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\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

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