Derivative of Natural Logarithm Function/Proof 2

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Theorem

Let $\ln x$ be the natural logarithm function.

Then:

$\map {\dfrac \d {\d x} } {\ln x} = \dfrac 1 x$


Proof

This proof assumes the definition of the natural logarithm as the inverse of the exponential function, where the exponential function is defined as the limit of a sequence:

$e^x := \ds \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$

It also assumes the Laws of Logarithms.

\(\ds \map {\frac \d {\d x} } {\ln x}\) \(=\) \(\ds \lim_{\Delta x \mathop \to 0} \frac {\map \ln {x + \Delta x} - \ln x} {\Delta x}\) Definition of Derivative
\(\ds \) \(=\) \(\ds \lim_{\Delta x \mathop \to 0} \frac {\map \ln {\frac {x + \Delta x} x} } {\Delta x}\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \lim_{\Delta x \mathop \to 0} \paren {\frac 1 {\Delta x} \centerdot \map \ln {1 + \frac {\Delta x} x} }\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \lim_{\Delta x \mathop \to 0} \paren {\map \ln {\paren {1 + \frac {\Delta x} x}^{1 / \Delta x} } }\) Natural Logarithm of Power


Define $u$ as:

\(\ds u\) \(=\) \(\ds \dfrac {\Delta x} x\)
\(\ds \leadsto \ \ \) \(\ds \Delta x\) \(=\) \(\ds u x\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 {\Delta x}\) \(=\) \(\ds \frac 1 x \cdot \frac 1 u\)

Hence:

\(\ds \) \(=\) \(\ds \lim_{u \mathop \to 0} \paren {\map \ln {\paren {1 + u}^{\frac 1 u \cdot \frac 1 x} } }\) substituting $u x$ for $\Delta x$ in $(1)$
\(\ds \) \(=\) \(\ds \lim_{u \mathop \to 0} \paren {\frac 1 x \cdot \map \ln {1 + u}^{\frac 1 u} }\) Natural Logarithm of Power
\(\ds \) \(=\) \(\ds \frac 1 x \cdot \lim_{u \mathop \to 0} \paren {\map \ln {1 + u}^{\frac 1 u} }\) factoring out constants
\(\ds \) \(=\) \(\ds \frac 1 x \cdot \lim_{v \mathop \to +\infty} \paren {\map \ln {1 + \frac 1 v}^v}\) substituting $\dfrac 1 v$ for $u$
\(\ds \) \(=\) \(\ds \frac 1 x \cdot \ln e^1\) Limit of Composite Function, Limit definition of $e^x$, Real Natural Logarithm Function is Continuous
\(\ds \) \(=\) \(\ds \frac 1 x\) Exponential of Natural Logarithm

$\blacksquare$


Sources