What is the derivative of \ln 5 or other constants?

The derivative of any constant (including \ln 5) is zero, because constants don't change. The formula \frac{d}{dx}\ln x = \frac{1}{x} applies when x is the variable; if the argument is constant, there's nothing to differentiate with respect to x.

Why is the derivative of \ln x equal to \frac{1}{x}?

This comes from the definition of the derivative and the fact that the exponential function e^x is the inverse of \ln x. Since the derivatives of inverse functions are reciprocals, and \frac{d}{dx}e^x = e^x, we get \frac{d}{dx}\ln x = \frac{1}{e^{\ln x}} = \frac{1}{x}. The intuition is that as x grows larger, the logarithm grows more slowly, so its rate of change \frac{1}{x} decreases.

What's the difference between the derivative of \ln x and the derivative of \log x (base 10)?

The derivative of \ln x is \frac{1}{x}, while the derivative of \log_{10} x is \frac{1}{x \ln 10}. The difference comes from the change of base formula: any logarithm with base a has derivative \frac{1}{x \ln a}. Natural log is special because the \ln in the denominator equals 1 (since \ln e = 1), making the derivative especially clean.

Can you take the derivative of \ln x where x is negative?

No, the natural logarithm is only defined for positive x, so the derivative does not exist for x \leq 0. If you need to work with logarithms of negative numbers, use \ln|x|, which has derivative \frac{1}{x} for all x \neq 0. The absolute value lets the function 'see' both positive and negative inputs, though you must still exclude zero.

Derivative of ln x

The derivative of ln x is 1/x, giving you the rate of change of the natural logarithm — essential for calculus problems involving logarithmic functions.

\frac{d}{dx}\ln x = \frac{1}{x}

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What each symbol means

What Derivative of ln x takes

Symbol	Meaning
$x$	x represents the input to the natural logarithm; it must always be positive (x > 0), because the logarithm is undefined for zero or negative values. If you confuse x with the logarithm itself or think x can be any real number, the result will be wrong.

When to use it

Use this when you need to find the rate of change of a natural logarithmic function or simplify derivatives of expressions involving ln x.

Level

Usually taught in: Calculus I

Worked examples

1. Find the derivative of a simple logarithm

Problem

Find

\frac{d}{dx} \ln(3x)

$\frac{d}{dx} \ln(3x) = \frac{1}{3x} \cdot \frac{d}{dx}(3x)$
Apply the chain rule: the derivative of $\ln(u)$ is $\frac{1}{u}$ times the derivative of $u$ .
$\frac{d}{dx}(3x) = 3$
The derivative of $3x$ is $3$ .
$\frac{1}{3x} \cdot 3 = \frac{3}{3x}$
Multiply the parts together.
$\frac{3}{3x} = \frac{1}{x}$
Simplify by canceling the $3$ in numerator and denominator.

Answer: $\frac{1}{x}$

This shows the chain rule in action — we found the derivative of the outer function (ln) and multiplied by the derivative of the inner function ( $3x$ ). Even though we started with $\ln(3x)$ , the coefficient of $x$ cancels when we simplify.

2. Find the derivative of a logarithm with a polynomial

Problem

Find

\frac{d}{dx} \ln(x^2 - 4)

and evaluate at

x = 3

$\frac{d}{dx} \ln(x^2 - 4) = \frac{1}{x^2 - 4} \cdot \frac{d}{dx}(x^2 - 4)$
Use the chain rule with $u = x^2 - 4$ .
$\frac{d}{dx}(x^2 - 4) = 2x$
Apply the power rule to find the derivative of the polynomial.
$\frac{1}{x^2 - 4} \cdot 2x = \frac{2x}{x^2 - 4}$
Multiply the results together.
$\frac{2(3)}{3^2 - 4} = \frac{6}{9 - 4} = \frac{6}{5}$
Substitute $x = 3$ and simplify: $9 - 4 = 5$ .

Answer: $f'(3) = \frac{6}{5}$

This example shows that the chain rule is essential when the argument to ln is more complex than just $kx$ . We must be careful with the algebra — the $(x^2 - 4)$ in the denominator doesn't simplify with the $2x$ in the numerator, so we leave it as a fraction.

3. Analyze productivity growth in a construction schedule

Problem

A supervisor tracks crew productivity according to

P(t) = 12 \ln(t + 1)

units per day, where

t

is months of the project. How fast is productivity increasing at month 5?

$\frac{dP}{dt} = 12 \cdot \frac{d}{dt} \ln(t+1)$
Factor out the constant $12$ .
$\frac{d}{dt} \ln(t+1) = \frac{1}{t+1} \cdot \frac{d}{dt}(t+1)$
Apply the chain rule to the logarithm.
$\frac{d}{dt}(t+1) = 1$
The derivative of $(t+1)$ is $1$ .
$\frac{dP}{dt} = 12 \cdot \frac{1}{t+1} = \frac{12}{t+1}$
Combine the results.
$\frac{dP}{dt} = \frac{12}{5+1} = \frac{12}{6} = 2$
Substitute $t = 5$ and simplify: $5 + 1 = 6$ and $\frac{12}{6} = 2$ .

Answer: $\frac{dP}{dt}\bigg|_{t=5} = 2 \text{ units per day per month}$

This shows how the derivative of ln helps us measure the rate of change in real situations. The result tells us that at the 5-month mark, productivity is increasing by 2 units per day for each additional month of the project — a concrete rate of change.

Common mistakes

Where Derivative of ln x usually goes wrong

The mistake	Why it is wrong	The fix
$\frac{d}{dx}\ln(2x) = \frac{1}{2x}$	You applied the basic logarithm derivative rule but forgot to apply the chain rule; the chain rule requires multiplying by the derivative of $2x$ , which is $2$ .	$\frac{d}{dx}\ln(2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x}$
$\frac{d}{dx}\ln(x^2) = \frac{1}{x^2}$	You forgot the chain rule; you must multiply by the derivative of $x^2$ , which is $2x$ .	$\frac{d}{dx}\ln(x^2) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}$
$\frac{d}{dx}(\ln x)^2 = 2 \cdot \frac{1}{x}$	You confused $(\ln x)^2$ (the square of the logarithm) with $\ln(x^2)$ (the logarithm of the square); $(\ln x)^2$ is a composition requiring the chain rule applied to both functions, giving $2(\ln x) \cdot \frac{1}{x}$ .	$\frac{d}{dx}(\ln x)^2 = 2(\ln x) \cdot \frac{1}{x} = \frac{2\ln x}{x}$

Tips and when to use something else

Always check if you need the chain rule: if the argument to ln is not just $x$ , multiply by the derivative of that argument.
Remember that $\ln x$ is only defined for $x > 0$ ; if your problem involves negative values, use $\ln|x|$ and remember the domain restriction.
For more complex expressions, consider using Logarithmic Differentiation instead — it sometimes simplifies products or quotients before taking the derivative.
Watch out for the difference between $\ln(x^2)$ and $(\ln x)^2$ ; they have different derivatives and require different applications of the chain rule.

Related formulas

Chain RuleThe Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.Derivative of e^xThe derivative of e^x is e^x itself—the unique exponential function used to model continuous growth, radioactive decay, and other rates of change.Logarithmic DifferentiationSolve derivatives of functions like x^x where both base and exponent depend on x using logarithms and implicit differentiation.Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.

Frequently asked questions

What is the derivative of $\ln 5$ or other constants?: The derivative of any constant (including $\ln 5$ ) is zero, because constants don't change. The formula $\frac{d}{dx}\ln x = \frac{1}{x}$ applies when $x$ is the variable; if the argument is constant, there's nothing to differentiate with respect to $x$ .
Why is the derivative of $\ln x$ equal to $\frac{1}{x}$ ?: This comes from the definition of the derivative and the fact that the exponential function $e^x$ is the inverse of $\ln x$ . Since the derivatives of inverse functions are reciprocals, and $\frac{d}{dx}e^x = e^x$ , we get $\frac{d}{dx}\ln x = \frac{1}{e^{\ln x}} = \frac{1}{x}$ . The intuition is that as $x$ grows larger, the logarithm grows more slowly, so its rate of change $\frac{1}{x}$ decreases.
What's the difference between the derivative of $\ln x$ and the derivative of $\log x$ (base 10)?: The derivative of $\ln x$ is $\frac{1}{x}$ , while the derivative of $\log_{10} x$ is $\frac{1}{x \ln 10}$ . The difference comes from the change of base formula: any logarithm with base $a$ has derivative $\frac{1}{x \ln a}$ . Natural log is special because the $\ln$ in the denominator equals $1$ (since $\ln e = 1$ ), making the derivative especially clean.
Can you take the derivative of $\ln x$ where $x$ is negative?: No, the natural logarithm is only defined for positive $x$ , so the derivative does not exist for $x \leq 0$ . If you need to work with logarithms of negative numbers, use $\ln|x|$ , which has derivative $\frac{1}{x}$ for all $x \neq 0$ . The absolute value lets the function 'see' both positive and negative inputs, though you must still exclude zero.

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Reviewed 2026-09-18