# Derivative of Natural Logarithm Function

## Theorem

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

Then:

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

## Proof 1

 $\ds \ln x$ $:=$ $\ds \int_1^x \dfrac 1 t \rd t$ Definition 1 of Natural Logarithm $\ds \frac \d {\d x} \ln x$ $=$ $\ds \frac \d {\d x} \int_1^x \dfrac 1 t \rd t$ $\ds$ $=$ $\ds \frac 1 x$ Fundamental Theorem of Calculus

$\blacksquare$

## Proof 2

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

$\blacksquare$

## Proof 3

This proof assumes the definition of the natural logarithm as the inverse of the exponential function as defined by differential equation:

$y = \dfrac {\d y} {\d x}$
$y = e^x \iff \ln y = x$
 $\ds \frac {\d y} {\d x}$ $=$ $\ds y$ Definition of Exponential Function $\ds \int \frac 1 y \rd y$ $=$ $\ds \int \rd x$ Separation of Variables $\ds$ $=$ $\ds x + C_0$ Integral of Constant where that constant is $1$ $\ds$ $=$ $\ds \ln y + C_0$ Definition 2 of Natural Logarithm: $x = \ln y$

The result follows from the definition of the antiderivative and the defined initial condition:

$\tuple {x_0, y_0} = \tuple {0, 1}$

$\blacksquare$

## Proof 4

This proof assumes the definition of the natural logarithm as the limit of a sequence of real functions.

Let $\sequence {f_n}$ be the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:

$\map {f_n} x = n \paren {\sqrt [n] x - 1}$

Fix $x_0 \in \R_{>0}$.

Pick $k \in \N : x_0 \in J := \closedint {\dfrac 1 k} k$.

From definition of bounded interval, $J$ is bounded.

$\forall n \in \N : \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$

In particular:

$\forall n: f_n$ is differentiable on $J$

From Defining Sequence of Natural Logarithm is Convergent, $\sequence {\map {f_n} {x_0} }$ is convergent.

### Lemma

Let $\sequence {f_n}_n$ be the sequence of real functions $f_n: \R_{>0} \to \R$ defined as:

$\map {f_n} x = n \paren {\sqrt [n] x - 1}$

Let $k \in \N$.

Let $J = \closedint {\dfrac 1 k} k$.

Then the sequence of derivatives $\sequence { {f_n}'}_n$ converges uniformly to some real function $g: J \to \R$.

$\Box$

From the lemma, $\sequence { {f_n}'}$ converges uniformly to $\dfrac 1 x$ on $J$.

From Derivative of Uniformly Convergent Sequence of Differentiable Functions, $\map {f'} x = \dfrac 1 x$ on $J$

In particular:

$\map {f'} {x_0} = \dfrac 1 {x_0}$

Hence the result.

$\blacksquare$

## Examples

### Example: $\paren {x - 2}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$

Let $y = \paren {x - 2}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$.

Then:

$\dfrac {\d y} {\d x} = \paren {\dfrac 1 {3 \paren {x - 2} } + \dfrac 1 {2 \paren {x - 3} } + \dfrac 3 {2 x - 1} } \paren {x - 2}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$