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:

• 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:

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

$\blacksquare$