x) = 4 x + 1$ , $y^{'} = 4 x + 1$ and $\frac{d y}{d x} = 4 x + 1$ say one thing three ways.

$\frac{d y}{d x}$ is a reminder of where the derivative came from: a change in $y$ over a change in $x$ , taken in the limit. It is one symbol, not a fraction of two numbers.

The letters follow the function. If $V (t)$ is a volume at time $t$ , its derivative is $V^{'} (t)$ or $\frac{d V}{d t}$ . At one input, write $f^{'} (3)$ , or say $\frac{d y}{d x}$ at $x = 3$ .

A derivative's units are the output's units per unit of input, because it is a limit of change in output over change in input. If $y$ is in meters and $x$ in seconds, $\frac{d y}{d x}$ is in meters per second.

Quick check · Fill the statement

No calculator. Tap a blank, then tap an answer.

A block of ice melts on a counter. It weighs $y = M (t)$ grams $t$ minutes after it is set out. Its derivative can be written $M^{'} (t)$ , $\frac{d y}{d t}$ or , and it is measured in .

2c. The slope of the tangent line

Picture the secant lines of 2.1, each through $(a, f (a))$ and a second point on the graph. As the second point slides toward $a$ , the secants close in on one line through $(a, f (a))$ : the line tangent to the graph at that point. Its slope is the limit of the secant slopes, and that limit is the derivative. So $f^{'} (a)$ is the slope of the line tangent to the graph of $f$ at $x = a$ .

That gives the tangent line from two numbers: the point $(a, f (a))$ and the slope $f^{'} (a)$ . In point-slope form,

y - f (a) = f^{'} (a) (x - a) .

For $f (x) = 2 x^{2} + x$ at $x = 1$ : $f (1) = 3$ , and $f^{'} (1) = 4 (1) + 1 = 5$ , so the tangent line is $y - 3 = 5 (x - 1)$ , which is also $y = 5 x - 2$ .

The slope has to be the number $f^{'} (a)$ , not the formula $f^{'} (x)$ , because a line has one slope. Where $f^{'} (a) = 0$ , the tangent line is horizontal. In the top graph of the picture, the line at $x = 2$ is one.

Quick check · Multiple choice

No calculator. Let $f (x) = x^{2} - 2 x + 3$ , whose derivative is $f^{'} (x) = 2 x - 2$ . Which is an equation of the line tangent to the graph of $f$ at $x = 3$ ?

2d. One derivative, four ways to show it

The same derivative can be shown four ways, and a question can hand you any of them.

Analytically, as a formula: $f^{'} (x) = 4 x + 1$ for the function of 2a.
Numerically, as a table of values: at $x = 0$ , 1 and 2, $f^{'} (x)$ is 1, 5 and 9.
Graphically, as slopes: $f^{'} (a)$ is how steep the graph of $f$ is at $a$ , and the values of $f^{'}$ can be plotted as a graph of their own. In the picture at the top, each tangent slope on the graph of $f$ is a dot on the graph of $f^{'}$ below it.
Verbally, as a sentence with units: if $s (t)$ is a bike's distance in meters after $t$ seconds, $s^{'} (4) = 6$ says that at 4 seconds the distance is increasing at 6 meters per second.

Whichever way it comes, a value of $f^{'}$ is a rate, never an output. $s (4)$ is how far the bike has gone after 4 seconds; $s^{'} (4)$ is how fast that distance is growing at that instant.

Quick check · Match

No calculator. A town has $P (t)$ thousand residents $t$ years after the start of 2020. Tap a statement, then what it says in words.

Common mistakes

The mistake	Why it is wrong	Do this instead
Using $f (a)$ as the slope of the tangent line, or swapping $f (a)$ and $f^{'} (a)$ in $y - f (a) = f^{'} (a) (x - a)$ .	$f (a)$ is a height on the graph and fixes the point the line passes through. Only $f^{'} (a)$ is a slope.	Before writing the line, label the two numbers: the point $(a, f (a))$ and the slope $f^{'} (a)$ .
Writing the formula $f^{'} (x)$ as the slope of a tangent line.	A line has one slope. With $f^{'} (x)$ in it, the right side changes with $x$ in two places, and the equation is no longer a line.	Evaluate the derivative at the point first, $f^{'} (a)$ , and use that number.
Subtracting $f (x)$ without parentheses in $f (x + h) - f (x)$ , so only its first term changes sign.	The whole of $f (x)$ is subtracted. Missing parentheses leave a term that does not cancel, and the quotient then has no limit or the wrong one.	When you replace $f (x)$ by its formula, put the whole formula in parentheses, as in $- (x^{2} - 7 x)$ , then distribute the minus sign to every term: $- x^{2} + 7 x$ .

Part 3 Watch it done A derivative from the definition, then its tangent line

Watch each step, then do the last one yourself.

Let $f (x) = 3 x^{2} - 4 x$ . Use the definition of the derivative to find $f^{'} (x)$ . Then write an equation for the line tangent to the graph of $f$ at $x = 2$ .

Step 1 · Write the limit
$f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
Step 2 · Expand the top
$f (x + h) = 3 (x + h)^{2} - 4 (x + h) = 3 x^{2} + 6 x h + 3 h^{2} - 4 x - 4 h$ . Subtracting $f (x) = 3 x^{2} - 4 x$ leaves $6 x h + 3 h^{2} - 4 h$ .
Step 3 · Simplify the quotient
$\frac{6 x h + 3 h^{2} - 4 h}{h} = 6 x + 3 h - 4$ , for every $h \neq 0$ .
Step 4 · Take the limit
$f^{'} (x) = \lim_{h \to 0} (6 x + 3 h - 4) = 6 x - 4$ .
Why this step?
Why does $x$ stay in the answer while $h$ disappears?
The limit is about $h$ approaching 0, with $x$ held fixed at whatever input you choose. So every $h$ term goes to 0, and the $x$ terms are left as a formula: the derivative at each input $x$ .
Step 5 · Find the point
$f (2) = 3 (4) - 4 (2) = 4$ , so the tangent line passes through $(2, 4)$ .
Your step · Calculate

No calculator. Finish it. For $f (x) = 3 x^{2} - 4 x$ , $f^{'} (x) = 6 x - 4$ , and the line tangent to the graph at $x = 2$ is $y - 4 = m (x - 2)$ . What is $m$ ?

Your answer

Part 4 Try it A level tangent and a cubic

Try it · Decide and justify

No calculator. Let $f (x) = x^{2} - 6 x$ , so $f^{'} (x) = 2 x - 6$ . A classmate says the line tangent to the graph of $f$ at $x = 3$ is horizontal. Is the classmate right?

Which reason supports your answer?

Try it · Calculate

No calculator. Let $p (x) = x^{3} - 3 x^{2}$ , whose derivative is $p^{'} (x) = 3 x^{2} - 6 x$ . The line tangent to the graph of $p$ at $x = 3$ crosses the $y$ -axis. At what $y$ -value?

Your answer

Part 5 Explain it Write it out, then score it

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. Water flows into a tank. The volume of water in the tank, in liters, $t$ minutes after the water starts, is $V (t) = 200 + 30 t - 2 t^{2}$ for $0 \leq t \leq 7$ .

(a) Use the definition of the derivative to find $V^{'} (t)$ . Show the work that leads to your answer.

(b) Find $V^{'} (4)$ . Using correct units, interpret the meaning of $V^{'} (4)$ in the context of the problem.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $V^{'} (t) = \lim_{h \to 0} \frac{V (t + h) - V (t)}{h}$ . $V (t + h) = 200 + 30 t + 30 h - 2 t^{2} - 4 t h - 2 h^{2}$ , so $V (t + h) - V (t) = 30 h - 4 t h - 2 h^{2}$ . For $h \neq 0$ , $\frac{30 h - 4 t h - 2 h^{2}}{h} = 30 - 4 t - 2 h$ . So $V^{'} (t) = \lim_{h \to 0} (30 - 4 t - 2 h) = 30 - 4 t$ .

(b) $V^{'} (4) = 30 - 16 = 14$ . At time $t = 4$ minutes, the volume of water in the tank is increasing at a rate of 14 liters per minute.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $\lim_{h \to 0} \frac{V (t + h) - V (t)}{h}$ and expands the top to $30 h - 4 t h - 2 h^{2}$ .
Divides out $h$ to get $30 - 4 t - 2 h$ and takes the limit: $V^{'} (t) = 30 - 4 t$ .
Gets $V^{'} (4) = 14$ and says the volume is increasing at 14 liters per minute at $t = 4$ minutes, with units.
Finds $V (4) = 288$ and writes $y - 288 = 14 (t - 4)$ , or an equivalent equation.

Part 6 Check yourself A delivery drone's altitude

The three case questions use the same table. Answer them from it.

A delivery drone takes off from the ground. Its altitude is $A (t)$ meters, $t$ seconds after takeoff, and $A$ is differentiable. The table gives some values of $A$ and of its derivative $A^{'}$ .

Altitude of the drone and its rate of change
$t$ (seconds)	$A (t)$ (meters)	$A^{'} (t)$ (meters per second)
0	0	8
2	14	6
4	24	4
6	30	1
8	28	$- 2$

Case question 1 · Multiple choice

No calculator. Which is an equation of the line tangent to the graph of $A$ at $t = 4$ ?

Case question 2 · Calculate

No calculator. The line tangent to the graph of $A$ at $t = 2$ is extended back to $t = 0$ . What value does that line give at $t = 0$ ?

Your answer

Case question 3 · Decide and justify

No calculator. A classmate reads $A^{'} (8) = - 2$ from the table and says it means the drone came down 2 meters between $t = 7$ and $t = 8$ . Is the classmate right?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds $f^{'} (x)$ for $f (x) = x^{2} - 7 x$ from the definition. Tap the first line with a mistake.

Check yourself · Sort it

No calculator. Each card gives one fact about a function $f$ , with $y = f (x)$ . Tap a card, then tap whether it gives a value of $f$ or a value of $f^{'}$ .

A value of $f$

A value of $f^{'}$

Part 7 Keep it Bring back earlier ideas

This lesson stands on the difference quotient from 2.1 and the limit algebra of Unit 1. Pulling them back up is what makes them last.

From 2.1 Defining Average and Instantaneous Rates of Change at a Point · Calculate

No calculator. Let $f (x) = x^{3} - 2 x$ . What is the average rate of change of $f$ over the interval $[1, 3]$ ?

Your answer

From 1.6 Determining Limits Using Algebraic Manipulation · Calculate

No calculator. Find $\lim_{x \to 0} \frac{\frac{1}{x + 2} - \frac{1}{2}}{x}$ . Give an exact decimal.

Your answer

In three lines

The derivative of $f$ is the function $f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ , defined wherever that limit exists.
For $y = f (x)$ it is written $f^{'} (x)$ , $y^{'}$ or $\frac{d y}{d x}$ , and it can be shown as a formula, a table, a graph or in words.
$f^{'} (a)$ is the slope of the line tangent to the graph at $x = a$ , so that line is $y - f (a) = f^{'} (a) (x - a)$ .

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question asks for a tangent line, write the two numbers first, the point $(a, f (a))$ and the slope $f^{'} (a)$ , then the line in point-slope form. You can leave it there: every rearrangement is another chance for an arithmetic slip. When it asks for a derivative from the definition, write the limit on every line until you take it.

Questions students ask

What is the difference between

f^{'} (a)

and

f^{'} (x)

$f^{'} (x)$ is the derivative as a function, a rule that gives the rate of change at any input. $f^{'} (a)$ is one value of it, a number: the rate at the input $a$ , and the slope of the tangent line there.

\frac{d y}{d x}

a fraction?

No. It is one symbol for the derivative, the limit of $\frac{change in y}{change in x}$ . The notation is a reminder of that quotient. Treat it as a single symbol, not as one number divided by another. Later topics sometimes write $d y$ and $d x$ apart in a step, but $\frac{d y}{d x}$ itself is still the limit.

Do I have to use the definition every time I need a derivative?

When a question says to use the definition, yes. The later lessons of this unit build rules, such as the power rule, that give the same derivatives faster, and each of those rules comes from this limit.

Can a tangent line cross the graph?

Yes. Tangent describes the line at one point. The line tangent to the graph of $x^{3}$ at $x = 1$ , for example, crosses the graph again at $x = - 2$ .

What units does a derivative have?

The output's units per unit of input. If $V (t)$ is in liters and $t$ in minutes, $V^{'} (t)$ is in liters per minute, and a negative value means the volume is going down.

Course alignment, for teachers

AP Calculus AB topic 2.2, Unit 2: Differentiation: Definition and Fundamental Properties.

Skill 1.D: Identify an appropriate mathematical rule or procedure based on the relationship between concepts (e.g., rate of change and accumulation) or processes (e.g., differentiation and its inverse process, anti-differentiation) to solve problems.

This lesson maps all 4 essential knowledge statements of the topic to the part that teaches each one: 4 are checked by a scored question. The map is in the lesson file.