Derivative of Natural Logarithm Function

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

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

Proof 1
This proof assumes the definition of the natural logarithm as:

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.

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 $u > 1$. Note that because the domain of the natural logarithm is $\left({0 \,.\,.\, +\infty}\right)$, $x$ is positive.

Proof 3
This proof assumes the definition of the natural logarithm as the inverse of the exponential function as defined by differential equation:


 * $y = \dfrac {\mathrm dy}{\mathrm dx}$


 * $y = e^x \iff \ln y = x$

The result follows from the definition of the antiderivative and the defined initial condition:
 * $\left({x_0, y_0}\right) = \left({0, 1}\right)$