AP Calculus AB and BC

When the Inverse Derivative Formula Fails

Q: How do I find the matching point?

Solve f(x) = b. That solution is f^{-1}(b), and it is where the derivative of f gets evaluated. On the exam the value is usually handed to you in a table.

Q: Why does the arcsine derivative blow up at plus or minus one?

Because \sin has horizontal tangents at \pm\pi/2, which map to \pm 1. The reciprocal of zero is undefined, and 1/\sqrt{1-x^2} correctly runs to infinity there.

The formula for the derivative of an inverse function divides by the derivative of the original function. Where that derivative is zero the formula has nothing to return, and the inverse genuinely has a vertical tangent there. The cube root at the origin is the standard case.

Derivative of an inverse function

\left(f^{-1}\right)'(b) = \frac{1}{f'\!\left(f^{-1}(b)\right)}

If f is one-to-one and differentiable, and its derivative at the matching point is nonzero, then the inverse is differentiable at b with derivative equal to the reciprocal.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
The derivative of f at the matching point is nonzero
$f'\!\left(f^{-1}(b)\right) \ne 0$
The formula is a reciprocal, so a zero in the denominator is fatal, and it is fatal for a reason rather than by accident. Reflecting a horizontal tangent across the line $y = x$ produces a vertical one, and a vertical tangent means the derivative does not exist.
Drop it and the theorem fails
$\left(x^{1/3}\right)' = \frac{1}{3\,|x|^{2/3}}$
The derivative of the cube root, near x = 0
Take $f(x) = x^3$ , which is one-to-one and differentiable everywhere. Its inverse is the cube root. At $b = 0$ the matching point is $f^{-1}(0) = 0$ , and $f'(0) = 3 \cdot 0^2 = 0$ , so the formula asks for $1/0$ . The check confirms the inverse's derivative is unbounded near the origin: the cube root has a vertical tangent there, and no finite derivative exists to report.
2
f is one-to-one on the interval
$f(x_1) = f(x_2) \implies x_1 = x_2$
Without this there is no inverse function to differentiate. $f(x) = x^2$ on the whole real line is differentiable with a perfectly good derivative, and still has no inverse, because both 2 and $-2$ map to 4. The usual fix is to restrict the domain, and the restriction you choose is what determines which branch the derivative formula describes. There is no counterexample function here, because the failure happens before differentiation is even on the table.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Start from the defining identity $f\!\left(f^{-1}(x)\right) = x$ , valid wherever the inverse is defined.
Differentiate both sides using the chain rule: $f'\!\left(f^{-1}(x)\right) \cdot \left(f^{-1}\right)'(x) = 1$ .
Solve for the unknown factor to get the formula. The division is legitimate exactly when $f'\!\left(f^{-1}(x)\right) \ne 0$ , which is the hypothesis.
This argument assumes the inverse is differentiable to begin with; a full proof establishes that separately, and it too needs the derivative to be nonzero.

What it does not say

The derivative of the inverse is the reciprocal of the derivative.

Only after evaluating at the right place. It is $1/f'$ evaluated at $f^{-1}(b)$ , not at $b$ . Skipping the inner evaluation is the single most common error in this topic.

You need a formula for the inverse to use it.

You do not, and that is the point of the formula. If $f(x) = x^5 + x + 1$ then $f^{-1}$ has no elementary form, yet $f(1) = 3$ gives $\left(f^{-1}\right)'(3) = 1/f'(1) = 1/6$ immediately.

A vertical tangent on the inverse means something went wrong.

It means the original had a horizontal tangent there. Reflection across $y = x$ turns one into the other, so this is the expected behaviour rather than a defect.

Frequently asked questions

How do I find the matching point?

Solve $f(x) = b$ . That solution is $f^{-1}(b)$ , and it is where the derivative of $f$ gets evaluated. On the exam the value is usually handed to you in a table.

Why does the arcsine derivative blow up at plus or minus one?

Because $\sin$ has horizontal tangents at $\pm\pi/2$ , which map to $\pm 1$ . The reciprocal of zero is undefined, and $1/\sqrt{1-x^2}$ correctly runs to infinity there.