Power Rule for Derivatives/Natural Number Index/Proof by Binomial Theorem
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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
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: $\dbinom r 1 = r$ |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Standard Differential Coefficients