Power Rule for Derivatives/Fractional Index/Proof 1

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

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

From the definition of the power to a rational number, or alternatively from the definition of the root of a number, $f \left({x}\right)$ 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 $\left({0 \,.\,.\, \infty}\right)$, but not at $x = 0$ where it is continuous only on the right.

Let $y > x$.

From Inequalities Concerning Roots:
 * $\forall n \in \N_{>0}: X Y^{1/n} \ \left|{x - y}\right| \le n X Y \ \left|{x^{1/n} - y^{1/n}}\right| \le Y X^{1/n} \ \left|{x - y}\right|$

where $x, y \in \left[{X \,.\,.\, Y}\right]$.

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


 * $\displaystyle \frac 1 {n y} y^{1/n} \le \frac {y^{1/n} - x^{1/n}} {y - x} \le \frac 1 {n x} x^{1/n}$

From the Squeeze Theorem, it follows that:
 * $\displaystyle \lim_{y \to x^+} \ \frac {y^{1/n} - x^{1/n}} {y - x} = \frac 1 {n x} x^{1/n} = \frac 1 n x^{\frac 1 n - 1}$

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

Thus the result holds for $f \left({x}\right) = x^{1/n}$.