Unit 3 · Topic 3.6 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 3.6: Calculating Higher-Order Derivatives

Q: Does every function have a second derivative?

No. f''(a) exists only when f' is differentiable at a. For f(x) = x5/3, the first derivative f'(x) = 53x2/3 exists everywhere, but f''(x) = 109x-1/3 is not defined at 0, so f''(0) does not exist.

Q: How many derivatives can I take?

As many as exist. A polynomial of degree n has a constant nth derivative, and every derivative after that is 0. Functions such as ex and x can be differentiated again and again without end.

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

Unit 3: 5 to 10% of the multiple-choice section

What you'll be able to do

Find the second, third and higher derivatives of a function, and write each one in standard notation.

Key vocabulary

Term	What it means	The exam trap
Second derivative	The derivative of $f^{'}$ , written $f^{''}$ , which exists wherever $f^{'}$ is differentiable.	Its units are the units of the first derivative per unit of input, such as people per year per year.
Higher-order derivative	Any derivative found by differentiating more than once: the second, third, fourth and later derivatives.	Simplify each derivative before you differentiate it again; a tidy expression keeps the next step short.
Leibniz notation, $\frac{d^{n} y}{d x^{n}}$	The $n$ th derivative of $y$ with respect to $x$ ; $\frac{d^{2} y}{d x^{2}}$ is the second derivative.	In a written answer, write the whole symbol, top and bottom, with the raised number in its place on each.
Prime and parenthesis notation, $f^{''} (x)$ and $f^{(n)} (x)$	Notation for derivatives of $f$ : primes for the first few, as in $f^{''} (x)$ , and $f^{(n)} (x)$ for the $n$ th derivative in general.	Count differentiations from $f$ itself: $f^{'''}$ is three steps from $f$ , not three steps from $f^{'}$ .

Graph of f(x) = x cubed over 6 minus x, its derivative f'(x) = x squared over 2 minus 1, and its second derivative f''(x) = x, for x from -3 to 3. A dashed line at x = 2 meets the curves at -0.67, 1 and 2.

Part 1 Think first How fast is the rate changing?

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

Predict first

A balloon is being inflated so that its volume after $t$ seconds is $V (t) = t^{3}$ cubic inches, for $0 \leq t \leq 5$ . The rate at which the volume grows is not constant: it changes as the balloon fills. At the instant $t = 4$ , how fast is that rate of growth itself changing?

In short. A higher-order derivative is a derivative of a derivative: differentiating $f^{'}$ gives the second derivative $f^{''}$ , provided the derivative of $f^{'}$ exists, and repeating the process gives the third, fourth and later derivatives.

Each higher derivative uses the same rules as the first, applied to the expression the previous step produced.
For $y = f (x)$ , the second derivative is written $\frac{d^{2} y}{d x^{2}}$ , $f^{''} (x)$ or $y^{''}$ , and the $n$ th derivative is $\frac{d^{n} y}{d x^{n}}$ or $f^{(n)} (x)$ .
To get a value such as $f^{''} (a)$ , differentiate first and substitute $a$ last; putting $a$ in early turns $f^{'}$ into a constant, whose derivative is 0.

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

2a. Differentiate the derivative

The derivative $f^{'}$ is a function, so it can have a derivative of its own. Differentiating $f^{'}$ produces the second derivative of $f$ , written $f^{''}$ , provided the derivative of $f^{'}$ exists. Nothing new is needed: use the same rules on $f^{'}$ that you used on $f$ .

The process repeats. Differentiate $f^{''}$ to get the third derivative $f^{'''}$ , then differentiate again for the fourth, and so on. These are the higher-order derivatives of $f$ .

Take $f (x) = x^{4} - 2 x^{3}$ :

f^{'} (x) = 4 x^{3} - 6 x^{2}, f^{''} (x) = 12 x^{2} - 12 x, f^{'''} (x) = 24 x - 12.

The fourth derivative is the constant 24, so every derivative after it is 0. For a polynomial, each step lowers the degree by one until a constant is left.

To find a value such as $f^{''} (2)$ , find the formula for $f^{''} (x)$ first and substitute last: $f^{''} (2) = 12 (4) - 12 (2) = 24$ . Substituting early does not work. $f^{'} (2) = 32 - 24 = 8$ is a single number, and the derivative of a constant is 0, which says nothing about how $f^{'}$ is changing at 2.

Quick check · Calculate

No calculator. Let $g (x) = x^{4} - 5 x^{3} + 2 x$ . What is $g^{''} (2)$ ?

Your answer

2b. Names for higher derivatives

For $y = f (x)$ the second derivative has three common names:

\frac{d^{2} y}{d x^{2}}, f^{''} (x), y^{''} .

Read $\frac{d^{2} y}{d x^{2}}$ as $\frac{d}{d x} (\frac{d y}{d x})$ : the operation $\frac{d}{d x}$ applied twice to $y$ . That is why the 2 sits on the $d$ on top and on the $x$ underneath. It is not ${(\frac{d y}{d x})}^{2}$ , the first derivative squared. For $y = x^{3}$ , $\frac{d^{2} y}{d x^{2}} = 6 x$ , but ${(\frac{d y}{d x})}^{2} = 9 x^{4}$ .

Primes get hard to count after three, so the $n$ th derivative is written

\frac{d^{n} y}{d x^{n}} or f^{(n)} (x) .

The parentheses matter. $f^{(4)} (x)$ is the fourth derivative of $f$ , while $f^{4} (x)$ is usually read as ${(f (x))}^{4}$ , a power. So $f^{'''} (x)$ , $y^{'''}$ , $\frac{d^{3} y}{d x^{3}}$ and $f^{(3)} (x)$ all name the same thing: the third derivative.

Quick check · Match

No calculator. Let $y = f (x)$ . Tap a piece of notation, then what it means.

2c. Every rule, applied again

The second derivative uses the rules you already know, on whatever $f^{'}$ turns out to be. If $f^{'}$ contains a product, the next differentiation needs the product rule; if it contains a composite function, it needs the chain rule again.

Take $y = e^{3 x}$ . The chain rule gives $\frac{d y}{d x} = 3 e^{3 x}$ . Differentiating again, the 3 is a constant multiple and $e^{3 x}$ brings another factor of 3, so $\frac{d^{2} y}{d x^{2}} = 9 e^{3 x}$ . Each differentiation multiplies by 3, so $\frac{d^{n} y}{d x^{n}} = 3^{n} e^{3 x}$ .

Sine and cosine cycle. For $f (x) = \cos x$ : $f^{'} (x) = - \sin x$ , $f^{''} (x) = - \cos x$ , $f^{'''} (x) = \sin x$ , and $f^{(4)} (x) = \cos x$ , back where it started. From there the pattern repeats every four derivatives.

The common slip is to use the chain rule on the first step and forget it on the next. The inside function is still there in $f^{'}$ , so its derivative appears as a factor every time you differentiate.

Quick check · Multiple choice

No calculator. Let $f (x) = \sin (2 x)$ . Which expression is $f^{'''} (x)$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Substituting the input before differentiating again, as in finding $f^{'} (2)$ and then differentiating that number.	$f^{'} (2)$ is a constant, and the derivative of any constant is 0. It carries no information about how $f^{'}$ changes near 2.	Find the formula for each derivative in turn, and substitute the input only at the end.
Reading $\frac{d^{2} y}{d x^{2}}$ as ${(\frac{d y}{d x})}^{2}$ , or $f^{(4)} (x)$ as ${(f (x))}^{4}$ .	Both notations count differentiations. Squaring the first derivative, or raising the function to a power, is a different operation with a different result.	Read the raised number in $\frac{d^{n} y}{d x^{n}}$ , or the number in parentheses in $f^{(n)} (x)$ , as how many times to differentiate.
Dropping a chain rule or product rule factor on the second differentiation.	$f^{'}$ still contains the inner function or the product, so differentiating it needs the same rule again.	Before each differentiation, look at the form of the expression in front of you and pick the rule for that form.

Part 3 Watch it done A second and a third derivative with the product rule

Watch each step, then do the last one yourself.

Let $f (x) = x e^{2 x}$ . Find $f^{''} (x)$ , and then find $f^{'''} (0)$ .

Step 1 · Differentiate once with the product rule
With $u = x$ and $v = e^{2 x}$ , $f^{'} (x) = u^{'} v + u v^{'} = (1) e^{2 x} + x (2 e^{2 x}) = e^{2 x} + 2 x e^{2 x}$ . The 2 in $2 e^{2 x}$ is the chain rule factor from $2 x$ .
Step 2 · Differentiate $f'$ term by term
$\frac{d}{d x} (e^{2 x}) = 2 e^{2 x}$ . The term $2 x e^{2 x}$ is a product again, so $\frac{d}{d x} (2 x e^{2 x}) = (2) e^{2 x} + 2 x (2 e^{2 x}) = 2 e^{2 x} + 4 x e^{2 x}$ .
Why this step?
Why does the second differentiation need the product rule again?
Because $f^{'} (x)$ still contains the product $2 x e^{2 x}$ . Each derivative is found from the form of the expression in front of you, and that term is $2 x$ times $e^{2 x}$ .
Step 3 · Collect the second derivative
$f^{''} (x) = 2 e^{2 x} + 2 e^{2 x} + 4 x e^{2 x} = 4 e^{2 x} + 4 x e^{2 x}$ .
Your step · Calculate

No calculator. Finish it. For $f (x) = x e^{2 x}$ , the second derivative is $f^{''} (x) = 4 e^{2 x} + 4 x e^{2 x}$ . What is $f^{'''} (0)$ ?

Your answer

Part 4 Try it A new function, then a calculator

Try it · Spot the error

No calculator. A student finds $\frac{d^{3} y}{d x^{3}}$ at $x = 2$ for $y = \frac{1}{x}$ . Tap the first line with a mistake.

Try it · Calculate

Graphing calculator required. A function $g$ has derivative $g^{'} (x) = x \cos (x^{2})$ . With your calculator in radian mode, find $g^{''} (1.3)$ . Give your answer to three decimal places.

Your answer

Part 5 Explain it Write it the way the exam asks

Free response is handwritten on the exam, so work this on paper first and write the setup beside every calculator result. 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 1, from free-response Part A (graphing calculator required).

Free-response practice · Question 1 style

Graphing calculator required. Water flows into a tank. The volume of water in the tank is $V (t)$ liters, $t$ minutes after the flow starts, and its first derivative is $V^{'} (t) = 6 + 5 \sin (\frac{t^{2}}{8})$ liters per minute for $0 \leq t \leq 6$ . Use radian mode, and give decimal answers to three decimal places.

(a) Find $V^{''} (3)$ . Give the units of your answer.

(b) Write an expression for $\frac{d^{2} V}{d t^{2}}$ in terms of $t$ . Show the rule you use.

(c) The third derivative of $V$ is written $\frac{d^{3} V}{d t^{3}}$ . Find its value at $t = 2$ . Show the setup of your calculator work.

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

Model answer

(a) $V^{''} (3)$ is the derivative of $V^{'}$ at $t = 3$ . The calculator's numerical derivative of $6 + 5 \sin (\frac{t^{2}}{8})$ at $t = 3$ gives $V^{''} (3) = 1.617$ liters per minute per minute.

(b) By the chain rule, with inner function $\frac{t^{2}}{8}$ , whose derivative is $\frac{t}{4}$ : $\frac{d^{2} V}{d t^{2}} = 5 \cos (\frac{t^{2}}{8}) \cdot \frac{t}{4} = \frac{5 t}{4} \cos (\frac{t^{2}}{8})$ .

(c) $\frac{d^{3} V}{d t^{3}}$ at $t = 2$ is the derivative of $\frac{5 t}{4} \cos (\frac{t^{2}}{8})$ at $t = 2$ . The calculator's numerical derivative gives 0.498 liters per minute per minute per minute.

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

Writes $V^{''} (3)$ as the derivative of $V^{'}$ at $t = 3$ and gets 1.617 (1.616 truncated).
Gives the units of $V^{''} (3)$ as liters per minute per minute.
Writes $\frac{d^{2} V}{d t^{2}} = \frac{5 t}{4} \cos (\frac{t^{2}}{8})$ , with the chain rule factor $\frac{t}{4}$ .
Sets up $\frac{d^{3} V}{d t^{3}}$ at $t = 2$ as the derivative of the part (b) expression and gets 0.498 (0.497 truncated).

Part 6 Check yourself A function known only from a table

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

A function $g$ is twice differentiable for all real numbers. The table gives values of $g$ , its derivative $g^{'}$ and its second derivative $g^{''}$ at three inputs.

Values of g and its first two derivatives
$x$	$g (x)$	$g^{'} (x)$	$g^{''} (x)$
1	4	$- 2$	3
2	1	$- 3$	5
3	$- 2$	2	$- 1$

Case question 1 · Calculate

No calculator. Let $k (x) = x^{2} g (x)$ . What is $k^{''} (2)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Let $m (x) = g (3 x)$ . What is $m^{''} (1)$ ?

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

Check yourself · Decide and justify

No calculator. Let $f (x) = x | x |$ . Its derivative is $f^{'} (x) = 2 | x |$ for every $x$ , so $f^{'} (0) = 0$ . A classmate says $f^{''} (0) = 0$ , because $f^{'} (0) = 0$ . Is the classmate right?

Which reason supports your answer?

Check yourself · Fill the statement

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

For $y = f (x)$ , the second derivative can be written $\frac{d^{2} y}{d x^{2}}$ , $f^{''} (x)$ or . The fifth derivative can be written $\frac{d^{5} y}{d x^{5}}$ or .

Part 7 Keep it Bring back earlier ideas

Higher derivatives stand on the chain rule and the product rule. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Let $h (x) = (x^{2} + 1)^{4}$ . Which expression is $h^{'} (x)$ ?

From 2.8 The Product Rule · Calculate

No calculator. Let $h (x) = (x^{2} + 1) (3 x - 2)$ . What is $h^{'} (2)$ ?

Your answer

In three lines

Differentiating $f^{'}$ gives the second derivative $f^{''}$ , provided it exists, and repeating gives the higher-order derivatives.
The second derivative is written $\frac{d^{2} y}{d x^{2}}$ , $f^{''} (x)$ or $y^{''}$ , and the $n$ th is $\frac{d^{n} y}{d x^{n}}$ or $f^{(n)} (x)$ .
Each step uses the usual rules on the previous derivative, and a value is substituted only at the end.

Unit 3 check: 10 questions from every lesson in Unit 3

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 part asks for a second derivative at a point, write what you are evaluating, such as $V^{''} (3)$ as the derivative of $V^{'}$ at $t = 3$ , before the number. On a calculator part, the numerical derivative of a given $f^{'}$ gives a value of $f^{''}$ directly; without a calculator, find each derivative's formula and keep every chain rule factor on each step.

Questions students ask

Does every function have a second derivative?

No. $f^{''} (a)$ exists only when $f^{'}$ is differentiable at $a$ . For $f (x) = x^{5 / 3}$ , the first derivative $f^{'} (x) = \frac{5}{3} x^{2 / 3}$ exists everywhere, but $f^{''} (x) = \frac{10}{9} x^{- 1 / 3}$ is not defined at 0, so $f^{''} (0)$ does not exist.

How many derivatives can I take?

As many as exist. A polynomial of degree $n$ has a constant $n$ th derivative, and every derivative after that is 0. Functions such as $e^{x}$ and $\sin x$ can be differentiated again and again without end.

Can my graphing calculator find a second derivative?

It finds a numerical derivative of a function you enter. Enter $f^{'}$ and take its numerical derivative at a point to get a value of $f^{''}$ . If you are given only $f$ , find $f^{'}$ by hand first.

Why write

f^{(4)}

instead of four primes?

Four or more primes are hard to count at a glance. The number in parentheses says how many times to differentiate, and the parentheses keep it from being read as an exponent.

Where do second derivatives come up later in the course?

Unit 4 reads them in context, and Unit 5 uses them to describe the shape of a graph. This topic is about finding them and naming them.

Course alignment, for teachers

AP Calculus AB topic 3.6, Unit 3: Differentiation: Composite, Implicit, and Inverse Functions.

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.