Power Rule for Derivatives/Real Number Index/Proof 2

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Let $n \in \R$.

Let $f: \R \to \R$ be the real function defined as $\map f x = x^n$.


$\map {f'} x = n x^{n-1}$

everywhere that $\map f x = x^n$ is defined.

When $x = 0$ and $n = 0$, $\map {f'} x$ is undefined.


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

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

Then $y = x^n$.


\(\ds y\) \(=\) \(\ds x^n\)
\(\ds \leadsto \ \ \) \(\ds \size y\) \(=\) \(\ds \size {x^n}\) 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


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:

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