Power Rule for Derivatives/Natural Number Index

Theorem
Let $n \in \N$.

Let $f: \R \to \R$ be the real function defined as $f \left({x}\right) = x^n$.

Then:
 * $f^{\prime} \left({x}\right) = n x^{n-1}$

everywhere that $f \left({x}\right) = x^n$ is defined.

When $x = 0$ and $n = 0$, $f^{\prime} \left({x}\right)$ is undefined.

Proof by Binomial Theorem
Let $f \left({x}\right) = x^n$ for $x \in R, n \in N$.

By the definition of the derivative:
 * $\displaystyle \frac {\mathrm d}{\mathrm d x} f \left({x}\right) = \lim_{h \to 0} \frac{f \left({x + h}\right) - f \left({x}\right)} h = \lim_{h \to 0} \frac{(x+h)^n - x^n} h$

Using the binomial theorem this simplifies to:

Proof by Induction
We will use the notation $D f \left({x}\right) = f^{\prime} \left({x}\right)$ as it is convenient.

Let $n = 0$.

Then $\forall x \in \R: x^n = 1$.

Thus $f \left({x}\right)$ is the constant function $f_1 \left({x}\right)$ on $\R$.

Thus from Differentiation of a Constant, $D f \left({x}\right) = D \left({x^0}\right) = 0 x^{-1}$, except where $x = 0$.

So the result holds for $n = 0$.

Let $n = 1$.

Then $\forall x \in \R: f \left({x}\right) = x^n = x$.

Then from Differentiation of the Identity Function $D \left({x}\right) = 1 = 1 \cdot x^{1-1}$.

So the result holds for $n = 1$.

Now assume $D \left({x^k}\right) = k x^{k-1}$.

Then by the Product Rule for Derivatives, $D \left({x^{k+1}}\right) = D \left({x^k x}\right) = x^k D \left({x}\right) + D \left({x^k}\right) x = x^k \cdot 1 + k x^{k-1} x = \left({k+1}\right) x^k$.

The result follows by induction.