Derivative of Natural Logarithm Function

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Theorem

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

Then:

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


Proof 1

\(\displaystyle \ln x\) \(:=\) \(\displaystyle \int_1^x \dfrac 1 t \ \mathrm d t\) Definition of Natural Logarithm
\(\displaystyle \frac {\mathrm d}{\mathrm d x} \ln x\) \(=\) \(\displaystyle \frac {\mathrm d}{\mathrm d x} \int_1^x \dfrac 1 t \ \mathrm d t\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 x\) 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:

$e^x := \displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n$

It also assumes the Laws of Logarithms.

\(\displaystyle D_x \left({\ln x}\right)\) \(=\) \(\displaystyle \lim_{\Delta x \to 0} \frac {\ln \left({x + \Delta x}\right) - \ln \left({x}\right)} {\Delta x}\) Definition of Derivative
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\Delta x \to 0} \frac {\ln \left({\frac {x + \Delta x} x}\right)}{\Delta x}\) Laws of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\Delta x \to 0} \left ({\frac 1 {\Delta x} \centerdot \ln \left({1 + \frac {\Delta x} x}\right)}\right)\)

Define $u$ as:

$u = \left|{\dfrac x {\Delta x} }\right|$

Then $u \to +\infty$ as $\Delta x \to 0$.

Substitute $u$ into the above equation, and since $u \to +\infty$, set $u > 1$.

From Domain of Real Natural Logarithm, $\operatorname{Dom} \left({\ln}\right)$ is $\left({0 \,.\,.\, +\infty}\right)$.

That is, $x$ is positive.

\(\displaystyle \) \(=\) \(\displaystyle \lim_{u \to +\infty} \left ({\frac u x \centerdot \ln \left({1 + \frac 1 u}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{u \to +\infty} \left ({\frac 1 x \centerdot \ln \left({1 + \frac 1 u}\right)^u}\right)\) Laws of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 x \centerdot \lim_{u \to +\infty} \left ({\ln \left({1 + \frac 1 u}\right)^u}\right)\) factoring out constants
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 x \centerdot \ln e^1\) Limit of Composite Function, Limit definition of $e^x$, Natural Logarithm Function is Continuous
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 x\) 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:

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

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

$\left({x_0, y_0}\right) = \left({0, 1}\right)$

$\blacksquare$


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.

From Derivative of Nth Root and Combination Theorem for Sequences:

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

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.

$\blacksquare$


Sources