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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \ln x\)	\(=\)	\(\displaystyle \int_1^x \dfrac 1 t \ \mathrm dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm d}{\mathrm dx} \ln x\)	\(=\)	\(\displaystyle \frac {\mathrm d}{\mathrm dx} \int_1^x \dfrac 1 t \ \mathrm dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	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.

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

Define $u$ as

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

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

Substitute $u$ into the above equation, and since $u \to \infty$ we will set $u > 1$. Note that because the domain of the natural logarithm is $(0..\infty)$, $x$ is positive.

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

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\mathrm d y}{\mathrm d x}\)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of the exponential function
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \displaystyle \int \frac 1 y \mathrm d y\)	\(=\)	\(\displaystyle \displaystyle \int \mathrm dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Separation of Variables
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x + C_0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integration of a Constant where that constant is $1$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \ln y + C_0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	substituting $x = \ln y$

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

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

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $13.27$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 14.1$

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