Differentiating Inverse Functions: Study Guide (Calculus AB 3.3)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so derivatives are written in a compact form.

• f⁻¹ means the inverse function of f. It undoes f: if f(a) = b, then f⁻¹(b) = a.
• (f⁻¹)′(b) means “the derivative of the inverse function, evaluated at b”.
• f⁻¹ is not 1/f. The −1 is a label for “inverse”, not a power. If you want the reciprocal, write 1/f(x) or [f(x)]⁻¹.

On the exam the inverse often has its own name. A question may say “g is the inverse of f” or “k is the inverse of h”. The ideas are the same whatever the letters.

The idea: reflecting a graph swaps rise and run

A one-to-one function f has an inverse. The graph of the inverse is the graph of f reflected in the line y = x. Every point (a, b) on f becomes the point (b, a) on f⁻¹.

Now think about a tangent line at (a, b) with slope m. Slope is rise over run. Reflecting in y = x swaps the horizontal and vertical directions, so the rise becomes the run and the run becomes the rise. The reflected tangent line has slope run over rise, which is 1/m.

Notice two things in Figure 1.

1. The slopes are reciprocals at matching points. The slope of g at x = 2.25 is linked to the slope of f at x = 3, not at x = 2.25.
2. A steep part of f becomes a shallow part of g, and a shallow part of f becomes a steep part of g. Near the origin f is almost flat, so g is almost vertical.

Deriving the rule with the chain rule

The picture suggests the rule. The chain rule proves it.

Let g be the inverse of f. By the definition of an inverse,

f(g(x)) = x for every x in the domain of g.

Differentiate both sides with respect to x. The left side is a composition, so use the chain rule (Topic 3.1):

f′(g(x)) · g′(x) = 1

Divide by f′(g(x)), which is allowed only when it is not 0:

Rule. If g is the inverse of f, then g′(x) = 1/f′(g(x)), provided f′(g(x)) ≠ 0.

At a single point the rule is easier to use in this form:

Point form. If f(a) = b and f′(a) ≠ 0, then g′(b) = 1/f′(a).

The implicit route. You can reach the same rule with Topic 3.2. Write y = g(x). Then f(y) = x. Differentiate implicitly: f′(y) · dy/dx = 1, so dy/dx = 1/f′(y). Since y = g(x), this is the same rule. In Leibniz notation, dx/dy = 1/(dy/dx) at matching points.

A three-step method for “find the derivative of the inverse at x = b”:

1. Find the matching input. Solve f(a) = b for a. (Usually by inspection or from a table.)
2. Find f′(a). Differentiate f, then substitute a.
3. Take the reciprocal. g′(b) = 1/f′(a).

The most common error is skipping step 1 and calculating 1/f′(b). The input b belongs to g, not to f.

Checking the rule on functions you already know

The skill this topic practises is confirming that an answer is correct. A good habit is to test a new rule on cases where you already know the answer.

Exponential and logarithm. f(x) = eˣ and g(x) = ln x are inverses. The rule gives

g′(x) = 1/f′(g(x)) = 1/e^(ln x) = 1/x.

That matches the derivative of ln x from Unit 2.

Square and square root. f(x) = x² for x ≥ 0 and g(x) = √x are inverses. The rule gives

g′(x) = 1/f′(g(x)) = 1/(2√x), for x > 0.

That matches the power rule applied to x^(1/2).

When a question gives you an inverse you can write down, you can always check your answer this way. When you cannot write the inverse, check with a nearby value instead (see Worked example 1).

When the inverse is not differentiable

The rule has a condition: f′(a) ≠ 0. Two things must be true before you use it.

• f must be one-to-one on the interval you are working on, so that the inverse exists. A function whose derivative keeps one sign (always positive or always negative) on an interval is one-to-one there. If a function is not one-to-one, the question will restrict its domain (for example “x ≥ 0”).
• f′(a) must not be 0. If f′(a) = 0, the tangent to f at (a, b) is horizontal. Its reflection is vertical, so the inverse has a vertical tangent at (b, a) and no derivative there.

Example: f(x) = x³ is one-to-one with inverse g(x) = ∛x. Since f′(0) = 0, the cube-root graph has a vertical tangent at the origin, and g′(0) does not exist. Away from 0 the rule works: f(2) = 8 and f′(2) = 12, so g′(8) = 1/12. The power rule gives the same: g′(x) = (1/3)x^(−2/3), and at x = 8 this is (1/3)(1/4) = 1/12.

Worked example 1: a formula with no algebraic inverse

Question. Let f(x) = x³ + 2x − 5, and let g be the inverse of f. (a) Explain why g exists. (b) Find g′(7). (c) Write an equation for the tangent line to the graph of g at x = 7.

1. Inverse exists. f′(x) = 3x² + 2. Since 3x² ≥ 0, f′(x) ≥ 2 > 0 for all x. So f is always increasing, which makes it one-to-one, and g exists.
2. Matching input. You need a with f(a) = 7. Try small integers: f(2) = 8 + 4 − 5 = 7. So a = 2, which means g(7) = 2. (f is one-to-one, so this is the only solution.)
3. Derivative of f at the matching input. f′(2) = 3(4) + 2 = 14.
4. Reciprocal. g′(7) = 1/f′(2) = 1/14.
5. Tangent line. The point on g is (7, 2) and the slope is 1/14: y = 2 + (1/14)(x − 7)

Check with a nearby value. f(2.01) = 7.140601, so the point (7.140601, 2.01) lies on g. The slope from (7, 2) to that point is 0.01/0.140601 ≈ 0.0711. That is close to 1/14 ≈ 0.0714, so the answer is reasonable.

Check the sign and size. f is steep at x = 2 (slope 14), so g should be shallow at x = 7. A small positive slope makes sense.

Why not solve for g? Solving y = x³ + 2x − 5 for x needs the cubic formula. You do not need it. One matching pair of values is enough.

Worked example 2: an inverse from a table

Question. The function r is differentiable and increasing. Let s be the inverse of r. Selected values are shown.

x	−1	0	1	2	3
r(x)	−4	−1	2	6	11
r′(x)	4	2	3.5	4.5	6

(a) Find s′(2). (b) Find s′(6). (c) Write the tangent line to s at x = −1. (d) Let H(x) = s(3x). Find H′(2).

(a) Look for 2 in the r(x) row, not the x row. r(1) = 2, so s(2) = 1. Then s′(2) = 1/r′(1) = 1/3.5 = 2/7.

A tempting wrong answer is 1/r′(2) = 2/9. That uses the slope of r at x = 2, which is the wrong point.

(b) r(2) = 6, so s(6) = 2. Then s′(6) = 1/r′(2) = 1/4.5 = 2/9.

(c) r(0) = −1, so s(−1) = 0 and the point is (−1, 0). The slope is s′(−1) = 1/r′(0) = 1/2. The tangent line is

y = (1/2)(x + 1)

(d) H is a composition, so use the chain rule: H′(x) = s′(3x) · 3. At x = 2, H′(2) = 3 · s′(6) = 3 · (2/9) = 2/3.

Check. Every r′ value is positive, so every s′ value should be positive. All four answers are positive. Also, a large r′ should give a small s′: r′(2) = 4.5 > r′(1) = 3.5, and s′(6) = 2/9 < s′(2) = 2/7, as expected.

How to read a table for an inverse. To evaluate s at a number, find that number among the outputs of r. Then read the input above it. Use that input for r′.

Common misconceptions

• Using the wrong input. Writing g′(b) = 1/f′(b). The derivative of f must be taken at a, the value with f(a) = b. Draw an arrow from b in the output row to a in the input row before you calculate.
• Treating f⁻¹ as 1/f. The derivative of 1/f(x) is −f′(x)/[f(x)]², which is a different thing. The inverse function is not the reciprocal.
• Forgetting the reciprocal. The slope of the inverse at (b, a) is 1/f′(a), not f′(a).
• Changing the sign. Reflection keeps the sign of a slope. If f is decreasing with f′(a) = −4, then the inverse has slope −1/4 at the matching point, not +1/4.
• Ignoring f′(a) = 0. A horizontal tangent on f means a vertical tangent on the inverse. There is no derivative there, not a derivative of 0.
• Believing you must find the inverse formula. You need only the matching pair of values and f′ at the right input.
• Forgetting the chain rule in a composition. If H(x) = g(3x), then H′(x) = 3g′(3x). The factor 3 is easy to drop.

Where this leads

Topic 3.4 uses exactly this rule (or the implicit route) to find the derivatives of the inverse trigonometric functions, such as arcsin x and arctan x. Continue with Differentiating Inverse Trigonometric Functions. Later, in Unit 6, the idea of “undoing” returns in a different form: antidifferentiation undoes differentiation. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.