# 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:

$f' \left({x}\right) = n x^{n-1}$

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:

$\displaystyle \dfrac \d {\d x} f \left({x}\right) = \lim_{h \mathop \to 0} \dfrac {f \left({x + h}\right) - f \left({x}\right)} h = \lim_{h \mathop \to 0} \dfrac{(x + h)^n - x^n} h$

Using the binomial theorem this simplifies to:

 $\displaystyle$  $\displaystyle \lim_{h \mathop \to 0} \left({\frac { {n \choose 0} x^n + {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n-1} x h^{n-1} + {n \choose n} h^n - x^n} h}\right)$ $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \left({\frac { {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n - 1} x h^{n - 1} + {n \choose n} h^n} h}\right)$ $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \left({ {n \choose 1} x^{n - 1} + {n \choose 2} x^{n - 2} h^1 + \cdots + {n \choose n - 1} x h^{n - 2} + {n \choose n} h^{n - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle {n \choose 1} x^{n - 1}$ evaluating the limit $\displaystyle$ $=$ $\displaystyle n x^{n - 1}$ Binomial Coefficient with One: $\dbinom r 1 = r$

$\blacksquare$

## 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 Derivative of 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 Derivative of 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.

$\blacksquare$

## Proof by Difference of Two Powers

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 $n=0$ it is possible to do

$\displaystyle f^\prime \left({a}\right) = \lim_{x \to a} \frac {x^0-a^0}{x-a} = \lim_{x \to a} \frac{0}{x-a} = 0$

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

$\Box$

### Case II

For $n=1$ we have

$f^\prime \left({x}\right) = 1$

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

$\Box$

### Case III

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

 $\displaystyle f^\prime \left({a}\right) \ \$ $\displaystyle = \lim_{x \to a} \frac {x^n-a^n}{x-a}$ $=$ $\displaystyle \lim_{x \to a} \frac {\left({x-a}\right) \sum_{j \mathop = 0}^{n-1} x^{n-j-1} a^j}{x-a}$ by Difference of Two Powers $\displaystyle$ $=$ $\displaystyle \lim_{x \to a} \sum_{j \mathop = 0}^{n-1} x^{n-j-1} a^j$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^{n-1} a^{n-j-1} a^j$ by Polynomial is Continuous $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^{n-1} a^{n-1}$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 1}^n a^{n-1}$ shift the index from 0 to 1 $\displaystyle$ $=$ $\displaystyle na^{n-1}$

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

$\blacksquare$