Short on time? This page is the recap. For the graph, the derivation and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: f⁻¹ is the inverse function, not 1/f. (f⁻¹)′(b) is the derivative of the inverse at b.
Recap
- The graph of f⁻¹ is the graph of f reflected in y = x. The point (a, b) moves to (b, a).
- Reflection swaps rise and run, so the slope m becomes 1/m at the matching point.
- Derivation: differentiate f(g(x)) = x with the chain rule to get f′(g(x)) · g′(x) = 1.
- You do not need a formula for the inverse. One pair f(a) = b and the value f′(a) are enough.
- If f′(a) = 0, the inverse has a vertical tangent at (b, a) and is not differentiable there.
Key relationships
| Situation | What to use |
|---|---|
| g is the inverse of f | g′(x) = 1/f′(g(x)), provided f′(g(x)) ≠ 0 |
| You know f(a) = b | g′(b) = 1/f′(a) |
| Implicit route | y = g(x) means f(y) = x, so dy/dx = 1/f′(y) |
| Leibniz form | dx/dy = 1/(dy/dx) at matching points |
| Tangent to g at x = b | y = a + (1/f′(a))(x − b) |
| Composition, e.g. H(x) = g(kx) | H′(x) = k · g′(kx) |
| Table of f, f′ | Find b in the f(x) row, read a above it, then use f′(a) |
Assumptions behind the method
- f is one-to-one on the interval used (for example, f′ keeps one sign there), so the inverse exists.
- f is differentiable at a, and f′(a) ≠ 0.
- The value a is the one in the stated domain. If f(x) = b has two solutions, the domain restriction picks one.
Mistakes to avoid
- Using 1/f′(b) instead of 1/f′(a).
- Reading f⁻¹ as 1/f.
- Forgetting the reciprocal and answering f′(a).
- Flipping the sign. A decreasing function has a decreasing inverse.
- Saying “the derivative is 0” where f′(a) = 0. The inverse has a vertical tangent there and no derivative.
- Dropping the chain-rule factor in compositions such as g(3x).
Quick self-check
- f(x) = x⁵ + 3x + 1 and g is its inverse. Find g′(5). (1/8: f(1) = 5 and f′(1) = 8)
- f is decreasing, f(4) = 10 and f′(4) = −2. Find (f⁻¹)′(10). (−1/2)
- f(x) = x³. Is the inverse differentiable at x = 0? (No. f′(0) = 0, so ∛x has a vertical tangent at the origin.)
- Check the rule on eˣ and ln x. (1/e^(ln x) = 1/x, the known derivative of ln x)
Next: practice questions.