# Power Rule for Derivatives/Real Number Index/Proof 2

## Theorem

Let $n \in \R$.

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

Note this proof does not hold for $x = 0$.

Let $y$ = $\map f x$.

Then $y = x^n$.

Then:

 $\ds y$ $=$ $\ds x^n$ $\ds \leadsto \ \$ $\ds \size y$ $=$ $\ds \size {x^n}$ taking the absolute value of both sides $\ds$ $=$ $\ds \size x^n$ Absolute Value of Power $\ds \leadsto \ \$ $\ds \ln \size y$ $=$ $\ds \map \ln {\size x^n}$ taking the natural logarithm of both sides $\ds$ $=$ $\ds n \ln \size x$ Logarithm of Power

Using:

and taking the derivative of both sides with respect to $x$ gives:

$\dfrac 1 y \dfrac {\d y} {\d x} = n \dfrac 1 x$

Multiplying both sides of the equation by $y$ yields:

$\dfrac {\d y} {\d x} = n \dfrac y x$

Substituting $x^n$ for $y$:

$\dfrac {\d y} {\d x} = n \dfrac {x^n} x$

From Quotient of Powers:

$\dfrac {\d y} {\d x} = n x^{n - 1}$

$\blacksquare$