# Derivative of Natural Logarithm Function/Proof 4

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## Theorem

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

Then:

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

## Proof

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.

$\blacksquare$