### Theorem
If $f$ has an interval $I$ as domain and $f^{\prime}(x)$ exists and is never zero on $I$, then $f^{-1}$ is differentiable at every point in its domain (the range of $f)$.
The value of $\left(f^{-1}\right)^{\prime}$ at a point $b$ in the domain
of $f^{-1}$ is the reciprocal of the value of $f^{\prime}$ at the point $a=f^{-1}(b):$
$
\left(f^{-1}\right)^{\prime}(b)=\frac{1}{f^{\prime}\left(f^{-1}(b)\right)}=\frac{1}{f^{\prime}\left(a\right)}
$
or
$
\left.\frac{d f^{-1}}{d x}\right|_{x=b}=\frac{1}{\left.\frac{df}{dx}\right|_{x=a}}
$
#### Proof
![[derivative-of-inverse-diagram.svg]]
As long as $f$ is [[Bijective Function|bijective]] and differentiable, so is its inverse.
$
\begin{aligned} f\left(f^{-1}(x)\right) &=x \\ \frac{d}{d x} f\left(f^{-1}(x)\right) &=1 \\ f^{\prime}\left(f^{-1}(x)\right) \cdot \frac{d}{d x} f^{-1}(x) &=1 \\ \frac{d}{d x} f^{-1}(x) &=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)} \end{aligned}
$