Mathematics · Calculus

Derivative of the Natural Logarithm

The derivative of the natural logarithm ln x is the hyperbola 1/x, valid for all x > 0.

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Formula

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

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

Dimensionless (calculus)

Variables & units – Derivative of the Natural Logarithm

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 – Derivative of the Natural Logarithm

The natural logarithm is the inverse of the exponential function; the inverse function rule gives (ln x)′ = 1/x for x > 0. This closes the gap of the power rule: 1/x = x⁻¹ is the only power without a power antiderivative, its antiderivative is ln|x| + C. For general logarithms (log_b x)′ = 1/(x·ln b), and for compositions the chain rule gives (ln u(x))′ = 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 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)·(ln x)′ = 1.
2Since e^(ln x) = x, it follows that (ln x)′ = 1/x.

Rearrangements

Composition

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

Logarithmic differentiation, useful for products and powers.

General base

\frac{d}{d x} lo g_{b} x = \frac{1}{x l n b}

Follows from log_b x = ln x/ln b.

Antiderivative of 1/x

\int \frac{1}{x} d x = ln ∣ x ∣ + C

Closes the gap of the power rule at n = −1.

Task variant

Differentiate f(x) = ln(3x² + 1).

u = 3x² + 1, u′ = 6x. f′(x) = 6x/(3x² + 1). At x = 1: f′(1) = 6/4 = 1.5.

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

f′(x) = 1/x, so f′(1) = 1. Tangent: y = 1·(x − 1) + 0 = x − 1.

Common mistakes

Using (ln x)′ = 1/x also for compositions like ln(x²).

Chain rule: (ln(x²))′ = 2x/x² = 2/x.

Confusing the antiderivative of ln x with 1/x.

∫ln x dx = x·ln x − x + C (integration by parts); 1/x is the derivative.

Setting (log₁₀ x)′ = 1/x.

Only ln gives 1/x; in general it is 1/(x·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 1/x.

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Formula cluster

Logarithm and inverse functions

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

\frac{d}{d x} e^{x} = e^{x}, \frac{d}{d x} e^{k x} = k \cdot e^{k x}

Derivative of the Exponential Function

lo g_{b} (x \cdot y) = lo g_{b} x + lo g_{b} y, lo g_{b} (x^{r}) = r \cdot lo g_{b} x

Laws of Logarithms

Worked example

f(x) = ln x: f′(2) = 1/2 = 0.5. Composite: g(x) = ln(x² + 1) → g′(x) = 2x/(x² + 1), so g′(1) = 2/2 = 1.

Applications

Curve analysis of logarithmic functions, integration of 1/x, half-life and doubling-time calculations, logarithmic scales

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Which formula describes Derivative of the Natural Logarithm?

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How do you rearrange (ln x)' = 1/x for Composition?

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Which common mistake happens with Derivative of the Natural Logarithm?

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Scientific sources

[1]Forster, O. (2016). Analysis 1. Springer Vieweg.

Common notations & search queries

(ln x)'=1/xln x ableitenln ableitenAbleitung ln xln(x^2) ableitennatürlicher Logarithmus Ableitungderivative of ln xln Funktion Ableitung

Related formulas

\frac{d}{d x} e^{x} = e^{x}, \frac{d}{d x} e^{k x} = k \cdot e^{k x}

Derivative of the Exponential Function

lo g_{b} (x \cdot y) = lo g_{b} x + lo g_{b} y, lo g_{b} (x^{r}) = r \cdot lo g_{b} x

Laws of Logarithms

(f \circ g)^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)

Chain Rule of Differentiation

More Mathematics formulas

a^{2} + b^{2} = c^{2}

Pythagorean Theorem

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Quadratic Formula

e^{i φ} = cos φ + i \cdot sin φ

Euler's Formula

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

Product Rule of Differentiation

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Frequently asked questions about Derivative of the Natural Logarithm

Why is the derivative of ln x equal to 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)·(ln x)′ on the left and 1 on the right. Since e^(ln x) = x, this reads x·(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(x² + 1)?+

Use the chain rule: outer derivative 1/u, inner derivative u′, together (ln u(x))′ = u′(x)/u(x). For ln(x² + 1): u = x² + 1, u′ = 2x, so the derivative is 2x/(x² + 1); at x = 1 this gives 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. ∫2x/(x² + 1) dx = ln(x² + 1) + C. Common mistake: writing just 1/(x² + 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 1/x as the derivative of ln x also exists only for x > 0. When integrating, the perspective flips: 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|: ∫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 = 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 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 ≈ 2.3. Rule of thumb: first rewrite a foreign base in ln, then differentiate.

What is the antiderivative of ln x?+

Not 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·ln x: choose u = ln x and v′ = 1, so u′ = 1/x and v = x. Then ∫ln x dx = x·ln x − ∫x·(1/x) dx = x·ln x − ∫1 dx = x·ln x − x + C. Check by differentiating: (x·ln x − x)′ = ln x + x·(1/x) − 1 = ln x ✓. This task is an exam classic precisely because the confusion with 1/x is so tempting. Remember the directions: differentiating turns ln x into the fraction 1/x, integrating turns it into x·ln x − x.

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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(3x² + 1).
Solution path
u = 3x² + 1, u′ = 6x. f′(x) = 6x/(3x² + 1). At x = 1: f′(1) = 6/4 = 1.5.
2
Task
Find the tangent to f(x) = ln x at the point (1|0).
Solution path
f′(x) = 1/x, so f′(1) = 1. Tangent: y = 1·(x − 1) + 0 = x − 1.