Unit 2 · Topic 2.3 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 2 from earlier lessons

AP Calculus AB 2.3: Estimating Derivatives of a Function at a Point

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

Estimate the derivative of a function at a point from a table of values, from a graph, or with a graphing calculator.

Key vocabulary

Term	What it means	The exam trap
Derivative at a point, $f^{'} (a)$	The instantaneous rate of change of $f$ at $x = a$ : the limit of the difference quotient as the interval shrinks to $a$ , provided that limit exists.	A table or a graph gives only an estimate of $f^{'} (a)$ , so write $\approx$ in front of a value you get from one.
Difference quotient	A change in output divided by the change in input, such as $\frac{f (b) - f (c)}{b - c}$ , the average rate of change from $c$ to $b$ .	Write the quotient with its values before you simplify it, so a reader can see which inputs you chose.
Tangent line	Where $f^{'} (a)$ exists, the line through $(a, f (a))$ whose slope is $f^{'} (a)$ , the line the graph runs along near that point.	When the tangent line is drawn, read your two points from the line itself, not from the curve.
Numerical derivative	A graphing calculator's estimate of $f^{'} (a)$ , worked out from a difference quotient over a very small interval around $a$ .	Write the derivative you evaluated, such as $f^{'} (4) =$ and its value, rather than the keys you pressed.

Graph of g(x) = 8 - 0.5(x - 1) squared for x from 0 to 5, with its tangent line at the point (3, 6). The tangent line also passes through (5, 2). Between the two points a run of 2 and a rise of -4 are marked, so the tangent line's slope is -2.

Part 1 Think first How fast is the battery charging?

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

Predict first

A phone is plugged in at $t = 0$ . The table gives its battery charge $B (t)$ , in percent, $t$ minutes later.

\begin{matrix} t (minutes) & 0 & 10 & 20 & 30 & 40 \\ B (t) (percent) & 12 & 31 & 45 & 57 & 66 \end{matrix}

About how fast is the charge rising at $t = 25$ minutes?

In short. When you cannot find $f^{'} (a)$ exactly, estimate it: from a table or a graph with the slope between two points close to $a$ , or from a formula with a graphing calculator's numerical derivative.

From a table, use the inputs closest to $a$ on each side, and write the difference quotient with their values before the number.
From a graph, use the slope of the tangent line at $a$ , or the slope between two points on the curve close to $a$ .
A calculator's numerical derivative is a difference quotient over a tiny interval, so it is an estimate, and it means something only where $f^{'} (a)$ exists.

Part 2 Learn it Three ways to estimate a derivative

2a. From a table: the closest values on each side

A table gives a function at only a few inputs, so it cannot give $f^{'} (a)$ exactly. It can estimate it. The derivative is the limit of average rates over intervals that shrink to $a$ , so the best the table can do is the average rate over the shortest interval it gives around $a$ .

When $a$ falls between two inputs in the table, use those two. That is what Think first did:

B^{'} (25) \approx \frac{B (30) - B (20)}{30 - 20} = \frac{57 - 45}{10} = 1.2.

When $a$ is itself in the table, the inputs just before and just after it give an interval centered on it: $B^{'} (20) \approx \frac{B (30) - B (10)}{30 - 10} = \frac{26}{20} = 1.3$ .

Write the difference quotient with the table's values before the number, as above. It shows which values you used and why the number is a rate, and on a free-response question the number alone does not earn the point. The units are the output's units per unit of input, so read the estimate back in words: about 25 minutes after it is plugged in, the charge is rising at about 1.2 percent per minute.

Quick check · Calculate

No calculator. The table gives values of a function $f$ .

\begin{matrix} x & 1 & 1.5 & 2 & 2.5 & 3 \\ f (x) & 4.0 & 5.2 & 6.1 & 6.7 & 7.0 \end{matrix}

Use the table to estimate $f^{'} (1.75)$ .

Your answer

2b. From a graph: the slope of the tangent line

On a graph, $g^{'} (a)$ is the slope of the line tangent to the graph at $x = a$ . When the tangent line is drawn, pick two points on it that you can read clearly, not too close together, and divide the rise by the run.

On the graph at the top, the tangent line at $x = 3$ passes through $(3, 6)$ and $(5, 2)$ , so

g^{'} (3) \approx \frac{2 - 6}{5 - 3} = \frac{- 4}{2} = - 2.

The slope is negative because the line falls from left to right. Read both points from the tangent line, not from the curve: away from $x = 3$ the curve bends away from the line, and at $x = 5$ the curve is at 0, not 2.

When no tangent line is drawn, read two points on the curve close to $x = a$ , one on each side, and find the slope between them. That is the same difference quotient a table gives, and the closer the two points, the better the estimate, as long as you can still read them accurately.

Quick check · Multiple choice

No calculator. An oven is cooling. On the graph of its temperature $T (t)$ , in degrees Fahrenheit, $t$ minutes after it is turned off, the points $(9, 356)$ and $(11, 344)$ lie on the curve, and no tangent line is drawn. Which estimate of $T^{'} (10)$ do these two points give?

2c. With a graphing calculator: the numerical derivative

When a function is given by a formula, a graphing calculator can estimate $f^{'} (a)$ for you. Its numerical derivative works out a difference quotient over a very small interval around $a$ , so the result is an estimate, and where the derivative exists it is a very close one.

For $f (x) = 2^{x}$ at $x = 3$ , the quotient over the interval from 2.999 to 3.001 is

\frac{2^{3.001} - 2^{2.999}}{0.002} \approx 5.5452,

and the calculator's numerical derivative gives $f^{'} (3) = 5.545$ to three decimal places.

On a free-response question, the expression you evaluated is part of the answer, so write the derivative and its value, $f^{'} (3) = 5.545$ , not the keys you pressed. Give three decimal places, rounded or truncated, as the question asks. For a function with sine or cosine, put the calculator in radian mode first.

One caution. A numerical derivative works out one difference quotient and never takes a limit, so it can show a number even where the derivative does not exist. At a sharp corner, such as the graph of $| x - 2 |$ at $x = 2$ , there is no derivative, but a quotient centered on 2 gives 0, and a calculator that uses one shows 0.

Quick check · Calculate

Graphing calculator required. In radian mode, let $f (x) = x \cos x$ . Use the calculator's numerical derivative to find $f^{'} (2)$ . Give your answer to three decimal places.

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Writing only the number for an estimate from a table.	The number alone does not show which values you used, and the difference quotient is what makes it an estimate of the derivative.	Write $f^{'} (a) \approx$ and the difference quotient with the table's values, then the number with its units.
Using an interval much wider than the table requires, such as its first and last inputs.	A wide interval averages the rate over a long stretch where it can change a lot.	Use the inputs closest to $a$ on each side.
Trusting a calculator's numerical derivative at a point where the graph has a sharp corner.	A numerical derivative works out one difference quotient and never takes the limit, so it can show a number even when $f^{'} (a)$ does not exist.	Check that the derivative exists at $a$ before you report the calculator's value.

Part 3 Watch it done An estimate from a table with uneven gaps

Watch each step, then do the last one yourself.

A gate is opened at the bottom of a reservoir. The table gives the depth of the water, $D (t)$ feet, $t$ hours later.

\begin{matrix} t (hours) & 0 & 3 & 5 & 9 & 12 \\ D (t) (feet) & 16.2 & 14.6 & 13.8 & 12.9 & 12.5 \end{matrix}

Use the table to estimate $D^{'} (4)$ . Show the work, and give units.

Step 1 · Find the inputs around 4
4 is not in the table. The closest input before it is 3 and the closest after it is 5. Other inputs, such as 0 and 9, also lie on either side of 4, but they are farther away.
Why this step?
Why use 3 and 5 rather than 3 and 9, which also lie on either side of 4?
The derivative is the limit of average rates as the interval shrinks to 4. From 3 to 5 is the shortest interval around 4 the table gives, so its average rate is the closest the data come to that limit. From 3 to 9 averages over six hours, while the rate keeps changing.
Step 2 · Write the difference quotient
$D^{'} (4) \approx \frac{D (5) - D (3)}{5 - 3} = \frac{13.8 - 14.6}{5 - 3}$
Your step · Calculate

No calculator. Finish it. Evaluate $\frac{13.8 - 14.6}{5 - 3}$ to estimate $D^{'} (4)$ , in feet per hour.

Your answer

Part 4 Try it A formula and a corner

Try it · Calculate

Graphing calculator required. The concentration of a medicine in a patient's blood is $C (t) = 20 t e^{- 0.5 t}$ milligrams per liter, $t$ hours after it is taken. Find $C^{'} (3)$ , in milligrams per liter per hour. Give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. Let $f (x) = | x - 2 |$ . A student enters $f$ in a graphing calculator, and its numerical derivative at $x = 2$ shows 0. The student concludes that $f^{'} (2) = 0$ . Is the student right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper with your graphing calculator ready. 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 2, from free-response Part A (graphing calculator required).

Free-response practice · Question 2 style

Graphing calculator required. A pot of soup is heating on a stove. Its temperature, in degrees Fahrenheit, is given by a differentiable function $S$ of time $t$ , in minutes, after the burner is turned on. Values of $S (t)$ at selected times are given in the table.

\begin{matrix} t (minutes) & 0 & 2 & 5 & 8 & 10 \\ S (t) (degrees Fahrenheit) & 70.0 & 100.3 & 134.1 & 157.6 & 169.2 \end{matrix}

(a) Use the data in the table to estimate $S^{'} (6.5)$ . Show the computation that leads to your answer. Using correct units, interpret the meaning of $S^{'} (6.5)$ in the context of the problem.

(b) For $0 \leq t \leq 10$ , the temperature of the soup is modeled by $M (t) = 212 - 142 e^{- 0.12 t}$ degrees Fahrenheit. Find $M^{'} (6.5)$ . Give your answer to three decimal places.

(c) According to the model, is the temperature of the soup increasing faster at $t = 2$ minutes or at $t = 8$ minutes? Justify your answer.

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

Model answer

(a) $S^{'} (6.5) \approx \frac{S (8) - S (5)}{8 - 5} = \frac{157.6 - 134.1}{3} = \frac{23.5}{3} \approx 7.833$ degrees Fahrenheit per minute. At $t = 6.5$ minutes after the burner is turned on, the temperature of the soup is increasing at about 7.833 degrees Fahrenheit per minute.

(b) $M^{'} (6.5) = 7.811$ degrees Fahrenheit per minute, from the calculator's numerical derivative of $M$ at $t = 6.5$ .

(c) $M^{'} (2) = 13.404$ and $M^{'} (8) = 6.524$ degrees Fahrenheit per minute. Both are positive, so the temperature is increasing at both times, and since $M^{'} (2) > M^{'} (8)$ , it is increasing faster at $t = 2$ minutes.

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

(a) Writes the difference quotient $\frac{S (8) - S (5)}{8 - 5}$ with the table's values, and gets about 7.833.
(a) Says that at $t = 6.5$ minutes the temperature is increasing at about 7.833 degrees Fahrenheit per minute, with units.
(b) Writes $M^{'} (6.5)$ and gets 7.811.
(c) Gives $M^{'} (2) = 13.404$ and $M^{'} (8) = 6.524$ , each as the derivative it evaluates.
(c) Concludes the temperature is increasing faster at $t = 2$ because $M^{'} (2) > M^{'} (8)$ .

Part 6 Check yourself A runner on a straight path

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

A runner sets off along a straight path. Her distance from the start, $s (t)$ meters, $t$ seconds later, is given at selected times.

Distance of the runner from the start
$t$ (seconds)	$s (t)$ (meters)
0	0
4	22
10	61
12	74
20	128

Case question 1 · Calculate

No calculator. Use the table to estimate $s^{'} (11)$ , in meters per second.

Your answer

Case question 2 · Multiple choice

No calculator. Which expression gives the best estimate of $s^{'} (15)$ that the table supports?

Case question 3 · Multiple choice

No calculator. From the table, $s^{'} (2) \approx \frac{s (4) - s (0)}{4 - 0} = 5.5$ . Which statement does this estimate support?

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

Check yourself · Spot the error

No calculator. A table gives $h (1) = 4.2$ and $h (2) = 3.1$ . A student uses it to estimate $h^{'} (1.5)$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. Each line is tangent to the graph of a function at the first point named, and passes through the second. Tap a tangent line, then the derivative it estimates at its point of tangency.

Part 7 Keep it Bring back earlier ideas

This lesson stands on the average rate of change from 2.1 and on estimating from a table, which you first did for limits in Unit 1.

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

No calculator. Let $g (x) = \frac{1}{x}$ . What is the value of $\frac{g (2 + h) - g (2)}{h}$ when $h = 2$ ? Give an exact decimal.

Your answer

From 1.4 Estimating Limit Values from Tables · Calculate

No calculator. The table gives values of a function $g$ near $x = 2$ .

\begin{matrix} x & g (x) \\ 1.9 & 11.41 \\ 1.99 & 11.9401 \\ 1.999 & 11.994001 \\ 2.001 & 12.006001 \\ 2.01 & 12.0601 \\ 2.1 & 12.61 \end{matrix}

Use the table to estimate $\lim_{x \to 2} g (x)$ .

Your answer

In three lines

From a table, estimate $f^{'} (a)$ with the difference quotient over the shortest interval around $a$ , written out before the number.
From a graph, use the slope of the tangent line at $a$ , or of a line through two points on the curve close to $a$ .
A graphing calculator's numerical derivative estimates $f^{'} (a)$ closely, but only where the derivative exists.

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 gives a table and asks you to estimate a derivative, write the difference quotient with the table's values before the number, and give units. When the work needs a calculator, write the derivative you are evaluating, such as $f^{'} (a) =$ , and its value to three decimal places.

Questions students ask

Why use the closest values in the table?

The derivative is the limit of average rates over intervals that shrink to the point. The shortest interval the table gives around the point is as close to that limit as the data allow, so its average rate is the best estimate the table supports.

What if the point I need is one of the inputs in the table?

Then the inputs just before and just after it give an interval centered on it, as for $B^{'} (20)$ in the lesson. If a question tells you which values to use, use those.

Is a calculator's numerical derivative exact?

No. It is a difference quotient over a tiny interval, so it is an estimate. Where the derivative exists, it is usually accurate well past three decimal places, which is all a question here asks for.

Should I round or truncate a calculator answer?

Follow the question. The questions here ask for three decimal places and accept the value rounded or truncated there. Keep the full value stored in the calculator while you work, and shorten it only for the final answer.

Can I estimate a derivative from a graph with no tangent line drawn?

Yes. Read two points on the curve close to the point you want, one on each side, and find the slope between them. It is the same difference quotient you would use with a table.

Course alignment, for teachers

AP Calculus AB topic 2.3, Unit 2: Differentiation: Definition and Fundamental Properties.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first How fast is the battery charging?

Part 2 Learn it Three ways to estimate a derivative

2a. From a table: the closest values on each side

2b. From a graph: the slope of the tangent line

2c. With a graphing calculator: the numerical derivative

Part 3 Watch it done An estimate from a table with uneven gaps

Part 4 Try it A formula and a corner

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A runner on a straight path

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic