Power Rule for Derivatives/Natural Number Index/Proof by Difference of Two Powers

Proof
Let $f \left({x}\right) = x^n$ for $x \in \R, n \in \N$, and let $a \in \R$.

By definition of the derivative:
 * $\displaystyle f^\prime \left({a}\right) = \lim_{x \to a} \frac {f \left({x}\right) - f \left({a}\right)}{x-a} = \lim_{x \to a} \frac {x^n-a^n}{x-a}$

Case I
For $n=1$ we have
 * $f^\prime \left({x}\right) = 1$

From Derivative of Identity Function

$1 \cdot x^{1-1} = 1$, so $f^\prime \left({x}\right) = nx^{n-1}$ for $n=1$

Case II
For $n=0$, let $a \ne 0$, then
 * $\displaystyle f^\prime \left({a}\right) = \lim_{x \to a} \frac {x^0-a^0}{x-a} = \lim_{x \to a} \frac{0}{x-a}$

From Case I, Sum Rule for Derivatives, and Derivative of Constant it is seen that $x-a$ is differentiable everywhere, so using L'Hopital's Rule yeilds


 * $\displaystyle f^\prime \left({a}\right) = \frac {0}{1} = 0$

$0 \cdot x^{0-1} = 0$ for all $x \ne 0$, so $f^\prime \left({x}\right) = nx^{n-1}$

Case III
$\R$ is a commutative ring, so for $n \ge 2$ it is possible to do

This holds for all $a \in \R$, so $f^\prime \left({x}\right) = nx^{n-1}$.