Concept preview

What you will be able to do

1. 01State the inverse-derivative rule $\left(f^{-1}\right)'(x)=\dfrac{1}{f'\!\left(f^{-1}(x)\right)}$ and explain what each symbol contributes to the slope.
2. 02Derive the rule by implicit differentiation of $f\!\left(f^{-1}(x)\right)=x$, and connect it to the reflection of a graph across the line $y=x$, which makes the slopes of $f$ and $f^{-1}$ reciprocals.
3. 03Read a point $(a,\,b)$ on $f$ to obtain the slope $\dfrac{1}{f'(a)}$ on $f^{-1}$ at $(b,\,a)$, computing values such as $\left(f^{-1}\right)'(2)$ for $f(x)=x^3+x$ while evaluating $f'$ at the correct input.

Suppose $f$ is a one-to-one function with inverse $f^{-1}$. By the very definition of an inverse, feeding a number into $f^{-1}$ and then into $f$ returns the number unchanged: $f\!\left(f^{-1}(x)\right)=x$ for every $x$ in the domain of $f^{-1}$. This identity is the seed of everything. It ties $f$ and $f^{-1}$ together in one equation, and differentiating that equation will let the known derivative $f'$ pin down the unknown derivative $\left(f^{-1}\right)'$.