Mathematics · Calculus

Derivative of the Natural Logarithm

Q: How do I differentiate ln(u(x)), for example $\ln{}(x^{2} + 1)$?

Use the chain rule: outer derivative $\frac{1}{u}$, inner derivative u′, together (ln u(x))′ = u′$\frac{(x)}{u}(x)$. For $\ln{}(x^{2} + 1)$: $u = x^{2} + 1$, u′ = 2x, so the derivative is $\frac{2x}{(x^{2} + 1)}$; at $x = 1$ this gives $\frac{2}{2} = 1$. The quotient u′/u is so important that it has its own name: logarithmic derivative. Read backwards it is an integration tool: if the numerator of a fraction is exactly the derivative of the denominator, the antiderivative is ln|denominator|, e.g. $\frac{\int{}2x}{(x^{2} + 1)} dx = \ln{}(x^{2} + 1) + C$. Common mistake: writing just $\frac{1}{(x^{2} + 1)}$ and omitting the inner derivative 2x.

Q: How do you differentiate log₁₀(x) or log₂(x)?

Via the change of base: $\log{}_{b} x =\frac{ \ln{} x}{\ln{}} b$, where ln b is a constant. Differentiating the constant multiple gives (log_b x)′ = 1/(x·ln b). For the base-10 logarithm this means (lg x)′ = 1/(x·ln 10) ≈ 1/(2.303·x), for the base-2 logarithm (log₂ x)′ = 1/(x·ln 2) ≈ 1/(0.693·x). Only the natural logarithm has the clean derivative $\frac{1}{x}$ without a prefactor, which makes it the standard choice in calculus. The typical mistake is writing (lg x)′ = 1/x; the result is then too large by the factor $\ln{} 10 \approx{} 2.3$. Rule of thumb: first rewrite a foreign base in ln, then differentiate.

Q: What is the antiderivative of ln x?

Not $\frac{1}{x}$, that is the derivative! You find the antiderivative with integration by parts and the trick of reading ln x as the product $1\cdot{}\ln{} x$: choose $u = \ln{} x$ and v′ = 1, so u′ = 1/x and $v = x$. Then $\int{}\ln{} x dx = x\cdot{}\ln{} x - \int{}x\cdot{}(\frac{1}{x}) dx = x\cdot{}\ln{} x - \int{}1 dx = x\cdot{}\ln{} x - x + C$. Check by differentiating: $(x\cdot{}\ln{} x - x)$′ = ln x + x·(1/x) − 1 = ln x ✓. This task is an exam classic precisely because the confusion with $\frac{1}{x}$ is so tempting. Remember the directions: differentiating turns ln x into the fraction $\frac{1}{x}$, integrating turns it into $x\cdot{}\ln{} x - x$.

The derivative of the natural logarithm ln x is the hyperbola $\frac{1}{x}$ , valid for all $x > 0$ .

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Formula

\frac{d}{d x} ln x = \frac{1}{x} (x > 0)

Dimensionless (calculus)

Variables & units

Variables & units
Symbol	Meaning	Unit
ln x	Natural logarithm with base e	dimensionless
$x$	Argument of the logarithm $(x > 0)$	dimensionless
$1/x$	Derivative function, decreasing for growing x	dimensionless

LaTeX

\frac{d}{dx}\, \ln x = \frac{1}{x} \quad (x > 0)

QUANTA STUDYquanta-study.de

Graph of the natural logarithm with two tangents, at x = 1 with slope 1 and at x = 4 with slope 1/4. — The tangent slope of ln x at x is 1/x: steep near 0, ever flatter for large x.

Derivation & background

The natural logarithm is the inverse of the exponential function; the inverse function rule gives $(\ln x)' = \frac{1}{x}$ for $x > 0$ . This closes the gap of the power rule: $\frac{1}{x} = x^{-1}$ is the only power without a power antiderivative, its antiderivative is $\ln|x| + C$ . For general logarithms $(\log_b x)' = \frac{1}{x\ln b}$ , and for compositions the chain rule gives $(\ln u(x))' = \frac{u'(x)}{u(x)}$ .

Exam blueprint

Validity range

Holds for $x > 0$ , the domain of ln x. On all of R without 0, (ln|x|)′ = 1/x, important when integrating $\frac{1}{x}$ .

Derivation steps

Inverse function rule: ln x is the inverse of eˣ.

1From $e^{(\ln{} x)} = x$ , differentiating with the chain rule gives $e^{(\ln{} x)}\cdot{}(\ln{} x)$ ′ = 1.
2Since $e^{(\ln{} x)} = x$ , it follows that (ln x)′ = 1/x.

Rearrangements

Composition

\frac{d}{dx}\, \ln u(x) = \frac{u'(x)}{u(x)}

Logarithmic differentiation, useful for products and powers.

General base

\frac{d}{dx}\, \log_{b} x = \frac{1}{x \ln b}

Follows from $\log{}_{b} x =\frac{ \ln{} x}{\ln{}} b$ .

Antiderivative of $\frac{1}{x}$

\int \frac{1}{x} \, dx = \ln|x| + C

Closes the gap of the power rule at $n = -1$ .

Task variant

Differentiate $f(x) = \ln{}(3x^{2} + 1)$ .

$u = 3x^{2} + 1$ , u′ = 6x. f′ $(x) =\frac{ 6x}{(3x^{2} + 1)}$ .
At $x = 1$ : f′ $(1) =\frac{ 6}{4} = 1.5$ .

Find the tangent to $f(x) = \ln{} x$ at the point (1|0).

f′ $(x) =\frac{ 1}{x}$ , so f′ $(1) = 1$ .
Tangent: $y = 1\cdot{}(x - 1) + 0 = x - 1$ .

Common mistakes

Using (ln x)′ = 1/x also for compositions like $\ln{}(x^{2})$ .

Chain rule: $(\ln{}(x^{2}))$ ′ = 2x/x² = 2/x.

Confusing the antiderivative of ln x with $\frac{1}{x}$ .

∫ln x dx = x·ln x − x + C (integration by parts); $\frac{1}{x}$ is the derivative.

Setting (log₁₀ x)′ = 1/x.

Only ln gives $\frac{1}{x}$ ; in general it is $\frac{1}{(x\cdot{}\ln{} b)}$ .

Ignoring the domain $x > 0$ .

ln x exists only for positive x; the behaviour as x → 0⁺ is part of curve analysis.

Exam context

Curve analysis of logarithmic functions, tangents, domains and integrals involving $\frac{1}{x}$ .

Formula cluster

Logarithm and inverse functions

ln and the exponential are mirror images; their derivatives determine each other.

(eˣ)' = eˣ

Derivative of the Exponential Function

log(x·y) = log x + log y

Laws of Logarithms

Worked example

$f(x) = \ln{} x$ : f′ $(2) =\frac{ 1}{2} = 0.5$ . Composite: $g(x) = \ln{}(x^{2} + 1) \to{} g$ ′ $(x) =\frac{ 2x}{(x^{2} + 1)}$ , so g′ $(1) =\frac{ 2}{2} = 1$ .

Applications

Curve analysis of logarithmic functions, integration of $\frac{1}{x}$ , half-life and doubling-time calculations, logarithmic scales

How do you calculate with Derivative of the Natural Logarithm?

Here is how to work through a typical Derivative of the Natural Logarithm ((ln x)' = 1/x) task step by step:

1
Task
Differentiate $f(x) = \ln{}(3x^{2} + 1)$ .
Solution path
1. $u = 3x^{2} + 1$ , u′ = 6x. f′ $(x) =\frac{ 6x}{(3x^{2} + 1)}$ .
2. At $x = 1$ : f′ $(1) =\frac{ 6}{4} = 1.5$ .
2
Task
Find the tangent to $f(x) = \ln{} x$ at the point (1|0).
Solution path
1. f′ $(x) =\frac{ 1}{x}$ , so f′ $(1) = 1$ .
2. Tangent: $y = 1\cdot{}(x - 1) + 0 = x - 1$ .

Frequently asked questions about Derivative of the Natural Logarithm

Why is the derivative of ln x equal to

\frac{1}{x}

The most elegant route goes via the inverse function. We have $e^{(\ln{} x)} = x$ for all $x > 0$ . Differentiating both sides, the chain rule gives $e^{(\ln{} x)}\cdot{}(\ln{} x)$ ′ on the left and 1 on the right. Since $e^{(\ln{} x)} = x$ , this reads $x\cdot{}(\ln{} x)$ ′ = 1, so (ln x)′ = 1/x. Intuitively: the ln curve is the exponential curve mirrored at the diagonal; where the exponential becomes steep, its inverse becomes flat. That is why the slope of ln x keeps decreasing as x grows: at $x = 1$ it is 1, at $x = 2$ only 0.5, at $x = 100$ just 0.01. The logarithm grows without bound, but ever more slowly.

How do I differentiate ln(u(x)), for example

\ln{}(x^{2} + 1)

Use the chain rule: outer derivative $\frac{1}{u}$ , inner derivative u′, together (ln u(x))′ = u′ $\frac{(x)}{u}(x)$ . For $\ln{}(x^{2} + 1)$ : $u = x^{2} + 1$ , u′ = 2x, so the derivative is $\frac{2x}{(x^{2} + 1)}$ ; at $x = 1$ this gives $\frac{2}{2} = 1$ . The quotient u′/u is so important that it has its own name: logarithmic derivative. Read backwards it is an integration tool: if the numerator of a fraction is exactly the derivative of the denominator, the antiderivative is ln|denominator|, e.g. $\frac{\int{}2x}{(x^{2} + 1)} dx = \ln{}(x^{2} + 1) + C$ . Common mistake: writing just $\frac{1}{(x^{2} + 1)}$ and omitting the inner derivative 2x.

Why does (ln x)′ = 1/x hold only for

x > 0

, and what does ln|x| mean?+

The natural logarithm is defined only for positive numbers, because eˣ takes only positive values; ln x answers the question "e to what power gives x?". So the derivative $\frac{1}{x}$ as the derivative of ln x also exists only for $x > 0$ . When integrating, the perspective flips: $\frac{1}{x}$ is defined for negative x as well, and there ln(−x) is an antiderivative, since the chain rule gives (ln(−x))′ = (−1)/(−x) = 1/x. Both cases are combined in ln|x|: $\frac{\int{}1}{x} dx = \ln{}|x| + C$ , valid on every interval that does not contain 0. Even so, you must never integrate across the pole at $x = 0$ .

How do you differentiate log₁₀(x) or log₂(x)?+

Via the change of base: $\log{}_{b} x =\frac{ \ln{} x}{\ln{}} b$ , where ln b is a constant. Differentiating the constant multiple gives (log_b x)′ = 1/(x·ln b). For the base-10 logarithm this means (lg x)′ = 1/(x·ln 10) ≈ 1/(2.303·x), for the base-2 logarithm (log₂ x)′ = 1/(x·ln 2) ≈ 1/(0.693·x). Only the natural logarithm has the clean derivative $\frac{1}{x}$ without a prefactor, which makes it the standard choice in calculus. The typical mistake is writing (lg x)′ = 1/x; the result is then too large by the factor $\ln{} 10 \approx{} 2.3$ . Rule of thumb: first rewrite a foreign base in ln, then differentiate.

What is the antiderivative of ln x?+

Not $\frac{1}{x}$ , that is the derivative! You find the antiderivative with integration by parts and the trick of reading ln x as the product $1\cdot{}\ln{} x$ : choose $u = \ln{} x$ and v′ = 1, so u′ = 1/x and $v = x$ . Then $\int{}\ln{} x dx = x\cdot{}\ln{} x - \int{}x\cdot{}(\frac{1}{x}) dx = x\cdot{}\ln{} x - \int{}1 dx = x\cdot{}\ln{} x - x + C$ . Check by differentiating: $(x\cdot{}\ln{} x - x)$ ′ = ln x + x·(1/x) − 1 = ln x ✓. This task is an exam classic precisely because the confusion with $\frac{1}{x}$ is so tempting. Remember the directions: differentiating turns ln x into the fraction $\frac{1}{x}$ , integrating turns it into $x\cdot{}\ln{} x - x$ .

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Calculus Volume 1, §3.9: Derivatives of Exponential and Logarithmic Functions

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(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

Chain Rule of Differentiation

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e^{i\varphi} = \cos\varphi + i \cdot \sin\varphi

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(f \cdot g)' = f' \cdot g + f \cdot g'

Product Rule of Differentiation

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n

Taylor Series

\frac{d}{dx}(x^n) = n \cdot x^{n-1}

Power Rule of Differentiation

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Derivative of the Natural Logarithm

The derivative of the natural logarithm ln x is the hyperbola

\frac{1}{x}

, valid for all

x > 0

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

ln x

Natural logarithm with base e

dimensionless

x

Argument of the logarithm

(x > 0)

dimensionless

1/x

Derivative function, decreasing for growing x

dimensionless

Derivation & background

Exam blueprint

Validity range

Holds for $x > 0$ , the domain of ln x. On all of R without 0, (ln|x|)′ = 1/x, important when integrating $\frac{1}{x}$ .

Derivation steps

Inverse function rule: ln x is the inverse of eˣ.

1From $e^{(\ln{} x)} = x$ , differentiating with the chain rule gives $e^{(\ln{} x)}\cdot{}(\ln{} x)$ ′ = 1.
2Since $e^{(\ln{} x)} = x$ , it follows that (ln x)′ = 1/x.

How do you calculate with Derivative of the Natural Logarithm?

Here is how to work through a typical Derivative of the Natural Logarithm ((ln x)' = 1/x) task step by step:

Task

Differentiate $f(x) = \ln{}(3x^{2} + 1)$ .

Solution path

$u = 3x^{2} + 1$ , u′ = 6x. f′ $(x) =\frac{ 6x}{(3x^{2} + 1)}$ .
At $x = 1$ : f′ $(1) =\frac{ 6}{4} = 1.5$ .

Task

Find the tangent to $f(x) = \ln{} x$ at the point (1|0).

Solution path

f′ $(x) =\frac{ 1}{x}$ , so f′ $(1) = 1$ .
Tangent: $y = 1\cdot{}(x - 1) + 0 = x - 1$ .

Frequently asked questions about Derivative of the Natural Logarithm

Why is the derivative of ln x equal to

\frac{1}{x}

How do I differentiate ln(u(x)), for example

\ln{}(x^{2} + 1)

Why does (ln x)′ = 1/x hold only for

x > 0

, and what does ln|x| mean?+

How do you differentiate log₁₀(x) or log₂(x)?+

What is the antiderivative of ln x?+