Derivative of Natural Logarithm Function

From ProofWiki
Jump to: navigation, search

Theorem

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

Then:

$D_x \left({\ln x}\right) = \dfrac 1 x$


Proof 1

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

$\blacksquare$


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.

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

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.

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

$\blacksquare$


Proof 3

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

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

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 functions.

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

$f_n \left({ x }\right) = n \left({ \sqrt[n]{ x } - 1 }\right)$

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

Pick $k \in \N : x_0 \in J := \left[{\dfrac 1 k , \,.\,.\, k}\right]$.

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

From Derivative of Nth Root and Combination Theorem for Sequences:

$\forall n \in \N : \forall x \in J : D_x f_n \left({x}\right) = \dfrac {\sqrt [n] x} x$

In particular:

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

From Defining Sequence of Natural Logarithm is Convergent, $\left\langle{ f_n \left({ x_0 }\right) }\right\rangle$ is convergent.


Lemma

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

$f_n \left({x}\right) = n \left({\sqrt [n] x - 1}\right)$

Let $k \in \N$.

Let $J = \left[{\frac 1 k \,.\,.\, k}\right]$.


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

$\Box$


From the lemma, $\left\langle{f_n'}\right\rangle$ converges uniformly to $\dfrac 1 x$ on $J$.

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

In particular, $f' \left({ x_0 }\right) = \dfrac 1 {x_0}$

Hence the result.

$\blacksquare$


Sources