Unit 3Derivative of an Inverse Function Formula

Concept 4.92Built for May 2027 AP Exam

Derivative of an Inverse Function Formula

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The derivative of an inverse function is the reciprocal of the original derivative, evaluated at the matching input. If $f(b) = a$ , then $\left(f^{-1}\right)'(a) = \frac{1}{f'(b)}$ . The point swaps its coordinates and the slope inverts, which is why the formula needs two pieces of information rather than one. The most common failure is evaluating $f'$ at the output $a$ instead of at the input $b$ , which produces a number that looks reasonable and is wrong. The formula also fails outright when $f'(b) = 0$ , because a horizontal tangent on the original graph becomes a vertical tangent on the inverse, where no finite slope exists.

On the AP Exam

This one lives in multiple choice, and the shape of the work explains why: the whole answer is a lookup plus a reciprocal, with nothing left to argue for nine points. The released free-response sets from 2017 to 2026 never ask for $\left(f^{-1}\right)'(a)$ , so what you buy on this page is speed on a distractor family, not credit on a page of writing.

What you’ll see: A table of $f$ and $f'$ values, or a formula for $f$ plus a request for $\left(f^{-1}\right)'(a)$ .
What you’ll do: Find the input $b$ with $f(b) = a$ , evaluate $f'(b)$ , then take the reciprocal.
Memory hook: Swap the point, invert the slope. The swap happens first, and the inversion happens afterwards.
Use it on the exam: For MCQs: The distractor built from $f'(a)$ rather than $f'(b)$ is always present, so check which input you used before checking the arithmetic. For FRQs: In nine released exams no scored part asks for it. The nearest thing is 2026 Q2, where an inverse is introduced only to switch the variable of integration, and no derivative of an inverse is ever requested.

Concept Review

The Reciprocal Comes From the Swap

Figure 1 — The inverse graph exchanges the two axes, and a slope measured against exchanged axes is the reciprocal of the original. The point moves; the slope inverts.

INVERSE SLOPES ARE RECIPROCALS

Reading the Formula

Given	Quantity	Value
$f(2) = 5$ , $f'(2) = 3$	$\left(f^{-1}\right)'(5)$	$\frac{1}{3}$
$f(1) = 4$ , $f'(1) = -2$	$\left(f^{-1}\right)'(4)$	$-\frac{1}{2}$
$f(3) = 0$ , $f'(3) = 6$	$\left(f^{-1}\right)'(0)$	$\frac{1}{6}$
$f(3) = 3$ , $f'(3) = 0$	$\left(f^{-1}\right)'(3)$	Undefined

The last row is the case the exam likes, because a zero derivative on the original curve means the inverse has a vertical tangent, and the reciprocal of zero does not exist.

Two Steps, in This Order

Find the input. Solve $f(b) = a$ for $b$ , using the table, the graph, or the formula.
Reciprocate. Evaluate $f'(b)$ and take its reciprocal.

Common Misconception Check

Myth 1: "The derivative of the inverse is the inverse of the derivative as a function." Truth: It is the reciprocal of the derivative's value at a specific input, not a new function obtained by algebra.
Myth 2: " $\left(f^{-1}\right)'(a) = \frac{1}{f'(a)}$ ." Truth: The two inputs differ. The formula evaluates $f'$ at $b$ , where $f(b) = a$ , not at $a$ .
Myth 3: "A zero slope on the original graph gives a zero slope on the inverse." Truth: It gives a vertical tangent on the inverse, so the derivative does not exist.

Topic-Specific Study Method

Find the Input First

Borrowed idea: Why does the slope of an inverse graph come out as a reciprocal? Because the inverse swaps the roles of the two axes, and a slope measured against swapped axes is the reciprocal of the original. The point $(b, a)$ on the inverse corresponds to $(a, b)$ on the original function. The slope at $a$ on the inverse is therefore the reciprocal of the slope at $b$ on the original. What has to be found first? The input $b$ , because the formula evaluates $f'$ at the original input rather than at the output.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A table	Values of $f$ and $f'$ at several inputs	A search for the row where $f$ equals the target	Locate $b$ , read $f'(b)$ , reciprocate
A formula	$f(x) = x^3 + 1$ , ask for $\left(f^{-1}\right)'(9)$	Solve $f(b) = 9$ first	$b = 2$ , then $\frac{1}{f'(2)}$
A graph	The curve of $f$ with marked points	The output is a height, the input is a location	Read across, then read the slope
A zero slope	$f'(b) = 0$ in the table	A vertical tangent on the inverse	State that the derivative does not exist

① Take a table of five $f$ values and, for each output, write down which input produced it. The search is the skill, and it takes about a minute to rehearse.

② For $f(x) = x^3 + 1$ , find $\left(f^{-1}\right)'(9)$ and then verify the answer by writing $f^{-1}$ explicitly. Seeing the two routes agree is what makes the formula believable.

③ Whenever a table gives $f'$ at every input, check whether the required output happens to equal one of the inputs. That coincidence is the most common trap in this topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f$ be differentiable with $f(2) = 5$ and $f'(2) = 3$ . Find $\left(f^{-1}\right)'(5)$ .

Standard Execution:

Name the target output. The value $5$ is an output of $f$ .

Find the input. Since $f(2) = 5$ , the matching input is $b = 2$ .

Evaluate the original slope there. $f'(2) = 3$ .

Reciprocate. $\left(f^{-1}\right)'(5) = \frac{1}{3}$ .

Stop. The answer is a slope on the inverse graph.

Exam Variation (Context / Parameter Shift): Let $f(x) = x^3 + x$ . Find $\left(f^{-1}\right)'(2)$ .

Solution: First find $b$ with $f(b) = 2$ : trying $b = 1$ gives $1 + 1 = 2$ , so $b = 1$ . Then $f'(x) = 3x^2 + 1$ , so $f'(1) = 4$ and $\left(f^{-1}\right)'(2) = \frac{1}{4}$ . The cubic is not invertible in closed form, which is exactly why the formula exists; attempting to solve for $f^{-1}$ explicitly is a dead end that the exam is designed to punish.

Common Traps & Failure Signals:

The wrong input (Signal: a reciprocal of $f'(a)$ rather than $f'(b)$ ): the output was used where the input belonged.
The unreciprocated answer (Signal: a slope of $3$ where $\frac{1}{3}$ was wanted): the swap happened but the inversion did not.
The zero reciprocal (Signal: an answer written as $\frac{1}{0}$ or as $0$ ): the vertical tangent was not recognised.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Input-Output Mix $\rightarrow$ Write " $f(b) = a$ " before touching the table.
Missed Inversion: You found the correct input and the correct slope, then reported the slope itself instead of its reciprocal, which happens because the search step feels like the whole problem and the final transformation gets skipped.
Zero Blindness: You reciprocated a zero slope without noticing.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The formula is judged in multiple choice; in nine released exams no free-response part asks for $\left(f^{-1}\right)'(a)$
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part, and the word inverse never appears in a scored stem
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Reading a Table

The differentiable function $f$ has $f(2) = 5$ and $f'(2) = 3$ . What is $\left(f^{-1}\right)'(5)$ ?

MCQ 2

A Cubic

Let $f(x) = x^3 + x$ . What is $\left(f^{-1}\right)'(2)$ ?

MCQ 3

The Zero Slope

A differentiable function satisfies $f(3) = 3$ and $f'(3) = 0$ . What can be said about $\left(f^{-1}\right)'(3)$ ?

FRQ 4

A Table and Two Inverses

The function $f$ is differentiable and strictly increasing. Selected values are given.

$x$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$5$	$8$	$12$
$f'(x)$	$2$	$4$	$6$	$9$

(a) Find $\left(f^{-1}\right)'(8)$ .

(b) Find $\left(f^{-1}\right)'(5)$ .

(c) Let $g(x) = f^{-1}(x)$ . Find $g'(12)$ and explain what the value means in terms of the graph of $f$ .

(d) A classmate computes $\left(f^{-1}\right)'(8) = \frac{1}{f'(8)}$ and stops, saying $f'(8)$ is not in the table. Explain the error.

Cheatsheet

The Inverse Function Derivative

	Use
$f(b) = a$	The matching input is $b$
$\left(f^{-1}\right)'(a)$	$\frac{1}{f'(b)}$
$f'(b) = 0$	The inverse derivative does not exist
$f$ strictly increasing	$f^{-1}$ is differentiable on its interior
A table of values	Search the $f$ row, then read the $f'$ row

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Start Topic Practice Next Topic: 3.6 Derivatives of Inverse Trigonometric Functions

Unit 3Derivative of an Inverse Function Formula

Concept 4.92Built for May 2027 AP Exam

Derivative of an Inverse Function Formula

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A table of $f$ and $f'$ values, or a formula for $f$ plus a request for $\left(f^{-1}\right)'(a)$ .
What you’ll do: Find the input $b$ with $f(b) = a$ , evaluate $f'(b)$ , then take the reciprocal.
Memory hook: Swap the point, invert the slope. The swap happens first, and the inversion happens afterwards.
Use it on the exam: For MCQs: The distractor built from $f'(a)$ rather than $f'(b)$ is always present, so check which input you used before checking the arithmetic. For FRQs: In nine released exams no scored part asks for it. The nearest thing is 2026 Q2, where an inverse is introduced only to switch the variable of integration, and no derivative of an inverse is ever requested.

Concept Review

The Reciprocal Comes From the Swap

Figure 1 — The inverse graph exchanges the two axes, and a slope measured against exchanged axes is the reciprocal of the original. The point moves; the slope inverts.

INVERSE SLOPES ARE RECIPROCALS

Reading the Formula

Given	Quantity	Value
$f(2) = 5$ , $f'(2) = 3$	$\left(f^{-1}\right)'(5)$	$\frac{1}{3}$
$f(1) = 4$ , $f'(1) = -2$	$\left(f^{-1}\right)'(4)$	$-\frac{1}{2}$
$f(3) = 0$ , $f'(3) = 6$	$\left(f^{-1}\right)'(0)$	$\frac{1}{6}$
$f(3) = 3$ , $f'(3) = 0$	$\left(f^{-1}\right)'(3)$	Undefined

The last row is the case the exam likes, because a zero derivative on the original curve means the inverse has a vertical tangent, and the reciprocal of zero does not exist.

Two Steps, in This Order

Find the input. Solve $f(b) = a$ for $b$ , using the table, the graph, or the formula.
Reciprocate. Evaluate $f'(b)$ and take its reciprocal.

Common Misconception Check

Myth 1: "The derivative of the inverse is the inverse of the derivative as a function." Truth: It is the reciprocal of the derivative's value at a specific input, not a new function obtained by algebra.
Myth 2: " $\left(f^{-1}\right)'(a) = \frac{1}{f'(a)}$ ." Truth: The two inputs differ. The formula evaluates $f'$ at $b$ , where $f(b) = a$ , not at $a$ .
Myth 3: "A zero slope on the original graph gives a zero slope on the inverse." Truth: It gives a vertical tangent on the inverse, so the derivative does not exist.

Topic-Specific Study Method

Find the Input First

Exam disguise	What you see on the page	What your brain must translate	Required execution
A table	Values of $f$ and $f'$ at several inputs	A search for the row where $f$ equals the target	Locate $b$ , read $f'(b)$ , reciprocate
A formula	$f(x) = x^3 + 1$ , ask for $\left(f^{-1}\right)'(9)$	Solve $f(b) = 9$ first	$b = 2$ , then $\frac{1}{f'(2)}$
A graph	The curve of $f$ with marked points	The output is a height, the input is a location	Read across, then read the slope
A zero slope	$f'(b) = 0$ in the table	A vertical tangent on the inverse	State that the derivative does not exist

① Take a table of five $f$ values and, for each output, write down which input produced it. The search is the skill, and it takes about a minute to rehearse.

② For $f(x) = x^3 + 1$ , find $\left(f^{-1}\right)'(9)$ and then verify the answer by writing $f^{-1}$ explicitly. Seeing the two routes agree is what makes the formula believable.

③ Whenever a table gives $f'$ at every input, check whether the required output happens to equal one of the inputs. That coincidence is the most common trap in this topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f$ be differentiable with $f(2) = 5$ and $f'(2) = 3$ . Find $\left(f^{-1}\right)'(5)$ .

Standard Execution:

Name the target output. The value $5$ is an output of $f$ .

Find the input. Since $f(2) = 5$ , the matching input is $b = 2$ .

Evaluate the original slope there. $f'(2) = 3$ .

Reciprocate. $\left(f^{-1}\right)'(5) = \frac{1}{3}$ .

Stop. The answer is a slope on the inverse graph.

Exam Variation (Context / Parameter Shift): Let $f(x) = x^3 + x$ . Find $\left(f^{-1}\right)'(2)$ .

Solution: First find $b$ with $f(b) = 2$ : trying $b = 1$ gives $1 + 1 = 2$ , so $b = 1$ . Then $f'(x) = 3x^2 + 1$ , so $f'(1) = 4$ and $\left(f^{-1}\right)'(2) = \frac{1}{4}$ . The cubic is not invertible in closed form, which is exactly why the formula exists; attempting to solve for $f^{-1}$ explicitly is a dead end that the exam is designed to punish.

Common Traps & Failure Signals:

The wrong input (Signal: a reciprocal of $f'(a)$ rather than $f'(b)$ ): the output was used where the input belonged.
The unreciprocated answer (Signal: a slope of $3$ where $\frac{1}{3}$ was wanted): the swap happened but the inversion did not.
The zero reciprocal (Signal: an answer written as $\frac{1}{0}$ or as $0$ ): the vertical tangent was not recognised.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Input-Output Mix $\rightarrow$ Write " $f(b) = a$ " before touching the table.
Missed Inversion: You found the correct input and the correct slope, then reported the slope itself instead of its reciprocal, which happens because the search step feels like the whole problem and the final transformation gets skipped.
Zero Blindness: You reciprocated a zero slope without noticing.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The formula is judged in multiple choice; in nine released exams no free-response part asks for $\left(f^{-1}\right)'(a)$
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part, and the word inverse never appears in a scored stem
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Reading a Table

The differentiable function $f$ has $f(2) = 5$ and $f'(2) = 3$ . What is $\left(f^{-1}\right)'(5)$ ?

MCQ 2

A Cubic

Let $f(x) = x^3 + x$ . What is $\left(f^{-1}\right)'(2)$ ?

MCQ 3

The Zero Slope

A differentiable function satisfies $f(3) = 3$ and $f'(3) = 0$ . What can be said about $\left(f^{-1}\right)'(3)$ ?

FRQ 4

A Table and Two Inverses

The function $f$ is differentiable and strictly increasing. Selected values are given.

$x$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$5$	$8$	$12$
$f'(x)$	$2$	$4$	$6$	$9$

(a) Find $\left(f^{-1}\right)'(8)$ .

(b) Find $\left(f^{-1}\right)'(5)$ .

(c) Let $g(x) = f^{-1}(x)$ . Find $g'(12)$ and explain what the value means in terms of the graph of $f$ .

(d) A classmate computes $\left(f^{-1}\right)'(8) = \frac{1}{f'(8)}$ and stops, saying $f'(8)$ is not in the table. Explain the error.

Cheatsheet

The Inverse Function Derivative

	Use
$f(b) = a$	The matching input is $b$
$\left(f^{-1}\right)'(a)$	$\frac{1}{f'(b)}$
$f'(b) = 0$	The inverse derivative does not exist
$f$ strictly increasing	$f^{-1}$ is differentiable on its interior
A table of values	Search the $f$ row, then read the $f'$ row

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Start Topic Practice Next Topic: 3.6 Derivatives of Inverse Trigonometric Functions

Unit 3Derivative of an Inverse Function Formula

Concept 4.92Built for May 2027 AP Exam

Derivative of an Inverse Function Formula

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A table of $f$ and $f'$ values, or a formula for $f$ plus a request for $\left(f^{-1}\right)'(a)$ .
What you’ll do: Find the input $b$ with $f(b) = a$ , evaluate $f'(b)$ , then take the reciprocal.
Memory hook: Swap the point, invert the slope. The swap happens first, and the inversion happens afterwards.
Use it on the exam: For MCQs: The distractor built from $f'(a)$ rather than $f'(b)$ is always present, so check which input you used before checking the arithmetic. For FRQs: In nine released exams no scored part asks for it. The nearest thing is 2026 Q2, where an inverse is introduced only to switch the variable of integration, and no derivative of an inverse is ever requested.

Concept Review

The Reciprocal Comes From the Swap

Figure 1 — The inverse graph exchanges the two axes, and a slope measured against exchanged axes is the reciprocal of the original. The point moves; the slope inverts.

INVERSE SLOPES ARE RECIPROCALS

Reading the Formula

Given	Quantity	Value
$f(2) = 5$ , $f'(2) = 3$	$\left(f^{-1}\right)'(5)$	$\frac{1}{3}$
$f(1) = 4$ , $f'(1) = -2$	$\left(f^{-1}\right)'(4)$	$-\frac{1}{2}$
$f(3) = 0$ , $f'(3) = 6$	$\left(f^{-1}\right)'(0)$	$\frac{1}{6}$
$f(3) = 3$ , $f'(3) = 0$	$\left(f^{-1}\right)'(3)$	Undefined

The last row is the case the exam likes, because a zero derivative on the original curve means the inverse has a vertical tangent, and the reciprocal of zero does not exist.

Two Steps, in This Order

Find the input. Solve $f(b) = a$ for $b$ , using the table, the graph, or the formula.
Reciprocate. Evaluate $f'(b)$ and take its reciprocal.

Common Misconception Check

Myth 1: "The derivative of the inverse is the inverse of the derivative as a function." Truth: It is the reciprocal of the derivative's value at a specific input, not a new function obtained by algebra.
Myth 2: " $\left(f^{-1}\right)'(a) = \frac{1}{f'(a)}$ ." Truth: The two inputs differ. The formula evaluates $f'$ at $b$ , where $f(b) = a$ , not at $a$ .
Myth 3: "A zero slope on the original graph gives a zero slope on the inverse." Truth: It gives a vertical tangent on the inverse, so the derivative does not exist.

Topic-Specific Study Method

Find the Input First

Exam disguise	What you see on the page	What your brain must translate	Required execution
A table	Values of $f$ and $f'$ at several inputs	A search for the row where $f$ equals the target	Locate $b$ , read $f'(b)$ , reciprocate
A formula	$f(x) = x^3 + 1$ , ask for $\left(f^{-1}\right)'(9)$	Solve $f(b) = 9$ first	$b = 2$ , then $\frac{1}{f'(2)}$
A graph	The curve of $f$ with marked points	The output is a height, the input is a location	Read across, then read the slope
A zero slope	$f'(b) = 0$ in the table	A vertical tangent on the inverse	State that the derivative does not exist

① Take a table of five $f$ values and, for each output, write down which input produced it. The search is the skill, and it takes about a minute to rehearse.

② For $f(x) = x^3 + 1$ , find $\left(f^{-1}\right)'(9)$ and then verify the answer by writing $f^{-1}$ explicitly. Seeing the two routes agree is what makes the formula believable.

③ Whenever a table gives $f'$ at every input, check whether the required output happens to equal one of the inputs. That coincidence is the most common trap in this topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f$ be differentiable with $f(2) = 5$ and $f'(2) = 3$ . Find $\left(f^{-1}\right)'(5)$ .

Standard Execution:

Name the target output. The value $5$ is an output of $f$ .

Find the input. Since $f(2) = 5$ , the matching input is $b = 2$ .

Evaluate the original slope there. $f'(2) = 3$ .

Reciprocate. $\left(f^{-1}\right)'(5) = \frac{1}{3}$ .

Stop. The answer is a slope on the inverse graph.

Exam Variation (Context / Parameter Shift): Let $f(x) = x^3 + x$ . Find $\left(f^{-1}\right)'(2)$ .

Solution: First find $b$ with $f(b) = 2$ : trying $b = 1$ gives $1 + 1 = 2$ , so $b = 1$ . Then $f'(x) = 3x^2 + 1$ , so $f'(1) = 4$ and $\left(f^{-1}\right)'(2) = \frac{1}{4}$ . The cubic is not invertible in closed form, which is exactly why the formula exists; attempting to solve for $f^{-1}$ explicitly is a dead end that the exam is designed to punish.

Common Traps & Failure Signals:

The wrong input (Signal: a reciprocal of $f'(a)$ rather than $f'(b)$ ): the output was used where the input belonged.
The unreciprocated answer (Signal: a slope of $3$ where $\frac{1}{3}$ was wanted): the swap happened but the inversion did not.
The zero reciprocal (Signal: an answer written as $\frac{1}{0}$ or as $0$ ): the vertical tangent was not recognised.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Input-Output Mix $\rightarrow$ Write " $f(b) = a$ " before touching the table.
Missed Inversion: You found the correct input and the correct slope, then reported the slope itself instead of its reciprocal, which happens because the search step feels like the whole problem and the final transformation gets skipped.
Zero Blindness: You reciprocated a zero slope without noticing.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The formula is judged in multiple choice; in nine released exams no free-response part asks for $\left(f^{-1}\right)'(a)$
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part, and the word inverse never appears in a scored stem
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

Reading a Table

The differentiable function $f$ has $f(2) = 5$ and $f'(2) = 3$ . What is $\left(f^{-1}\right)'(5)$ ?

MCQ 2

A Cubic

Let $f(x) = x^3 + x$ . What is $\left(f^{-1}\right)'(2)$ ?

MCQ 3

The Zero Slope

A differentiable function satisfies $f(3) = 3$ and $f'(3) = 0$ . What can be said about $\left(f^{-1}\right)'(3)$ ?

FRQ 4

A Table and Two Inverses

The function $f$ is differentiable and strictly increasing. Selected values are given.

$x$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$5$	$8$	$12$
$f'(x)$	$2$	$4$	$6$	$9$

(a) Find $\left(f^{-1}\right)'(8)$ .

(b) Find $\left(f^{-1}\right)'(5)$ .

(c) Let $g(x) = f^{-1}(x)$ . Find $g'(12)$ and explain what the value means in terms of the graph of $f$ .

(d) A classmate computes $\left(f^{-1}\right)'(8) = \frac{1}{f'(8)}$ and stops, saying $f'(8)$ is not in the table. Explain the error.

Cheatsheet

The Inverse Function Derivative

	Use
$f(b) = a$	The matching input is $b$
$\left(f^{-1}\right)'(a)$	$\frac{1}{f'(b)}$
$f'(b) = 0$	The inverse derivative does not exist
$f$ strictly increasing	$f^{-1}$ is differentiable on its interior
A table of values	Search the $f$ row, then read the $f'$ row

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