Power Rule for Derivatives/Real Number Index

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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.


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Proof 1

We are going to prove that $\map {f'} x = n x^{n - 1}$ holds for all real $n$.

To do this, we compute the limit $\ds \lim_{h \mathop \to 0} \frac {\paren {x + h}^n - x^n} h$:


\(\ds \frac {\paren {x + h}^n - x^n} h\) \(=\) \(\ds \frac {x^n} h \paren {\paren {1 + \frac h x}^n - 1}\)
\(\ds \) \(=\) \(\ds \frac{x^n} h \paren {e^{n \map \ln {1 + \frac h x} } - 1}\)
\(\ds \) \(=\) \(\ds x^n \cdot \frac {e^{n \map \ln {1 + \frac h x} } - 1} {n \map \ln {1 + \frac h x} } \cdot \frac {n \map \ln {1 + \frac h x} } {\frac h x} \cdot \frac 1 x\)


Now we use the following results:

$\ds \lim_{x \mathop \to 0} \frac {\exp x - 1} x = 1$ from Derivative of Exponential at Zero
$\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$ from Derivative of Logarithm at One

to obtain:

$x^n \cdot \dfrac {e^{n \map \ln {1 + \frac h x} } - 1} {n \map \ln {1 + \dfrac h x} } \cdot \dfrac {n \map \ln {1 + \dfrac h x}} {\dfrac h x} \cdot \dfrac 1 x \to n x^{n - 1}$ as $h \to 0$

Hence the result.

$\blacksquare$


Proof 2

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:

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

$\blacksquare$

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