Elementary power rule of derivatives/Proof/Binomial Theorem method

$\frac{d}{dx} x^{n} = nx^{n-1}$ , for $n$ is a natural number.

Prerequisites

• Binomial Theorem, $(x+y)^n=\sum_{k=0}^n\dfrac{n!}{k!(n-k)!}x^ky^{n-k}$ , for $n$ is a natural number.
• Limit definition of the derivative

Proof

Let $f(x) = x^n$ . Then:

$\begin{align} f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to0}\frac{(x+h)^n-x^n}{h}\\ &=\lim_{h\to0}\dfrac{\sum\limits_{k=0}^n\dfrac{n!}{k!(n-k)!}x^kh^{n-k}-x^n}{h}\\ &=\lim_{h\to0}\frac{x^n+\sum\limits_{k=0}^{n-1}\dfrac{n!}{k!(n-k)!}x^kh^{n-k}-x^n}{h}=\lim_{h\to0}\frac{\sum\limits_{k=0}^{n-1}\dfrac{n!}{k!(n-k)!}x^kh^{n-k}}{h}\\ &=\lim_{h\to0}\sum_{k=0}^{n-1}\frac{n!}{k!(n-k)!h}x^kh^{n-k}=\lim_{h\to0}\sum_{k=0}^{n-1}\frac{n!}{k!(n-k)!}x^kh^{n-k-1}\\ &=\lim_{h\to0}\frac{n!}{(n-1)!(n-(n-1))!}x^{n-1}h^0+\sum_{k=0}^{n-2}\frac{n!}{k!(n-k)!h}x^kh^{n-k}=nx^{n-1}+0 \end{align}$

$f'(x)=nx^{n-1}$

QED