Derivative of Nth Root

Theorem
Let $n \in \N_{>0}$.

Let $f: \R \to \R$ be the real function defined as $\map f x = \sqrt [n] x$.

Then:
 * $\map {f'} x = \dfrac 1 {n \paren {\sqrt [n] x}^{n - 1} }$

everywhere that $\map f x = \sqrt [n] x$ is defined.