AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/3/3-3)
Unit 3 · Topic 3.3
3.3 Differentiating Inverse Functions
An inverse function undoes the original, so its graph is the reflection over y = x, and its slopes are reciprocals. If f(a) = b, then the slope of f⁻¹ at x = b is 1/f′(a), as long as f′(a) ≠ 0.
Key terms
- inverse function
- f⁻¹(x)
- reciprocal slope
- reflection over y = x
The formula
If f is differentiable and has an inverse g = f⁻¹, and f(a) = b, then g′(b) = 1 / f′(a), provided f′(a) ≠ 0. Written another way, g′(x) = 1 / f′(g(x)).
It comes from the chain rule. Since f(g(x)) = x for every x in the domain of g, differentiate both sides: f′(g(x))·g′(x) = 1. Divide: g′(x) = 1/f′(g(x)).
Why the slopes are reciprocals
Reflecting a graph over y = x swaps x and y. So the point (a, b) on f becomes (b, a) on f⁻¹, and a rise of 3 over a run of 1 becomes a rise of 1 over a run of 3. Slope m on f turns into slope 1/m on f⁻¹ at the matching point.
If f′(a) = 0, the tangent line to f is horizontal there, and the reflected tangent line is vertical. That's why the formula needs f′(a) ≠ 0: the inverse has no derivative at that point.
Graph version: if the graph of f passes through (2, 5) with tangent slope 4, then the graph of f⁻¹ passes through (5, 2) with tangent slope 1/4.
The method, step by step
Most problems give you b, an output of f. You need to work backward to the input a:
- You want (f⁻¹)′(b). Find the a that makes f(a) = b. This is the matching point on f.
- Compute f′(a).
- Flip it: (f⁻¹)′(b) = 1/f′(a).
Finding a when you can't solve for it
Often f can't be inverted with algebra, like f(x) = x³ + 2x − 1. You don't need the inverse formula. You only need one point: try small integers until f(a) equals the target value. Exam problems are designed so a nice value works.
With a table, look in the f(x) row for the target value b, then read the x above it.
Where this shows up
This idea explains the derivative of ln x. Since ln x is the inverse of eˣ, its slope at x is 1 / (slope of eʸ at y = ln x) = 1/e^(ln x) = 1/x. It also produces the inverse trig derivatives in 3.4.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Inverse slope from a formula
f(x) = x³ + 2x − 1, and g is the inverse of f. Find g′(2).
Show the solutionHide the solution
- Step 1: Find a with f(a) = 2: try a = 1: 1 + 2 − 1 = 2. So f(1) = 2, which means g(2) = 1.
- Step 2: f′(x) = 3x² + 2, so f′(1) = 5.
- Step 3: g′(2) = 1/f′(g(2)) = 1/f′(1).
Answer: g′(2) = 1/5
- Example 2
Trap: using the wrong input in a table
f is differentiable and invertible with g = f⁻¹. The table shows f(3) = 7, f′(3) = 2, f(7) = 10 and f′(7) = 5. Find g′(7).
Show the solutionHide the solution
- Step 1: You need the point on f whose output is 7. The table shows f(3) = 7, so g(7) = 3.
- Step 2: g′(7) = 1/f′(g(7)) = 1/f′(3) = 1/2.
- Step 3: The tempting wrong answer is 1/f′(7) = 1/5, which uses the input 7 instead of the output 7.
Answer: g′(7) = 1/2
- Example 3
Inverse slope at a point given by a value
f(x) = x⁵ + x + 4. Find (f⁻¹)′(6).
Show the solutionHide the solution
- Step 1: Find a with f(a) = 6: a⁵ + a + 4 = 6, so a⁵ + a = 2. a = 1 works.
- Step 2: f′(x) = 5x⁴ + 1, so f′(1) = 6.
- Step 3: (f⁻¹)′(6) = 1/f′(1).
Answer: (f⁻¹)′(6) = 1/6
Common mistakes
- Computing 1/f′(b) instead of 1/f′(a). The derivative of f must be taken at the input a that produces b.
- Confusing f⁻¹(x) with 1/f(x). The −1 means inverse, not reciprocal.
- Trying to find a formula for f⁻¹ when it isn't needed. You only need the matching point.
On the exam
- This topic appears often in multiple choice, frequently with a table where the right and wrong rows are both listed. Find the row where f(x) equals the given value.
- If an answer choice is 1/f′(b), it's almost certainly the trap.
Connected topics
Videos
Check yourself
4 questions on 3.3 Differentiating Inverse Functions. Pick an answer to see if you got it, and why.
The function f is differentiable and one-to-one, with f(2) = 5, f′(2) = −3, f(5) = 7 and f′(5) = 4. What is the value of (f⁻¹)′(5)?
Let f be a function with f′(x) > 0 for all x, and let g be the inverse function of f. Which of the following statements must be true? I. g is increasing. II. If f(1) = 3 and f′(1) = 2, then g′(3) = 1/2. III. g′(x) = 1/f′(x) for all x in the domain of g.
Let f(x) = x³ + 2x + 1. Because f′(x) = 3x² + 2 is positive for all x, f is one-to-one and has an inverse function g.
Described function
What is the value of g′(4)?
Which of the following is an equation of the line tangent to the graph of g at x = 4?
0 of 4 answered