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

## Proof by Induction

### Prerequisites

• The product rule of derivatives, $\frac{d}{dx}(uv_=uv'+vu'$

### Proof

The proof proceeds by mathematical induction. Take the base case k=0. Then:

$\frac{d}{dx}x^0=\frac{d}{dx}1=0=0x^{-1}$

The induction hypothesis is that the rule is true for n=k:

$\frac{d}{dx}(x^k)=kx^{k-1}$

We must now show that it is true for n=k+1:

\begin{align} \frac{d}{dx}(x^{k+1})&=\frac{d}{dx}xx^k\\ &=x\frac{d}{dx}x^k+x^k\\ &=xkx^{k-1}+x^k=kx^k+x^k\\ &=(k+1)x^k \end{align}

Since the power rule is true for k=0 and given k is true, k+1 follows, the power rule is true for any natural number.

QED

## Proof by Exponentiation

This proof is validates the power rule for all real numbers $n$ such that the derivative $\frac{d}{dx}(x^n)=nx^{n-1}$

### Prerequisites

• The exponential rule of derivatives, $\frac{d}{dx}(e^x)=e^x$
• The chain rule of derivatives, $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

### Proof

\begin{align} \frac{d}{dx}(x^n)&=\frac{d}{dx}\left(e^{\ln(x)}\right)^n\\ &=\frac{d}{dx}\left(e^{n\ln(x)}\right)\\ &=e^{n\ln(x)}\cdot\frac{d}{dx}\big(n\ln(x)\big)\\ &=x^n\cdot n\cdot\frac1x\\ &=nx^{n-1} \end{align}

## Proof by Binomial Expansion

This proof is only valid for positive real integer $n$ exponents, $x^n$

### Prerequisites

• The Binomial Expansion:
$(a+b)^n=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cdot a^{n-k}\cdot b^k$
• The first principle limit definition of the derivative
$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

### Proof

\begin{align} \frac{d}{dx}(x^n)&=\lim_{h\to0}\frac{(x+h)^n-x^n}{h}\\ &=\lim_{h\to0}\frac{\displaystyle\left(\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cdot x^{n-k}\cdot h^k\right)-x^n}{h}\\ &=\lim_{h\to0}\frac{\displaystyle\left(x^n+nx^{n-1}h+\sum_{k=2}^n\frac{n!}{k!(n-k)!}\cdot x^{n-k}\cdot h^k\right)-x^n}{h}\\ &=\lim_{h\to0}\frac{\displaystyle nx^{n-1}h+\sum_{k=2}^n\frac{n!}{k!(n-k)!}\cdot x^{n-k}\cdot h^k}{h}\\ &=\lim_{h\to0}\left[nx^{n-1}+\sum_{k=2}^n\frac{n!}{k!(n-k)!}\cdot x^{n-k}\cdot h^{k-1}\right]\\ &=nx^{n-1}+\lim_{h\to0}\sum_{k=2}^n\frac{n!}{k!(n-k)!}\cdot x^{n-k}\cdot h^{k-1}\\ &=nx^{n-1}+0\\ &=nx^{n-1} \end{align}