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

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## 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:

- Derivative of Composite Function
- Derivative of Constant Multiple
- Corollary to Primitive of Reciprocal

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:

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

$\blacksquare$