Derivative of Natural Logarithm Function

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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 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.

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.

$\blacksquare$


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:

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


Sources

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