Differentiating Inverse Functions AB & BC
The graph of $f^{-1}$ is the graph of $f$ reflected across the line $y = x$, and reflecting exchanges rise with run. Slopes of a function and its inverse are therefore reciprocals, taken at points that are themselves reflections of one another: $\left(f^{-1}\right)'(x) = \frac{1}{f'\left(f^{-1}(x)\right)}$. The formula follows in one step from differentiating $f\left(f^{-1}(x)\right) = x$ with the chain rule.
The dominant mistake is feeding the outer derivative the wrong number, writing $\frac{1}{f'(a)}$ where the formula calls for $\frac{1}{f'\left(f^{-1}(a)\right)}$. The input to the inverse lives in the original function's range, so it has to be converted back to the original's domain before $f'$ can accept it. With a table this means reading a row in reverse, and with a pair of coordinates it means keeping straight which one belongs to $f$ and which to its inverse.
The work
3 ways in · any order
Lesson
Differentiating Inverse Functions
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Derives the reciprocal-slope relationship from the reflection across y = x, shows the one-line implicit derivation, and drills the step everyone skips: converting the input before evaluating.
Diagnostic
10-item topic check
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Ten items spanning the two failure modes of this topic: the outer derivative evaluated at the wrong input, and the algebra of solving and evaluating afterward. Take it cold to find which one is yours, or after the lesson to confirm it is not.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.