# Derivative of Natural Logarithm Function/Proof 2

## 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

$\blacksquare$