Slopes of an inverse AB & BC

If $(a, b)$ lies on $f$, then $(b, a)$ lies on $f^{-1}$: that is what inverting does. Reflecting across $y = x$ turns a rise of $\Delta y$ over a run of $\Delta x$ into a rise of $\Delta x$ over a run of $\Delta y$. The slope inverts.

So if $f$ has slope 4 at the point $(2, 7)$, then $f^{-1}$ has slope $\frac{1}{4}$ at the point $(7, 2)$. Notice that the two statements are anchored at different points, and the $x$-coordinate of one is the $y$-coordinate of the other. Every error in this topic is a version of losing track of that swap.

Written as a formula:

$$\left(f^{-1}\right)'(x) = \frac{1}{f'\left(f^{-1}(x)\right)}.$$

The denominator is the whole message. It is not $f'(x)$. The input $x$ is a value in the inverse's domain, which is the original function's range, so it cannot be handed to $f'$ directly. It has to be converted back first, and $f^{-1}(x)$ is what does the converting.

To compute $\left(f^{-1}\right)'(a)$:

1. Find the number $c$ with $f(c) = a$. That is $c = f^{-1}(a)$, the point on the original curve that reflects onto the one you care about.
2. Compute $f'(c)$.
3. Take the reciprocal.

Step 1 is the step that gets skipped. Writing $\left(f^{-1}\right)'(a) = \frac{1}{f'(a)}$ feeds $a$ to the wrong function, and it produces a number that is usually plausible and always wrong. It happens to be right only in the coincidental case $f^{-1}(a) = a$, where the curve meets the line $y = x$.

Given a table, step 1 is a lookup in reverse: you scan the $f(x)$ column for the value $a$ and read off the $x$ beside it. Scanning the $x$ column instead is the same error wearing a different hat.

The formula is implicit differentiation in one line. Start from the statement that the inverse undoes $f$:

$$f\left(f^{-1}(x)\right) = x.$$

Differentiate both sides with respect to $x$. The left side is a composition, so the chain rule applies: the outer derivative $f'$ evaluated at the inside, times the inside's derivative. The right side is 1.

$$f'\left(f^{-1}(x)\right)\cdot\left(f^{-1}\right)'(x) = 1.$$

Dividing gives the formula. The argument $f^{-1}(x)$ is not a convention someone chose; it is simply what was already sitting inside $f$ when the chain rule reached it. That is worth remembering, because it means the formula cannot be misremembered as $1/f'(x)$ without also misremembering the chain rule.

If $f'(c) = 0$, the formula divides by zero and $\left(f^{-1}\right)'$ is undefined at the reflected point. Geometrically this is honest: a horizontal tangent on $f$ reflects to a vertical tangent on $f^{-1}$, and vertical tangents have no slope.

This also explains the requirement that $f$ be strictly monotonic on the interval in question. A function that turns around fails the horizontal line test, so no inverse function exists there to differentiate. Checking monotonicity first is not bookkeeping; it is what makes the question well posed.