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.

Proof

\(\ds \map f x\)

\(=\)

\(\ds \sqrt [n] x\)

\(\ds \)

\(=\)

\(\ds x^{1 / n}\)

Definition of $n$th Root

\(\ds \leadsto \ \ \)

\(\ds \map {f'} x\)

\(=\)

\(\ds \frac 1 n x^{\paren {1 / n} - 1}\)

Power Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \frac 1 n x^{\paren {1 / n} \paren {1 - n} }\)

rearrangement

\(\ds \)

\(=\)

\(\ds \frac 1 {n x^{\paren {1 / n} \paren {n - 1} } }\)

Exponent Combination Laws: Negative Power

\(\ds \)

\(=\)

\(\ds \frac 1 {n \paren {\sqrt [n] x}^{n - 1} }\)

Definition of $n$th Root

$\blacksquare$