Power Rule for Derivatives/Fractional Index

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $f: \R \to \R$ be the real function defined as $\map f x = x^{1 / n}$.


Then:

$\map {f'} x = n x^{n - 1}$

everywhere that $\map f x = x^n$ is defined.


When $x = 0$ and $n = 0$, $\map {f'} x$ is undefined.


Proof 1

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

Thus, let $\map f x = x^{1 / n}$.

From the definition of the power to a rational number, or alternatively from the definition of the root of a number, $\map f x$ is defined when $x \ge 0$.

(However, see the special case where $x = 0$.)

From Continuity of Root Function, $f \left({x}\right)$ is continuous over the open interval $\openint 0 \infty$, but not at $x = 0$ where it is continuous only on the right.


Let $y > x$.

From Inequalities Concerning Roots:

$\forall n \in \N_{>0}: X Y^{1 / n} \, \size {x - y} \le n X Y \, \size {x^{1 / n} - y^{1 / n} } \le Y X^{1 / n} \, \size {x - y}$

where $x, y \in \closedint X Y$.

Setting $X = x$ and $Y = y$, this reduces (after algebra) to:

$\dfrac 1 {n y} y^{1 / n} \le \dfrac {y^{1 / n} - x^{1 / n} } {y - x} \le \dfrac 1 {n x} x^{1 / n}$

From the Squeeze Theorem, it follows that:

$\displaystyle \lim_{y \mathop \to x^+} \dfrac {y^{1 / n} - x^{1 / n} } {y - x} = \dfrac 1 {n x} x^{1 / n} = \dfrac 1 n x^{\dfrac 1 n - 1}$


A similar argument shows that the left hand limit is the same.


Thus the result holds for $\map f x = x^{1 / n}$.

$\blacksquare$


Proof 2

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

Thus, let $f \left({x}\right) = y = x^{1/n}$.

Thus $f^{-1} \left({y}\right) = x = y^n$ from the definition of root.

So:

\(\displaystyle D x^{1/n}\) \(=\) \(\displaystyle \frac 1 {D y^n}\) Derivative of Inverse Function‎
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n y^{n-1} }\) Power Rule for Derivatives: Integer Index
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n \left({x^{1/n} }\right)^{n-1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 n x^{\frac 1 n - 1}\)

$\blacksquare$