AP® Calculus BC · Unit 3 · Topic 3.3
Differentiating Inverse Functions
The derivative of an inverse function at a point equals 1 divided by the derivative of the original function, evaluated at the matching input — not at the point you plugged into the inverse.
An inverse function undoes another function, and its graph is the original reflected across the line y = x. That reflection swaps every x and y, which means it swaps rise and run too: wherever the original curve had slope rise/run, the inverse has slope run/rise. So slopes become reciprocals. This gives the rule (f−1)'(f(c)) = 1/f'(c). To use it, you need a c with f(c) equal to the output you care about — then you plug that c, not the output, into f'.
The catch is matching inputs to outputs correctly. If you want (f−1)'(4), you first solve f(c) = 4 for c, then compute f'(c), then take the reciprocal. Skipping the solving step and plugging 4 straight into f' gives a number, but it answers a different question — the slope of f at x = 4, not the slope of f inverse at output 4. These only agree when c happens to equal 4.
This rule connects directly to where a function's slope hits zero. If f'(c) = 0 at some point, then 1/f'(c) is undefined, so f inverse has a vertical tangent there and no derivative. That flat spot on the original curve is exactly the kind of critical point you study later when hunting for local maxes and mins — here it shows up as the place where the inverse's slope blows up instead of existing.
Key terms
- Inverse function
- A function that reverses another, so f−1(f(x)) = x for every x in the domain.
- Reciprocal slope rule
- The fact that (f−1)'(f(c)) = 1/f'(c), since reflecting across y = x swaps rise and run.
- Matching input c
- The x-value satisfying f(c) equals the output you are evaluating the inverse's derivative at.
- Critical point
- An x-value where a function's derivative is zero or undefined, often marking a local max or min.
Watch it explained
- How it worksHow does reflecting a graph across the line y = x swap rise and run?
- Worked exampleFind the derivative of the inverse of f(x) = x3 + 2x + 1 at the output value 4.
- Common mistakeWhy isn't the inverse's derivative just 1 over f'(x) at the same x value?
- Ties togetherHow does the original curve going flat connect to that x-value being a critical point in Unit 5's analysis?
Check yourself
End of topic.