Power Rule for Derivatives/Natural Number Index
Theorem
Let $n \in \N$.
Let $f: \R \to \R$ be the real function defined as $\map f x = x^n$.
Then:
- $\map {f'} x = n x^{n-1}$
everywhere that $\map f x = x^n$ is defined.
When $x = 0$ and $n = 0$, $\map {f'} x$ is undefined.
Proof by Binomial Theorem
Let $\map f x = x^n$ for $x \in \R, n \in \N$.
By the definition of the derivative:
- $\ds \dfrac \d {\d x} \map f x = \lim_{h \mathop \to 0} \dfrac {\map f {x + h} - \map f x} h = \lim_{h \mathop \to 0} \dfrac {\paren {x + h}^n - x^n} h$
Using the binomial theorem this simplifies to:
\(\ds \) | \(\) | \(\ds \lim_{h \mathop \to 0} \paren {\frac {\dbinom n 0 x^n + \dbinom n 1 x^{n - 1} h + \dbinom n 2 x^{n - 2} h^2 + \cdots + \dbinom n {n - 1} x h^{n - 1} + \dbinom n n h^n - x^n} h}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \paren {\frac {\dbinom n 1 x^{n - 1} h + \dbinom n 2 x^{n - 2} h^2 + \cdots + \dbinom n {n - 1} x h^{n - 1} + \dbinom n n h^n} h}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \paren {\dbinom n 1 x^{n - 1} + \dbinom n 2 x^{n - 2} h^1 + \cdots + \dbinom n {n - 1} x h^{n - 2} + \dbinom n n h^{n - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom n 1 x^{n - 1}\) | evaluating the limit | |||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n - 1}\) | Binomial Coefficient with One: $\dbinom r 1 = r$ |
$\blacksquare$
Proof by Induction
We will use the notation $D \map f x = \map {f'} x$ as it is convenient.
Let $n = 0$.
Then:
- $\forall x \in \R: x^n = 1$
Thus $\map f x$ is the constant function $\map {f_1} x$ on $\R$.
Thus from Derivative of Constant, $D \map f x = \map D {x^0} = 0 x^{-1}$, except where $x = 0$.
So the result holds for $n = 0$.
Let $n = 1$.
Then:
- $\forall x \in \R: \map f x = x^n = x$
Then from Derivative of Identity Function:
- $\map D x = 1 = 1 \cdot x^{1 - 1}$
So the result holds for $n = 1$.
Now assume $\map D {x^k} = k x^{k - 1}$.
Then by the Product Rule for Derivatives:
- $\map D {x^{k + 1} } = \map D {x^k x} = x^k \map D x + \map D {x^k} x = x^k \cdot 1 + k x^{k - 1} x = \paren {k + 1} x^k$
The result follows by induction.
$\blacksquare$
Proof by Difference of Two Powers
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:
\(\ds \map {f'} a\) | \(=\) | \(\ds \lim_{x \mathop \to a} \frac {x^0 - a^0} {x - a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \frac 0 {x - a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
We have that:
- $0 \cdot x^{0 - 1} = 0$
for all $x \ne 0$.
So in this case:
- $\map {f'} x = n x^{n - 1}$
$\Box$
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 $n = 1$:
- $\map {f'} x = n x^{n - 1}$
$\Box$
Case $\text {III}$
$\R$ is a commutative ring, so for $n \ge 2$ it is possible to do:
\(\ds \map {f'} a\) | \(=\) | \(\ds \lim_{x \mathop \to a} \frac {x^n - a^n} {x - a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \frac {\paren {x - a} \sum_{j \mathop = 0}^{n - 1} x^{n - j - 1} a^j} {x - a}\) | Difference of Two Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \sum_{j \mathop = 0}^{n - 1} x^{n - j - 1} a^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} a^j\) | Real Polynomial Function is Continuous | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 0}^{n - 1} a^{n - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n a^{n - 1}\) | shifting the index from $0$ to $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n a^{n - 1}\) |
This holds for all $a \in \R$, so:
- $\map {f'} x = n x^{n - 1}$
$\blacksquare$