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

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