Power Rule for Derivatives/Real Number Index

Theorem
Let $n \in \R$.

Let $f: \R \to \R$ be the real function defined as $f \left({x}\right) = x^n$.

Then:
 * $f^{\prime} \left({x}\right) = n x^{n-1}$

everywhere that $f \left({x}\right) = x^n$ is defined.

When $x = 0$ and $n = 0$, $f^{\prime} \left({x}\right)$ is undefined.

Proof 1
We are going to prove that $f^{\prime}(x) = n x^{n-1}$ holds for all real $n$.

To do this, we compute the limit $\displaystyle \lim_{h \to 0} \frac{\left({x + h}\right)^n - x^n} h$:

Now we use the following results:
 * $\displaystyle \lim_{x \to 0} \frac {\exp x - 1} x = 1$ from Derivative of Exponent at Zero
 * $\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} x = 1$ from Derivative of Logarithm at One

... to obtain:
 * $\displaystyle \frac {e^{n \ln \left({1 + \frac h x}\right)} - 1} {n \ln \left( {1 + \frac h x}\right)} \cdot \frac {n \ln \left({1 + \frac h x}\right)} {\frac h x} \cdot \frac 1 x \to n x^{n-1}$ as $h \to 0$

Hence the result.

Proof 2
This proof only holds for $x^n > 0$.

Let $y$ = $f(x)$.

Then $y = x^n$.

Taking the natural logarithm of both sides we have:


 * $\ln y = \ln x^n$

Using Logarithms of Powers this becomes:


 * $\ln y = n \ln x$

Using: and taking the derivative of both sides with respect to $x$ gives us:
 * Derivative of a Composite Function
 * Derivative of Constant Multiple
 * Derivative of Natural Logarithm Function.


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

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


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

Substituting $x^n$ for $y$:


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

From Exponent Combination Laws/Difference of Powers:


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