Is (f-1)'(x) the same as 1f'(x)?

Only by accident. The rule is (f-1)'(x) = 1f'(f-1(x)): f' is evaluated at f-1(x), the input that f sends to x, not at x itself.

Unit 3 · Topic 3.3 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 3.3: Differentiating Inverse Functions

Q: Do I need a formula for the inverse to find its derivative?

No. You need f, f' and the matching input a with f(a) = b. That is why the rule is useful: many inverses, like the inverse of x5 + 5x + 1, have no formula you can write.

Q: Why is the inverse's slope 1 over the original slope?

The graph of f-1 is the graph of f reflected across y = x, which swaps every rise with its run. A tangent line with slope riserun becomes one with slope runrise.

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

Calculate the derivative of an inverse function, at a point or as an expression, from the chain rule and the definition of an inverse, and recognize where that derivative does not exist.

Key vocabulary

Term	What it means	The exam trap
Inverse function, $f^{- 1}$	The function that reverses $f$ : $f^{- 1} (b) = a$ exactly when $f (a) = b$ , so $f (f^{- 1} (x)) = x$ for every $x$ in the domain of $f^{- 1}$ .	Read the $- 1$ in $f^{- 1}$ as the name of the inverse, never as an exponent.
Chain rule	The rule for differentiating a composition: the derivative of $f (g (x))$ is $f^{'} (g (x)) \cdot g^{'} (x)$ .	It applies to every composition, including one whose inner function is an inverse with no formula.
Derivative of an inverse function	For the inverse $g$ of a differentiable function $f$ , the rate of change $g^{'} (x) = \frac{1}{f^{'} (g (x))}$ , provided the derivative exists.	The rule involves two different inputs, one for $g$ and one for $f^{'}$ ; decide which number goes where before you substitute.

Graph of f(x) = x cubed over 16 plus x over 2 for x from 0 to 4 and of its inverse for x from 0 to 6, reflected across the dashed line y = x, with tangent lines at the point (2, 1.5) on f and at its mirror point (1.5, 2) on the inverse.

Part 1 Think first Minutes per kilometer from kilometers per minute

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

Predict first

Twelve minutes into a ride, a cyclist has covered 5 kilometers and is riding at 0.4 kilometer per minute. Her bike computer can also show the time on the clock as a function of the distance covered. At the 5 kilometer mark, how fast is that time changing, in minutes per kilometer?

In short. If $g$ is the inverse of a differentiable function $f$ , applying the chain rule to $f (g (x)) = x$ gives $g^{'} (x) = \frac{1}{f^{'} (g (x))}$ , provided the derivative of $g$ exists.

Differentiating both sides of $f (g (x)) = x$ gives $f^{'} (g (x)) \cdot g^{'} (x) = 1$ , and no formula for $g$ is needed.
At a point, if $f (a) = b$ , then $g^{'} (b) = \frac{1}{f^{'} (a)}$ , and the input $a$ comes from solving $f (a) = b$ .
Where $f^{'} (a) = 0$ , the inverse has a vertical tangent at $(b, a)$ and no derivative there.

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

3a. Differentiate the equation that defines an inverse

Suppose $f$ has an inverse function $g$ , so $g = f^{- 1}$ . By the definition of an inverse, $g$ undoes $f$ and $f$ undoes $g$ :

f (g (x)) = x

for every $x$ in the domain of $g$ .

Both sides are functions of $x$ , so their derivatives are equal. The left side is a composition, so it takes the chain rule: the outer derivative evaluated at the inner function, times the inner derivative. The right side has derivative 1.

f^{'} (g (x)) \cdot g^{'} (x) = 1

Provided the derivative of $g$ exists, divide by $f^{'} (g (x))$ :

g^{'} (x) = \frac{1}{f^{'} (g (x))} .

You never needed a formula for $g$ . That matters, because a function such as $f (x) = x^{3} + x$ has an inverse whose formula is far too messy to use, and its derivative is still within reach.

You can confirm the rule on a pair you already know. The natural logarithm is the inverse of $e^{x}$ , so take $f (x) = e^{x}$ and $g (x) = \ln x$ . Then $f^{'} (x) = e^{x}$ , and the rule gives $g^{'} (x) = \frac{1}{e^{\ln x}} = \frac{1}{x}$ , the derivative of $\ln x$ you learned before.

Quick check · Fill the statement

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

Let $g$ be the inverse function of $f$ , so $f (g (x)) = x$ . By the chain rule, the derivative of the left side is times $g^{'} (x)$ . The derivative of the right side is . So, provided the derivative of $g$ exists, $g^{'} (x)$ is .

3b. At a point, the input and the output trade places

Most questions ask for the derivative of an inverse at one number. Say $f (a) = b$ . Then $g (b) = a$ , because $g$ sends $b$ back to $a$ . Put $x = b$ into the rule:

g^{'} (b) = \frac{1}{f^{'} (g (b))} = \frac{1}{f^{'} (a)} .

So the slope of $g$ at $b$ is 1 over the slope of $f$ at $a$ . On the graph at the top, the point $(2, 1.5)$ on $f$ and the point $(1.5, 2)$ on its inverse are mirror images across $y = x$ , and so are their tangent lines. The tangent to $f$ at $(2, 1.5)$ has slope 1.25, and the tangent to the inverse at $(1.5, 2)$ has slope 0.8, which is $\frac{1}{1.25}$ . Reflecting swaps rise and run, which is why one slope is 1 over the other.

The real work is finding $a$ . You are handed $b$ , an output of $f$ , so solve $f (a) = b$ .

Take $f (x) = x^{3} + x$ and find $g^{'} (10)$ . Solve $a^{3} + a = 10$ : $a = 2$ works, since $8 + 2 = 10$ . Then $f^{'} (x) = 3 x^{2} + 1$ , so $f^{'} (2) = 13$ and $g^{'} (10) = \frac{1}{13}$ .

Notice what was never computed: $f^{'} (10)$ . That is the slope of $f$ where its input is 10, a different point of the graph, far from $(2, 10)$ .

Quick check · Calculate

No calculator. Let $f (x) = x^{3} + 2 x - 1$ , and let $g$ be the inverse function of $f$ . Find $g^{'} (2)$ . Give an exact decimal.

Your answer

3c. Where the inverse has no derivative

The rule divides by $f^{'} (a)$ , so it says nothing where $f^{'} (a) = 0$ . In fact the inverse has no derivative there. If $g^{'} (b)$ existed, the chain rule step would give $f^{'} (a) \cdot g^{'} (b) = 1$ , and 0 times any number is 0, never 1.

The graph shows the same thing. Where $f$ has a horizontal tangent at $(a, b)$ , its mirror image across $y = x$ has a vertical tangent at $(b, a)$ , and a vertical line has no slope.

Take $f (x) = x^{3}$ . It is increasing, so it has an inverse, $g (x) = \sqrt[3]{x}$ . But $f^{'} (0) = 0$ , so $g$ has no derivative at $f (0) = 0$ : its graph stands straight up at the origin. At every other input $f^{'} (a) = 3 a^{2}$ is positive, and $g^{'} (a^{3}) = \frac{1}{3 a^{2}}$ exists.

That is what the condition "provided the derivative exists" protects you from. Before you divide, check the slope you are dividing by. For a differentiable function with an inverse, the derivative of the inverse at $b$ exists exactly when $f^{'} (a) \neq 0$ , where $f (a) = b$ .

Quick check · Decide and justify

No calculator. Let $f (x) = x^{3} - 3 x^{2} + 3 x$ . Then $f^{'} (x) = 3 (x - 1)^{2}$ , $f$ is increasing, and $f (1) = 1$ , so $f$ has an inverse function $g$ with $g (1) = 1$ . A classmate says $g^{'} (1) = 0$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Evaluating $f^{'}$ at the number you were given: writing $g^{'} (b) = \frac{1}{f^{'} (b)}$ .	$b$ is an output of $f$ . The slope of $g$ at $b$ matches the slope of $f$ at the point where $f$ outputs $b$ , which is $x = a$ .	Solve $f (a) = b$ first, then use $g^{'} (b) = \frac{1}{f^{'} (a)}$ .
Reading $f^{- 1}$ as $\frac{1}{f}$ , so writing $g^{'} (b) = \frac{1}{f (a)}$ or $g (b) = \frac{1}{f (b)}$ .	The $- 1$ names the inverse function; it is not an exponent. The rule takes 1 over the derivative $f^{'}$ , not 1 over $f$ .	Read $g (b)$ as the input that $f$ sends to $b$ , and put $f^{'}$ in the denominator.
Using the rule where $f^{'} (a) = 0$ and reporting a slope of 0 or of $\frac{1}{0}$ .	A horizontal tangent of $f$ reflects to a vertical tangent of the inverse, so the derivative of the inverse does not exist there.	Check that $f^{'} (a) \neq 0$ before you divide, and say the derivative does not exist when it is 0.

Part 3 Watch it done The derivative of an inverse with no formula

Watch each step, then do the last one yourself.

Let $p (x) = x^{5} + 5 x + 1$ . The function $p$ is increasing, so it has an inverse function $q$ . Find $q^{'} (7)$ .

Step 1 · Check the slope you will divide by
$p^{'} (x) = 5 x^{4} + 5$ , which is at least 5 for every $x$ . So $p^{'}$ is never 0, and the rule for $q^{'}$ never divides by 0.
Step 2 · Write the rule from the chain rule
$p (q (x)) = x$ , so $p^{'} (q (x)) \cdot q^{'} (x) = 1$ and $q^{'} (7) = \frac{1}{p^{'} (q (7))}$ , provided $p^{'} (q (7)) \neq 0$ . Since $p^{'} (x) \geq 5$ , that holds.
Step 3 · Find the matching input
$q (7)$ is the input $a$ with $p (a) = 7$ : $a^{5} + 5 a + 1 = 7$ , so $a^{5} + 5 a = 6$ . Trying $a = 1$ gives $1 + 5 = 6$ , so $q (7) = 1$ . It is the only solution, because $p$ is increasing.
Why this step?
Why not put 7 into $p^{'}$ ?
7 is an output of $p$ , not an input. The tangent line of $q$ at $(7, 1)$ is the mirror image of the tangent line of $p$ at $(1, 7)$ , so the slope that matters is $p^{'} (1)$ . $p^{'} (7)$ is the slope at $(7, p (7))$ , a different point.
Your step · Calculate

No calculator. Finish it. For $p (x) = x^{5} + 5 x + 1$ with inverse function $q$ , you have $q^{'} (7) = \frac{1}{p^{'} (q (7))}$ and $q (7) = 1$ . What is $q^{'} (7)$ ? Give an exact decimal.

Your answer

Part 4 Try it A table and a new function

Try it · Match

No calculator. The function $r$ is differentiable and increasing, so it has an inverse function $r^{- 1}$ . Some values of $r$ and $r^{'}$ are in the table.

\begin{matrix} x & 1 & 2 & 3 & 4 \\ r (x) & 2 & 5 & 9 & 14 \\ r^{'} (x) & 3 & 4 & 5 & 6 \end{matrix}

Tap an expression, then its value.

Try it · Calculate

No calculator. Let $f (x) = 3 x + \ln x$ for $x > 0$ , and let $g$ be the inverse function of $f$ . Find $g^{'} (3)$ . Give an exact decimal.

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 with your graphing calculator ready, and write the equation you solve 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 2, from free-response Part A (graphing calculator required).

Free-response practice · Question 2 style

Graphing calculator required. Water is pumped into an empty tank. The volume of water in the tank, in gallons, $t$ minutes after pumping starts is $W (t) = 6 e^{0.08 t} + 0.5 t - 6$ for $0 \leq t \leq 25$ . $W$ is increasing on this interval, so it has an inverse function $T$ : $T (v)$ is the time, in minutes, at which the tank holds $v$ gallons. Give decimal answers to three decimal places.

(a) Find $W^{'} (t)$ .

(b) Find $T (25)$ . Show the setup of your calculator work.

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

Model answer

(a) By the chain rule, $W^{'} (t) = 6 \cdot 0.08 e^{0.08 t} + 0.5 = 0.48 e^{0.08 t} + 0.5$ .

(b) $T (25)$ is the time $t$ at which $W (t) = 25$ . Solving $6 e^{0.08 t} + 0.5 t - 6 = 25$ on the calculator gives $T (25) = 16.626$ minutes (16.625 truncated).

(c) By the chain rule applied to $W (T (v)) = v$ , $T^{'} (25) = \frac{1}{W^{'} (T (25))}$ . With $T (25)$ stored from (b), $W^{'} (T (25)) = 2.315$ gallons per minute (2.314 truncated), so $T^{'} (25) = 0.432$ minute per gallon (0.431 truncated). When the tank holds 25 gallons, the time since pumping started is increasing at 0.432 minute per gallon of water.

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

(a) Finds $W^{'} (t) = 0.48 e^{0.08 t} + 0.5$ .
(b) Writes the equation $W (t) = 25$ as the setup and gets $T (25) = 16.626$ (or 16.625).
(c) Writes $T^{'} (25) = \frac{1}{W^{'} (T (25))}$ , evaluating $W^{'}$ at the time from (b), not at 25.
(c) Gets $T^{'} (25) = 0.432$ (or 0.431).
(c) Interprets it with units: when the tank holds 25 gallons, time is increasing at 0.432 minute per gallon.

Part 6 Check yourself A tank filling up

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

A tank is filling. The depth of the water is $D (t)$ centimeters, $t$ minutes after a valve opens, for $0 \leq t \leq 8$ . $D$ is differentiable and increasing, so it has an inverse function $T$ : $T (d)$ is the time, in minutes, at which the depth is $d$ centimeters. Some values of $D$ and $D^{'}$ are in the table.

Depth of the water and its rate of change
$t$ (minutes)	$D (t)$ (centimeters)	$D^{'} (t)$ (centimeters per minute)
0	10	2
2	16	4
4	28	8
6	40	5
8	46	1.25

Case question 1 · Calculate

No calculator. Find $T^{'} (28)$ , in minutes per centimeter. Give an exact decimal.

Your answer

Case question 2 · Multiple choice

No calculator. Which statement gives the value of $T^{'} (40)$ and its meaning, with correct units?

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

Check yourself · Spot the error

No calculator. A student finds $g^{'} (5)$ , where $g$ is the inverse function of $f (x) = x^{3} + 4 x$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. The function $f$ is differentiable and has an inverse function $f^{- 1}$ . The line tangent to the graph of $f$ at $x = 2$ is $y = 3 x - 1$ . Tap a blank, then tap an answer.

The line tangent to the graph of $f^{- 1}$ at $x = 5$ passes through the point and has slope .

Part 7 Keep it Bring back earlier ideas

This lesson runs on the chain rule, as implicit differentiation did, and on derivative rules from Unit 2. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. The point $(3, 4)$ is on the curve $x^{2} + y^{2} = 25$ . What is $\frac{d y}{d x}$ at $(3, 4)$ ?

Bring it back · Multiple choice

No calculator. Let $f (x) = 4 x^{3 / 2}$ . What is $f^{'} (4)$ ?

In three lines

Differentiating $f (g (x)) = x$ with the chain rule gives $g^{'} (x) = \frac{1}{f^{'} (g (x))}$ for the inverse $g$ of $f$ .
At a point, solve $f (a) = b$ first; then the derivative of the inverse at $b$ is $\frac{1}{f^{'} (a)}$ .
Where $f^{'} (a) = 0$ , the inverse has a vertical tangent at $(b, a)$ , and its derivative does not exist.

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 the derivative of an inverse at a value $b$ , write $g^{'} (b) = \frac{1}{f^{'} (g (b))}$ first, then find $g (b)$ by solving $f (x) = b$ . On a calculator part, write that equation as your setup before you report its solution, store the solution, and use the stored value in $f^{'}$ .

Questions students ask

Do I need a formula for the inverse to find its derivative?

No. You need $f$ , $f^{'}$ and the matching input $a$ with $f (a) = b$ . That is why the rule is useful: many inverses, like the inverse of $x^{5} + 5 x + 1$ , have no formula you can write.

(f^{- 1})^{'} (x)

the same as

\frac{1}{f^{'} (x)}

Only by accident. The rule is $(f^{- 1})^{'} (x) = \frac{1}{f^{'} (f^{- 1} (x))}$ : $f^{'}$ is evaluated at $f^{- 1} (x)$ , the input that $f$ sends to $x$ , not at $x$ itself.

How do I know a function has an inverse?

It must take each output only once. A function that is increasing on its whole domain, or decreasing on it, does. Showing that from the sign of the derivative comes in Unit 5; in this lesson, each function is given as increasing or as having an inverse.

Why is the inverse's slope 1 over the original slope?

The graph of $f^{- 1}$ is the graph of $f$ reflected across $y = x$ , which swaps every rise with its run. A tangent line with slope $\frac{rise}{run}$ becomes one with slope $\frac{run}{rise}$ .

What if solving

f (x) = b

has no algebraic solution?

On a calculator question, solve it numerically, write the equation as your setup, and store the result. On a no-calculator question, try simple inputs such as 0, 1 and 2 first, or look for the value in a table you were given.

Course alignment, for teachers

AP Calculus AB topic 3.3, Unit 3: Differentiation: Composite, Implicit, and Inverse Functions.

Skill 3.G: Confirm that solutions are accurate and appropriate.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is 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 Minutes per kilometer from kilometers per minute

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

3a. Differentiate the equation that defines an inverse

3b. At a point, the input and the output trade places

3c. Where the inverse has no derivative

Part 3 Watch it done The derivative of an inverse with no formula

Part 4 Try it A table and a new function

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A tank filling up

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic