Unit 3Derivatives of General Exponential and Logarithmic Functions

Concept 4.77Built for May 2027 AP Exam

Derivatives of General Exponential and Logarithmic Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A general exponential brings its base's logarithm as a factor. So $\frac{d}{dx}\left[a^x\right] = a^x\ln a$ , and the corresponding logarithm row is $\frac{d}{dx}\left[\log_a x\right] = \frac{1}{x\ln a}$ . The reason is a rewriting: $a^x = e^{x\ln a}$ , which converts the problem into the base- $e$ rule with a constant inner slope. The base $e$ is simply the case where $\ln e = 1$ , so the factor disappears and the familiar rule returns. Almost every error here is a missing or misplaced logarithm, not a misunderstanding of the exponential itself. The exam version usually hides the general base inside a product, so the logarithm factor has to survive that product rule as well. Handled that way, this topic is one extra factor rather than one extra rule.

On the AP Exam

These derivatives appear as single terms in larger expressions, usually inside a product or a sum where the base is deliberately not $e$ .

What you’ll see: $2^x$ , $5^{3x}$ , $\log_3 x$ , or a rewritten form such as $x \cdot 2^x$ that combines a polynomial with a general exponential.
What you’ll do: Attach $\ln a$ as a factor, and for a logarithm put $x\ln a$ in the denominator.
Memory hook: Every base carries its own logarithm. If the base is not $e$ , the logarithm is not $1$ .
Use it on the exam: For MCQs: The distractor without $\ln a$ is present in every version, and it is the option that looks like the base- $e$ answer. For FRQs: Writing $a^x = e^{x\ln a}$ first is a legitimate justification and often earns the setup point even when the final simplification is imperfect.

Concept Review

The Logarithm Arrives as a Factor

Figure 1 — Changing the base changes one thing: the logarithm of that base becomes a multiplicative factor. The base e is the case where that factor equals one and vanishes.

ANY BASE BRINGS ITS OWN LOGARITHM

The Four Forms Worth Knowing

Function	Derivative	Why
$e^x$	$e^x$	$\ln e = 1$ , so the factor disappears
$a^x$	$a^x\ln a$	Rewrite as $e^{x\ln a}$
$\ln x$	$\dfrac{1}{x}$	The base- $e$ logarithm
$\log_a x$	$\dfrac{1}{x\ln a}$	Change of base brings $\ln a$ into the denominator

The two rows on the right are mirror images of each other, and that symmetry is worth noticing: an exponential multiplies by $\ln a$ , while a logarithm divides by it.

Why the Rewriting Works

Starting from $a^x = e^{x\ln a}$ , the chain rule gives $e^{x\ln a} \cdot \ln a$ , which is $a^x \ln a$ . The inner slope is the constant $\ln a$ , so no variable factor appears. The same rewriting, run in reverse, produces the logarithm rule through the inverse-function formula.

Common Misconception Check

Myth 1: " $\frac{d}{dx}\left[2^x\right] = x2^{x-1}$ ." Truth: The power rule requires a constant exponent. Here the exponent is the variable, so the answer is $2^x\ln 2$ .
Myth 2: " $\frac{d}{dx}\left[\log_3 x\right] = \frac{1}{x}$ ." Truth: The base contributes $\ln 3$ , so the answer is $\frac{1}{x\ln 3}$ , which is smaller than $\frac{1}{x}$ because $\ln 3 > 1$ .
Myth 3: "The logarithm factor goes in the numerator for logarithms too." Truth: For $\log_a x$ the factor sits in the denominator, which is the structural difference between the two rows.

Topic-Specific Study Method

Rewrite, Then Differentiate

Borrowed idea: What changes when the base of an exponential is no longer $e$ ? One factor appears, and that factor is the natural logarithm of the base. So $\frac{d}{dx}\left[a^x\right] = a^x\ln a$ , and the logarithm row is its mirror image. Why does the logarithm of the base show up at all? Because $a^x = e^{x\ln a}$ , which turns the base into a chain-rule problem with a constant inner slope. The base $e$ is simply the case where that logarithm equals $1$ and the factor disappears.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A general exponential	$2^x$	Base $2$ , so $\ln 2$ appears	Multiply by $\ln 2$
A scaled exponent	$5^{3x}$	Two factors: $\ln 5$ and the inner $3$	Multiply by $3\ln 5$
A general logarithm	$\log_3 x$	The base belongs in the denominator	Write $\frac{1}{x\ln 3}$
A product with a general base	$x \cdot 2^x$	Product rule, then the exponential rule	Two terms, one carrying $\ln 2$
A quotient with a general base	$\frac{2^x}{x}$	Quotient rule, then the exponential rule	Watch the sign of the second term

① Rewrite $2^x$ , $3^x$ , and $10^x$ as base- $e$ expressions on one line, then differentiate each with the chain rule. Seeing the logarithm emerge from the rewriting is what makes the formula memorable.

② Take the four rows of the table and cover the right-hand column. Recite each derivative aloud, including the factor. The factor is the whole content of this topic.

③ On any product that mixes a polynomial with a general exponential, apply the product rule first and only then attach the logarithm factor. Attaching it to the wrong term is the most common structural error.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\frac{d}{dx}\left[5^{3x}\right]$ .

Standard Execution:

Identify the base. The base is $5$ , so the logarithm is $\ln 5$ .

Identify the inner slope. The exponent is $3x$ , with slope $3$ .

Multiply everything. The derivative is $5^{3x} \cdot \ln 5 \cdot 3$ .

Order the factors. Conventionally $3\ln 5 \cdot 5^{3x}$ .

Stop. No further simplification exists.

Exam Variation (Context / Parameter Shift): Let $f(x) = x \cdot 2^x$ . Find $f'(x)$ and $f'(1)$ .

Solution: The product rule gives $f'(x) = 1 \cdot 2^x + x \cdot 2^x\ln 2 = 2^x\left(1 + x\ln 2\right)$ . At $x = 1$ this is $2(1 + \ln 2) \approx 3.386$ . Factoring out $2^x$ is optional, but the logarithm must appear exactly once in the second term.

Common Traps & Failure Signals:

The power-rule reflex (Signal: an answer of the form $x2^{x-1}$ ): a variable exponent was treated as a constant one.
The missing logarithm (Signal: an answer identical to the base- $e$ result): the base was differentiated as though it were $e$ .
The misplaced factor in a logarithm (Signal: $\frac{\ln 3}{x}$ ): the factor belongs in the denominator.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Base Blindness $\rightarrow$ Read the base before choosing a rule.
Power Reflex: You have used $nx^{n-1}$ for years, so the pattern fires before the exponent is inspected, and the answer comes out with the variable in the exponent treated as a constant.
Factor Placement: You kept the logarithm but put it in the wrong position.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Six of the nine released exams score a derivative that passes through a logarithm, and one of them uses a base other than $e$
10-Year Total Count: 6 of 9 Exams · 6 FRQ Parts; Every one of these parts is a composite, so the logarithm rule and the chain rule are graded together
FRQ Years Appeared: 2017 (Q5a), 2021 (Q2c), 2022 (Q2b), 2024 (Q2b), 2025 (Q6D), 2026 (Q4A)

Practice Examples

MCQ 1

A General Exponential

What is the derivative of $f(x) = 2^x$ ?

MCQ 2

A General Logarithm

What is the derivative of $g(x) = \log_3 x$ ?

MCQ 3

A Scaled Exponent

What is the derivative of $h(x) = 5^{3x}$ ?

FRQ 4

Two Bases and a Tangent

Let $f(x) = 2^x$ and $g(x) = \log_2 x$ for $x > 0$ .

(a) Find $f'(x)$ .

(b) Find the equation of the line tangent to the graph of $f$ at $x = 1$ .

(c) Find $g'(x)$ and evaluate $g'(2)$ .

(d) Explain why $g'(x) = \frac{1}{f'(g(x))}$ . Use the two formulas you found to verify the identity at $x = 2$ .

Cheatsheet

General Exponential and Logarithmic Derivatives

	Derivative
$e^x$	$e^x$
$a^x$	$a^x\ln a$
$e^{u}$	$e^{u}u'$
$a^{u}$	$a^{u}\ln a \cdot u'$
$\ln x$	$\dfrac{1}{x}$
$\log_a x$	$\dfrac{1}{x\ln a}$

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Unit 3Derivatives of General Exponential and Logarithmic Functions

Concept 4.77Built for May 2027 AP Exam

Derivatives of General Exponential and Logarithmic Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

These derivatives appear as single terms in larger expressions, usually inside a product or a sum where the base is deliberately not $e$ .

What you’ll see: $2^x$ , $5^{3x}$ , $\log_3 x$ , or a rewritten form such as $x \cdot 2^x$ that combines a polynomial with a general exponential.
What you’ll do: Attach $\ln a$ as a factor, and for a logarithm put $x\ln a$ in the denominator.
Memory hook: Every base carries its own logarithm. If the base is not $e$ , the logarithm is not $1$ .
Use it on the exam: For MCQs: The distractor without $\ln a$ is present in every version, and it is the option that looks like the base- $e$ answer. For FRQs: Writing $a^x = e^{x\ln a}$ first is a legitimate justification and often earns the setup point even when the final simplification is imperfect.

Concept Review

The Logarithm Arrives as a Factor

Figure 1 — Changing the base changes one thing: the logarithm of that base becomes a multiplicative factor. The base e is the case where that factor equals one and vanishes.

ANY BASE BRINGS ITS OWN LOGARITHM

The Four Forms Worth Knowing

Function	Derivative	Why
$e^x$	$e^x$	$\ln e = 1$ , so the factor disappears
$a^x$	$a^x\ln a$	Rewrite as $e^{x\ln a}$
$\ln x$	$\dfrac{1}{x}$	The base- $e$ logarithm
$\log_a x$	$\dfrac{1}{x\ln a}$	Change of base brings $\ln a$ into the denominator

The two rows on the right are mirror images of each other, and that symmetry is worth noticing: an exponential multiplies by $\ln a$ , while a logarithm divides by it.

Why the Rewriting Works

Common Misconception Check

Myth 1: " $\frac{d}{dx}\left[2^x\right] = x2^{x-1}$ ." Truth: The power rule requires a constant exponent. Here the exponent is the variable, so the answer is $2^x\ln 2$ .
Myth 2: " $\frac{d}{dx}\left[\log_3 x\right] = \frac{1}{x}$ ." Truth: The base contributes $\ln 3$ , so the answer is $\frac{1}{x\ln 3}$ , which is smaller than $\frac{1}{x}$ because $\ln 3 > 1$ .
Myth 3: "The logarithm factor goes in the numerator for logarithms too." Truth: For $\log_a x$ the factor sits in the denominator, which is the structural difference between the two rows.

Topic-Specific Study Method

Rewrite, Then Differentiate

Exam disguise	What you see on the page	What your brain must translate	Required execution
A general exponential	$2^x$	Base $2$ , so $\ln 2$ appears	Multiply by $\ln 2$
A scaled exponent	$5^{3x}$	Two factors: $\ln 5$ and the inner $3$	Multiply by $3\ln 5$
A general logarithm	$\log_3 x$	The base belongs in the denominator	Write $\frac{1}{x\ln 3}$
A product with a general base	$x \cdot 2^x$	Product rule, then the exponential rule	Two terms, one carrying $\ln 2$
A quotient with a general base	$\frac{2^x}{x}$	Quotient rule, then the exponential rule	Watch the sign of the second term

② Take the four rows of the table and cover the right-hand column. Recite each derivative aloud, including the factor. The factor is the whole content of this topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\frac{d}{dx}\left[5^{3x}\right]$ .

Standard Execution:

Identify the base. The base is $5$ , so the logarithm is $\ln 5$ .

Identify the inner slope. The exponent is $3x$ , with slope $3$ .

Multiply everything. The derivative is $5^{3x} \cdot \ln 5 \cdot 3$ .

Order the factors. Conventionally $3\ln 5 \cdot 5^{3x}$ .

Stop. No further simplification exists.

Exam Variation (Context / Parameter Shift): Let $f(x) = x \cdot 2^x$ . Find $f'(x)$ and $f'(1)$ .

Solution: The product rule gives $f'(x) = 1 \cdot 2^x + x \cdot 2^x\ln 2 = 2^x\left(1 + x\ln 2\right)$ . At $x = 1$ this is $2(1 + \ln 2) \approx 3.386$ . Factoring out $2^x$ is optional, but the logarithm must appear exactly once in the second term.

Common Traps & Failure Signals:

The power-rule reflex (Signal: an answer of the form $x2^{x-1}$ ): a variable exponent was treated as a constant one.
The missing logarithm (Signal: an answer identical to the base- $e$ result): the base was differentiated as though it were $e$ .
The misplaced factor in a logarithm (Signal: $\frac{\ln 3}{x}$ ): the factor belongs in the denominator.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Base Blindness $\rightarrow$ Read the base before choosing a rule.
Power Reflex: You have used $nx^{n-1}$ for years, so the pattern fires before the exponent is inspected, and the answer comes out with the variable in the exponent treated as a constant.
Factor Placement: You kept the logarithm but put it in the wrong position.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Six of the nine released exams score a derivative that passes through a logarithm, and one of them uses a base other than $e$
10-Year Total Count: 6 of 9 Exams · 6 FRQ Parts; Every one of these parts is a composite, so the logarithm rule and the chain rule are graded together
FRQ Years Appeared: 2017 (Q5a), 2021 (Q2c), 2022 (Q2b), 2024 (Q2b), 2025 (Q6D), 2026 (Q4A)

Practice Examples

MCQ 1

A General Exponential

What is the derivative of $f(x) = 2^x$ ?

MCQ 2

A General Logarithm

What is the derivative of $g(x) = \log_3 x$ ?

MCQ 3

A Scaled Exponent

What is the derivative of $h(x) = 5^{3x}$ ?

FRQ 4

Two Bases and a Tangent

Let $f(x) = 2^x$ and $g(x) = \log_2 x$ for $x > 0$ .

(a) Find $f'(x)$ .

(b) Find the equation of the line tangent to the graph of $f$ at $x = 1$ .

(c) Find $g'(x)$ and evaluate $g'(2)$ .

(d) Explain why $g'(x) = \frac{1}{f'(g(x))}$ . Use the two formulas you found to verify the identity at $x = 2$ .

Cheatsheet

General Exponential and Logarithmic Derivatives

	Derivative
$e^x$	$e^x$
$a^x$	$a^x\ln a$
$e^{u}$	$e^{u}u'$
$a^{u}$	$a^{u}\ln a \cdot u'$
$\ln x$	$\dfrac{1}{x}$
$\log_a x$	$\dfrac{1}{x\ln a}$

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Unit 3Derivatives of General Exponential and Logarithmic Functions

Concept 4.77Built for May 2027 AP Exam

Derivatives of General Exponential and Logarithmic Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

These derivatives appear as single terms in larger expressions, usually inside a product or a sum where the base is deliberately not $e$ .

What you’ll see: $2^x$ , $5^{3x}$ , $\log_3 x$ , or a rewritten form such as $x \cdot 2^x$ that combines a polynomial with a general exponential.
What you’ll do: Attach $\ln a$ as a factor, and for a logarithm put $x\ln a$ in the denominator.
Memory hook: Every base carries its own logarithm. If the base is not $e$ , the logarithm is not $1$ .
Use it on the exam: For MCQs: The distractor without $\ln a$ is present in every version, and it is the option that looks like the base- $e$ answer. For FRQs: Writing $a^x = e^{x\ln a}$ first is a legitimate justification and often earns the setup point even when the final simplification is imperfect.

Concept Review

The Logarithm Arrives as a Factor

Figure 1 — Changing the base changes one thing: the logarithm of that base becomes a multiplicative factor. The base e is the case where that factor equals one and vanishes.

ANY BASE BRINGS ITS OWN LOGARITHM

The Four Forms Worth Knowing

Function	Derivative	Why
$e^x$	$e^x$	$\ln e = 1$ , so the factor disappears
$a^x$	$a^x\ln a$	Rewrite as $e^{x\ln a}$
$\ln x$	$\dfrac{1}{x}$	The base- $e$ logarithm
$\log_a x$	$\dfrac{1}{x\ln a}$	Change of base brings $\ln a$ into the denominator

The two rows on the right are mirror images of each other, and that symmetry is worth noticing: an exponential multiplies by $\ln a$ , while a logarithm divides by it.

Why the Rewriting Works

Common Misconception Check

Myth 1: " $\frac{d}{dx}\left[2^x\right] = x2^{x-1}$ ." Truth: The power rule requires a constant exponent. Here the exponent is the variable, so the answer is $2^x\ln 2$ .
Myth 2: " $\frac{d}{dx}\left[\log_3 x\right] = \frac{1}{x}$ ." Truth: The base contributes $\ln 3$ , so the answer is $\frac{1}{x\ln 3}$ , which is smaller than $\frac{1}{x}$ because $\ln 3 > 1$ .
Myth 3: "The logarithm factor goes in the numerator for logarithms too." Truth: For $\log_a x$ the factor sits in the denominator, which is the structural difference between the two rows.

Topic-Specific Study Method

Rewrite, Then Differentiate

Exam disguise	What you see on the page	What your brain must translate	Required execution
A general exponential	$2^x$	Base $2$ , so $\ln 2$ appears	Multiply by $\ln 2$
A scaled exponent	$5^{3x}$	Two factors: $\ln 5$ and the inner $3$	Multiply by $3\ln 5$
A general logarithm	$\log_3 x$	The base belongs in the denominator	Write $\frac{1}{x\ln 3}$
A product with a general base	$x \cdot 2^x$	Product rule, then the exponential rule	Two terms, one carrying $\ln 2$
A quotient with a general base	$\frac{2^x}{x}$	Quotient rule, then the exponential rule	Watch the sign of the second term

② Take the four rows of the table and cover the right-hand column. Recite each derivative aloud, including the factor. The factor is the whole content of this topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\frac{d}{dx}\left[5^{3x}\right]$ .

Standard Execution:

Identify the base. The base is $5$ , so the logarithm is $\ln 5$ .

Identify the inner slope. The exponent is $3x$ , with slope $3$ .

Multiply everything. The derivative is $5^{3x} \cdot \ln 5 \cdot 3$ .

Order the factors. Conventionally $3\ln 5 \cdot 5^{3x}$ .

Stop. No further simplification exists.

Exam Variation (Context / Parameter Shift): Let $f(x) = x \cdot 2^x$ . Find $f'(x)$ and $f'(1)$ .

Solution: The product rule gives $f'(x) = 1 \cdot 2^x + x \cdot 2^x\ln 2 = 2^x\left(1 + x\ln 2\right)$ . At $x = 1$ this is $2(1 + \ln 2) \approx 3.386$ . Factoring out $2^x$ is optional, but the logarithm must appear exactly once in the second term.

Common Traps & Failure Signals:

The power-rule reflex (Signal: an answer of the form $x2^{x-1}$ ): a variable exponent was treated as a constant one.
The missing logarithm (Signal: an answer identical to the base- $e$ result): the base was differentiated as though it were $e$ .
The misplaced factor in a logarithm (Signal: $\frac{\ln 3}{x}$ ): the factor belongs in the denominator.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Base Blindness $\rightarrow$ Read the base before choosing a rule.
Power Reflex: You have used $nx^{n-1}$ for years, so the pattern fires before the exponent is inspected, and the answer comes out with the variable in the exponent treated as a constant.
Factor Placement: You kept the logarithm but put it in the wrong position.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Six of the nine released exams score a derivative that passes through a logarithm, and one of them uses a base other than $e$
10-Year Total Count: 6 of 9 Exams · 6 FRQ Parts; Every one of these parts is a composite, so the logarithm rule and the chain rule are graded together
FRQ Years Appeared: 2017 (Q5a), 2021 (Q2c), 2022 (Q2b), 2024 (Q2b), 2025 (Q6D), 2026 (Q4A)

Practice Examples

MCQ 1

A General Exponential

What is the derivative of $f(x) = 2^x$ ?

MCQ 2

A General Logarithm

What is the derivative of $g(x) = \log_3 x$ ?

MCQ 3

A Scaled Exponent

What is the derivative of $h(x) = 5^{3x}$ ?

FRQ 4

Two Bases and a Tangent

Let $f(x) = 2^x$ and $g(x) = \log_2 x$ for $x > 0$ .

(a) Find $f'(x)$ .

(b) Find the equation of the line tangent to the graph of $f$ at $x = 1$ .

(c) Find $g'(x)$ and evaluate $g'(2)$ .

(d) Explain why $g'(x) = \frac{1}{f'(g(x))}$ . Use the two formulas you found to verify the identity at $x = 2$ .

Cheatsheet

General Exponential and Logarithmic Derivatives

	Derivative
$e^x$	$e^x$
$a^x$	$a^x\ln a$
$e^{u}$	$e^{u}u'$
$a^{u}$	$a^{u}\ln a \cdot u'$
$\ln x$	$\dfrac{1}{x}$
$\log_a x$	$\dfrac{1}{x\ln a}$

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