Derivative of ln(x): Differentiate the Natural Log on Its Domain

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | FAQ | How This Fits

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The Principle

The move: Differentiate the direct logarithmic form $\ln(x)$ by replacing it with $\frac{1}{x}$.

The invariant: This gives the exact derivative of the real-valued function $\ln(x)$ on its domain $x>0$.

Pattern: $\frac{d}{dx}\ln(x) \quad \longrightarrow \quad \frac{1}{x}$

Legal ✓	Illegal ✗
$\dfrac{d}{dx}\ln(x)\big\rvert_{x=2} = \dfrac{1}{2}$	$\dfrac{d}{dx}\ln(x)\big\rvert_{x=-2} \not\to -\dfrac{1}{2}$

Left: the condition $x>0$ holds, so the rule applies. Right: $\ln(x)$ is not a real-valued function at $x=-2$, so writing $\frac{1}{x}$ there invents a derivative at a point outside the domain. The contrast is about applicability: the expression looks familiar, but the condition fails.

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Conditions of Applicability

Condition: $x>0$

Before applying, check: confirm that the local pattern is the bare logarithm $\ln(x)$ rather than a composite form $\ln(g(x))$, and that the current input satisfies $x>0$.

If the condition is violated: $\ln(x)$ is not a real-valued differentiable function there, so substituting $\frac{1}{x}$ creates a derivative where the original function is not defined.

• The condition is a domain guard, not extra decoration. It tells you when the logarithm exists as a real-valued function.
• If the local expression is $\ln(g(x))$ instead of $\ln(x)$, chain rule governs the step: $\frac{d}{dx}\ln(g(x)) = \frac{g'(x)}{g(x)}$ on intervals where $g(x)>0$.
• At $x=0$ or any negative input, the direct rule is unavailable because $\ln(x)$ itself is unavailable there.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: treat every logarithm as though it were the bare pattern $\ln(x)$ → you either drop the chain factor in $\ln(g(x))$ or apply the rule where the domain condition $x>0$ fails.

Debug: ask two questions before differentiating: “Is this exactly $\ln(x)$?” and “Does the current input satisfy $x>0$?”

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does the derivative formula for $\ln(x)$ naturally produce $\frac{1}{x}$, and why is that output itself sensitive to the domain where the logarithm exists?
• What does the condition $x>0$ protect here: the algebraic rewrite, the function’s existence, or both?

For the Principle

• What is the fastest two-part check you can run before using the derivative of $\ln(x)$ in a larger derivative computation?
• How does your decision process change when the expression is $\ln(g(x))$ instead of the bare pattern $\ln(x)$?

Between Principles

• How does the rule for differentiating $\ln(x)$ connect to the derivative chain rule when the logarithm’s input is not just $x$?

Generate an Example

• Describe a derivative step that looks like a direct $\ln(x)$ case but fails because the condition $x>0$ is not satisfied.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Differentiate the direct logarithmic form ln(x) by replacing it with 1/x, provided x>0.

Write the canonical equation: _____$\frac{d}{dx}\ln(x) = \frac{1}{x}$

State the canonical condition: _____$x>0$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $\frac{d}{dx}\bigl(\ln(x) + \ln(x^2+1)\bigr)$, differentiate assuming $x>0$.

Step	Expression	Operation
0	$\frac{d}{dx}\bigl(\ln(x) + \ln(x^2+1)\bigr)$	—
1	$\frac{d}{dx}\ln(x) + \frac{d}{dx}\ln(x^2+1)$	Sum rule
2	$\frac{1}{x} + \frac{d}{dx}\ln(x^2+1)$	Direct $\ln(x)$ rule on the first term because the condition $x>0$ is assumed
3	$\frac{1}{x} + \frac{1}{x^2+1}\cdot\frac{d}{dx}(x^2+1)$	Chain rule on the composite logarithm
4	$\frac{1}{x} + \frac{1}{x^2+1}\cdot 2x$	Power rule and constant rule inside the composite term
5	$\frac{1}{x} + \frac{2x}{x^2+1}$	Simplify

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Drills

Forward step (Format A)

Apply the rule once. Assume $x>0$.

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

Reveal

This is the direct canonical pattern, so the rule applies immediately:

$\frac{d}{dx}\ln(x) = \frac{1}{x}$

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Does the direct $\ln(x)$ rule apply to the whole expression below? Assume $5x-1>0$.

$\frac{d}{dx}\ln(5x-1)$

Reveal

No. The logarithm is defined here because $5x-1>0$, but the local pattern is not the bare $\ln(x)$ form. It is a composite logarithm, so chain rule governs the step:

$\frac{d}{dx}\ln(5x-1) = \frac{1}{5x-1}\cdot 5 = \frac{5}{5x-1}$

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Differentiate this expression. Assume $x>0$.

$\frac{d}{dx}\bigl(5\ln(x)\bigr)$

Reveal

Use the constant multiple rule, then the direct $\ln(x)$ rule:

$\frac{d}{dx}\bigl(5\ln(x)\bigr) = 5\,\frac{d}{dx}\ln(x) = \frac{5}{x}$

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Which of the following derivative steps use the direct $\ln(x)$ rule as written? Assume every logarithm shown is defined.

$\frac{d}{dx}\ln(x) \qquad \frac{d}{dx}\ln(5x-1) \qquad \frac{d}{dx}\ln|x|$

Reveal

Only $\dfrac{d}{dx}\ln(x)$ uses the direct rule as written.

• $\dfrac{d}{dx}\ln(x)$: yes — direct pattern, with condition $x>0$
• $\dfrac{d}{dx}\ln(5x-1)$: no — composite logarithm, so chain rule is required
• $\dfrac{d}{dx}\ln|x|$: no — different function and different domain statement

This is the essential move-selection check: direct pattern versus neighboring log structures.

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Differentiate this expression. Assume $x>0$.

$\frac{d}{dx}\bigl(\ln(x) + x^3\bigr)$

Reveal

Differentiate term by term:

$\frac{d}{dx}\bigl(\ln(x) + x^3\bigr) = \frac{1}{x} + 3x^2$

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A student writes $\dfrac{d}{dx}\ln(x^2) = \dfrac{1}{x^2}$. What is wrong with that move?

Reveal

The step treats $\ln(x^2)$ as though it were the direct $\ln(x)$ pattern. It is not. The input is $g(x)=x^2$, so chain rule is required:

$\frac{d}{dx}\ln(x^2) = \frac{1}{x^2}\cdot 2x = \frac{2}{x}$

for $x \neq 0$. The tempting move misses the inner derivative and misidentifies the local structure.

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Action label (Format B)

What was done between these two steps? Assume $x>0$.

$\frac{d}{dx}\ln(x) \quad\longrightarrow\quad \frac{1}{x}$

Reveal

The direct $\ln(x)$ rule was applied. The move is valid because the canonical condition $x>0$ is assumed.

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What was done between these two steps? Assume $x>0$.

$\frac{d}{dx}\bigl(7\ln(x)\bigr) \quad\longrightarrow\quad \frac{7}{x}$

Reveal

Constant multiple rule preserved the factor $7$, then the direct $\ln(x)$ rule turned $\frac{d}{dx}\ln(x)$ into $\frac{1}{x}$.

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A student writes $\dfrac{d}{dx}\ln(3x+2) \longrightarrow \dfrac{1}{3x+2}$. What error occurred?

Reveal

The student used the direct $\ln(x)$ rule on a composite logarithm and dropped the inner derivative. The logarithm may be defined, but the local structure is $\ln(g(x))$, not bare $\ln(x)$. The correct derivative is

$\frac{d}{dx}\ln(3x+2) = \frac{1}{3x+2}\cdot 3 = \frac{3}{3x+2}$

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Name the rule used on each term in this derivative. Assume $x>0$.

$\frac{d}{dx}\bigl(\ln(x) + e^x\bigr) \quad\longrightarrow\quad \frac{1}{x} + e^x$

Reveal

• $\ln(x)$ term: the direct rule for differentiating $\ln(x)$
• $e^x$ term: derivative of $e^x$

The whole derivative also uses the sum rule to split the expression into two local derivative steps.

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Transition identification (Format C)

Which transition uses the direct $\ln(x)$ rule, and which one uses chain rule? Assume $x>0$.

$\frac{d}{dx}\bigl(\ln(x) + \ln(x^2+1)\bigr) \xrightarrow{(1)} \frac{d}{dx}\ln(x) + \frac{d}{dx}\ln(x^2+1) \xrightarrow{(2)} \frac{1}{x} + \frac{d}{dx}\ln(x^2+1) \xrightarrow{(3)} \frac{1}{x} + \frac{1}{x^2+1}\cdot\frac{d}{dx}(x^2+1) \xrightarrow{(4)} \frac{1}{x} + \frac{2x}{x^2+1}$

Reveal

Transition (2) uses the direct $\ln(x)$ rule. It changes $\frac{d}{dx}\ln(x)$ into $\frac{1}{x}$.

• Transition (1): sum rule
• Transition (2): direct $\ln(x)$ rule
• Transition (3): chain rule on the composite logarithm
• Transition (4): simplify after differentiating the inner polynomial

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In the chain below, which transition is invalid, and why?

$\frac{d}{dx}\ln(x^2+1) \xrightarrow{(1)} \frac{1}{x^2+1} \xrightarrow{(2)} \frac{1}{x^2+1}\cdot 2x$

Reveal

Transition (1) is invalid. It applies the direct $\ln(x)$ rule to a composite logarithm and skips the inner derivative.

The correct move is to use chain rule first, which is what Transition (2) starts to supply:

$\frac{d}{dx}\ln(x^2+1) = \frac{1}{x^2+1}\cdot 2x$

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Differentiate $f(x) = 2\ln(x) + \ln(x^2+4)$ for $x>0$.

Full solution

Step	Expression	Move
0	$\frac{d}{dx}\bigl(2\ln(x) + \ln(x^2+4)\bigr)$	—
1	$2\,\frac{d}{dx}\ln(x) + \frac{d}{dx}\ln(x^2+4)$	Sum rule and constant multiple rule
2	$2\cdot\frac{1}{x} + \frac{d}{dx}\ln(x^2+4)$	Direct $\ln(x)$ rule on the first term
3	$2\cdot\frac{1}{x} + \frac{1}{x^2+4}\cdot\frac{d}{dx}(x^2+4)$	Chain rule on the second logarithm
4	$2\cdot\frac{1}{x} + \frac{1}{x^2+4}\cdot 2x$	Power rule and constant rule inside the composite term
5	$\frac{2}{x} + \frac{2x}{x^2+4}$	Simplify

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FAQ

What is the derivative of $\ln(x)$?

The derivative of $\ln(x)$ is $\frac{1}{x}$, but the real-valued rule is used on the domain $x>0$.

Why does the condition say $x>0$?

Because the real-valued natural logarithm is defined only for positive inputs. The derivative rule inherits that domain guard instead of bypassing it.

Can I use this rule at $x=0$?

No. $\ln(0)$ is not defined as a real number, so this rule is not applicable there.

What if the expression is $\ln(g(x))$ instead of $\ln(x)$?

Then chain rule governs the step: $\frac{d}{dx}\ln(g(x)) = \frac{g'(x)}{g(x)}$ on intervals where $g(x)>0$. This rule covers only the bare pattern $\ln(x)$.

How is the derivative of $\ln(x)$ related to the derivative of $e^x$?

The functions $\ln(x)$ and $e^x$ are inverse functions on their real domains, so derivative of $e^x$ and the derivative of $\ln(x)$ sit next to each other in the calculus derivative family.

What is the derivative of $\ln|x|$?

For $x \neq 0$, the derivative is also $\frac{1}{x}$. But $\ln|x|$ is a different function from $\ln(x)$, so do not treat it as the same canonical direct pattern.

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How This Fits in Unisium

Within the calculus subdomain, the derivative of $\ln(x)$ is a compact example of why transformational fluency is more than memorizing a formula. You need the pattern, domain guard, and structure check all at once: bare $\ln(x)$ uses this rule, $\ln(g(x))$ calls for the derivative chain rule, and mixed expressions still depend on habits built through retrieval practice and self-explanation.

Explore further:

• Calculus Subdomain Map — Return to the calculus hub to see where the natural logarithm rule sits beside the other derivative families
• Derivative chain rule — extend the logarithm derivative to nested inputs
• Derivative of $e^x$ — compare the paired exponential rule
• Principle Structures — see where this rule lives in the calculus derivative map

Ready to practice it in context? See Unisium’s 14-day free trial or explore the full framework in Masterful Learning.

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