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

Proof
Let $\map f x = x^n$ for $x \in \R, n \in \N$.

Let $a \in \R$.

By definition of the derivative:
 * $\ds \map {f'} a = \lim_{x \mathop \to a} \frac {\map f x - \map f a} {x - a} = \lim_{x \mathop \to a} \frac {x^n - a^n} {x - a}$

Case $\text I$
For $n = 0$ it is possible to do:

We have that:
 * $0 \cdot x^{0 - 1} = 0$

for all $x \ne 0$.

So in this case:
 * $\map {f'} x = n x^{n - 1}$

Case $\text {II}$
For $n = 1$ we have:
 * $\map f x = x$

From Derivative of Identity Function:


 * $\map {f'} x = 1$

Then we note that:


 * $1 \cdot x^{1 - 1} = 1$

So for the case $n = 1$:
 * $\map {f'} x = n x^{n - 1}$

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

This holds for all $a \in \R$, so:
 * $\map {f'} x = n x^{n - 1}$