Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
An inverse function undoes the original, so its graph is the original reflected across y=x, and its slope is the original slope turned upside down. If f(a)=b and f′(a)=m, then (f−1)′(b)=m1. The formula is short, and the mistake it invites is a swap: the derivative of the inverse at b needs the original slope at a, not at b. Most exam items supply a table of values, so the work is a lookup followed by a reciprocal. One condition has to hold: f′(a) must be nonzero, since a horizontal tangent in the original becomes a vertical tangent in the inverse.
Figure 5 — A curve and its reflection across y=x. The point (2,5) becomes (5,2), and the slope 3 at the first point becomes 31 at the second.
The formula follows from the reflection. If the original rises 3 units for every 1 unit across, then the reflected curve rises 1 unit for every 3 units across, because rise and run have exchanged roles:
(f−1)′(b)=f′(a)1,where f(a)=b
| Given | Wanted | Use | Not this |
|---|---|---|---|
| f(2)=5, f′(2)=3 | (f−1)′(5) | f′(2)1=31 | f′(5), which is unknown |
| f(1)=4, f′(1)=−2 | (f−1)′(4) | −21 | −f′(4)1 |
| f(3)=1, f′(3)=0 | (f−1)′(1) | does not exist | any finite value |
The third row is the condition stated in words: a horizontal tangent in the original becomes a vertical tangent in the inverse, and a vertical tangent has no finite slope.
Most exam items give a table with rows for x, f(x), and f′(x). To find (f−1)′(k):
If k does not appear in the f(x) column, the item is telling you that f−1 is not defined there, and the slope question has no answer.
Differentiate f(f−1(x))=x with respect to x. The chain rule gives
f′(f−1(x))⋅(f−1)′(x)=1
and dividing by the first factor produces the formula. This derivation is worth keeping, because it also shows where the formula fails: the division is illegal when f′(f−1(x))=0.
Borrowed idea: Ask what the inverse actually does before differentiating it. It sends an output back to its input. So a slope of the inverse is a run over a rise in the original picture. Which quantity does the formula use, the input of the inverse or the input of the original? Why is that backwards reading the whole error? Write the point pair first, then take the reciprocal of the slope at the other point.
| Table row | f(a) | f′(a) | (f−1)′ at f(a) |
|---|---|---|---|
| a=1 | 4 | −2 | −21 |
| a=2 | 5 | 3 | 31 |
| a=0 | −1 | 6 | 61 |
① Build a small table of f(x)=x3+1 at x=1 and x=2, then compute the inverse slopes at the two outputs directly from the formula and again by solving for f−1.
② Take a function with a horizontal tangent, such as f(x)=x2 at x=0, and explain what happens to the inverse slope there.
③ Write the tangent line to f−1 at a point, then check that the line really does pass through the reflected point rather than the original one.
Core Archetype (Simplest Form): Let f(2)=5 and f′(2)=3. Find (f−1)′(5).
Standard Execution:
- Confirm the pair. f(2)=5 means the inverse sends 5 back to 2.
- Read the slope. f′(2)=3.
- Reciprocate. (f−1)′(5)=31.
- Check the geometry. The original is steep at x=2, so the inverse should be shallow at x=5, and 31 is shallow.
Exam Variation (Context / Parameter Shift): The differentiable function g has selected values g(1)=3, g′(1)=−4, g(3)=7, g′(3)=2. Find the slope of the tangent to g−1 at x=3 and write the tangent line there.
Solution: Since g(1)=3, the inverse passes through (3,1), and the slope is g′(1)1=−41. The tangent line is y−1=−41(x−3). The row with g(3)=7 is a distractor: its slope belongs to a different point pair.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f be a differentiable function with f(3)=8 and f′(3)=4. What is (f−1)′(8)?
MCQ 2
The function h satisfies h(1)=4 and h′(1)=−2. What is (h−1)′(4)?
MCQ 3
Let k be differentiable with k(0)=2 and k′(0)=0. Which statement about (k−1)′(2) is correct?
The differentiable function f has selected values in the table.
| x | 1 | 2 | 4 | 6 |
|---|---|---|---|---|
| f(x) | 3 | 5 | 8 | 10 |
| f′(x) | 4 | −2 | 21 | 6 |
(a) Find (f−1)′(5).
(b) Find the slope of the tangent to f−1 at x=8.
(c) Write an equation for the tangent line to f−1 at x=5.
(d) Explain why the value f′(6)=6 cannot be used to find (f−1)′(6).
| Formula | Condition | |
|---|---|---|
| Slope of f−1 at b | (f−1)′(b)=f′(a)1 | f(a)=b |
| Slope of f−1 at b | (f−1)′(b)=f′(f−1(b))1 | f′=0 |
| Point on the inverse | (b,a) | from f(a)=b |
| Tangent line | y−a=f′(a)1(x−b) | slope from above |
| Failure case | no finite slope | f′(a)=0 |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
An inverse function undoes the original, so its graph is the original reflected across y=x, and its slope is the original slope turned upside down. If f(a)=b and f′(a)=m, then (f−1)′(b)=m1. The formula is short, and the mistake it invites is a swap: the derivative of the inverse at b needs the original slope at a, not at b. Most exam items supply a table of values, so the work is a lookup followed by a reciprocal. One condition has to hold: f′(a) must be nonzero, since a horizontal tangent in the original becomes a vertical tangent in the inverse.
Figure 5 — A curve and its reflection across y=x. The point (2,5) becomes (5,2), and the slope 3 at the first point becomes 31 at the second.
The formula follows from the reflection. If the original rises 3 units for every 1 unit across, then the reflected curve rises 1 unit for every 3 units across, because rise and run have exchanged roles:
(f−1)′(b)=f′(a)1,where f(a)=b
| Given | Wanted | Use | Not this |
|---|---|---|---|
| f(2)=5, f′(2)=3 | (f−1)′(5) | f′(2)1=31 | f′(5), which is unknown |
| f(1)=4, f′(1)=−2 | (f−1)′(4) | −21 | −f′(4)1 |
| f(3)=1, f′(3)=0 | (f−1)′(1) | does not exist | any finite value |
The third row is the condition stated in words: a horizontal tangent in the original becomes a vertical tangent in the inverse, and a vertical tangent has no finite slope.
Most exam items give a table with rows for x, f(x), and f′(x). To find (f−1)′(k):
If k does not appear in the f(x) column, the item is telling you that f−1 is not defined there, and the slope question has no answer.
Differentiate f(f−1(x))=x with respect to x. The chain rule gives
f′(f−1(x))⋅(f−1)′(x)=1
and dividing by the first factor produces the formula. This derivation is worth keeping, because it also shows where the formula fails: the division is illegal when f′(f−1(x))=0.
Borrowed idea: Ask what the inverse actually does before differentiating it. It sends an output back to its input. So a slope of the inverse is a run over a rise in the original picture. Which quantity does the formula use, the input of the inverse or the input of the original? Why is that backwards reading the whole error? Write the point pair first, then take the reciprocal of the slope at the other point.
| Table row | f(a) | f′(a) | (f−1)′ at f(a) |
|---|---|---|---|
| a=1 | 4 | −2 | −21 |
| a=2 | 5 | 3 | 31 |
| a=0 | −1 | 6 | 61 |
① Build a small table of f(x)=x3+1 at x=1 and x=2, then compute the inverse slopes at the two outputs directly from the formula and again by solving for f−1.
② Take a function with a horizontal tangent, such as f(x)=x2 at x=0, and explain what happens to the inverse slope there.
③ Write the tangent line to f−1 at a point, then check that the line really does pass through the reflected point rather than the original one.
Core Archetype (Simplest Form): Let f(2)=5 and f′(2)=3. Find (f−1)′(5).
Standard Execution:
- Confirm the pair. f(2)=5 means the inverse sends 5 back to 2.
- Read the slope. f′(2)=3.
- Reciprocate. (f−1)′(5)=31.
- Check the geometry. The original is steep at x=2, so the inverse should be shallow at x=5, and 31 is shallow.
Exam Variation (Context / Parameter Shift): The differentiable function g has selected values g(1)=3, g′(1)=−4, g(3)=7, g′(3)=2. Find the slope of the tangent to g−1 at x=3 and write the tangent line there.
Solution: Since g(1)=3, the inverse passes through (3,1), and the slope is g′(1)1=−41. The tangent line is y−1=−41(x−3). The row with g(3)=7 is a distractor: its slope belongs to a different point pair.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f be a differentiable function with f(3)=8 and f′(3)=4. What is (f−1)′(8)?
MCQ 2
The function h satisfies h(1)=4 and h′(1)=−2. What is (h−1)′(4)?
MCQ 3
Let k be differentiable with k(0)=2 and k′(0)=0. Which statement about (k−1)′(2) is correct?
The differentiable function f has selected values in the table.
| x | 1 | 2 | 4 | 6 |
|---|---|---|---|---|
| f(x) | 3 | 5 | 8 | 10 |
| f′(x) | 4 | −2 | 21 | 6 |
(a) Find (f−1)′(5).
(b) Find the slope of the tangent to f−1 at x=8.
(c) Write an equation for the tangent line to f−1 at x=5.
(d) Explain why the value f′(6)=6 cannot be used to find (f−1)′(6).
| Formula | Condition | |
|---|---|---|
| Slope of f−1 at b | (f−1)′(b)=f′(a)1 | f(a)=b |
| Slope of f−1 at b | (f−1)′(b)=f′(f−1(b))1 | f′=0 |
| Point on the inverse | (b,a) | from f(a)=b |
| Tangent line | y−a=f′(a)1(x−b) | slope from above |
| Failure case | no finite slope | f′(a)=0 |
Strengthen this topic with AP-style practice, or continue to next topic.