# Elementary power rule of derivatives/Proof

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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:

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

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 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

$\frac{d}{dx} x^n= \frac{d}{dx} (e^{\ln(x)})^n$
$\frac{d}{dx} x^n= \frac{d}{dx} e^{n \cdot\ln(x)}$
$\frac{d}{dx} x^n= e^{n \cdot\ln(x)} \cdot \frac{d}{dx}(n \cdot \ln(x))$
$\frac{d}{dx} x^n= x^{n} \cdot n \cdot \frac{1}{x}$
$\frac{d}{dx} x^n= n\cdot x^{n-1}$

## Proof by Binomial Expansion

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

### Prerequisites

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

### Proof

$\frac{d}{dx} x^n = \lim_{h\to 0} \frac{(x+h)^n - x^n}{h}$
$\frac{d}{dx} x^n = \lim_{h\to 0} \frac{\left(\sum_{i=0}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^i\right) - x^n}{h}$
$\frac{d}{dx} x^n = \lim_{h\to 0} \frac{\left(x^n + nx^{n-1}h + \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^i\right) - x^n}{h}$
$\frac{d}{dx} x^n = \lim_{h\to 0} \frac{nx^{n-1}h + \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^i}{h}$
$\frac{d}{dx} x^n = \lim_{h\to 0} nx^{n-1} + \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^{i-1}$
$\frac{d}{dx} x^n = nx^{n-1} + \lim_{h\to 0} \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^{i-1}$
$\frac{d}{dx} x^n = nx^{n-1} + 0$
$\frac{d}{dx} x^n = nx^{n-1}$