Unit 3Differentiation of

f(g(x))

Concept 4.184Built for May 2027 AP Exam

Differentiation of $f(g(x))$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Knowing the chain rule formula is not the same as executing it on a function written in a form you did not choose. Execution means three moves: decide what the outermost operation is, differentiate it while the inside block stays frozen, and multiply by the derivative of that block. The block itself may be linear, a power, a product, or another composition. On the exam the inner block is usually not a single variable, and that is exactly where answers collapse: the outer derivative comes out clean and the trailing factor is missing. This section drills the move until the trailing factor appears without being remembered.

On the AP Exam

What you’ll see: A composite written in a form you did not choose: a root of a quadratic, a trig function of a product, an exponential of a quotient, a log of a power.
What you’ll do: Bracket the inside, differentiate outside, close the bracket, then multiply by the inside rate.
Memory hook: Bracket it, peel it, multiply by it. The three verbs match the three moves.
Use it on the exam: For MCQs: Check the exponent and the coefficient separately. Most wrong options get one of them right and the other wrong, so reading only the shape of the answer is usually enough. For FRQs: Write the inner rate on its own line before simplifying. Graders award the chain point on the presence of that factor, and a derivative that collapses it into a single number can cost the point even when the value is right.

Concept Review

Three Moves, Always in This Order

Figure 2 — Three numbered steps, then three compositions carried through them. The middle column keeps the inside exactly as it was; the right-hand column is the factor that most answers drop.

The notation $f'(g(x))$ is where students lose the thread. It does not mean "the derivative of $f$ , composed with the derivative of $g$ ." It means "the derivative of $f$ , evaluated at the number $g(x)$ ." For $f(u) = u^5$ and $g(x) = 3x+1$ , the value $f'(g(x))$ is $5(3x+1)^4$ , a number that depends on $x$ through the inner function only.

Three numbered steps for the chain rule and a table of three compositions showing the outer derivative, the untouched inside, the inside rate and the product

The Four Shapes That Cover Most Parts

Shape	Example	Outer derivative	Inner rate
Power	$(2x-7)^4$	$4(2x-7)^3$	$2$
Root	$\sqrt{x^2+9}$	$\frac{1}{2\sqrt{x^2+9}}$	$2x$
Trigonometric	$\cos(1-3x)$	$-\sin(1-3x)$	$-3$
Exponential or log	$e^{5x}$ , $\ln(x^2+1)$	$e^{5x}$ , $\frac{1}{x^2+1}$	$5$ , $2x$

Every row is the same sentence with different nouns. If you can fill this table from memory, the rest of the section is bookkeeping.

When the Inside Is Not Linear

A linear inside contributes a constant, which is why so many pupils believe the chain rule is optional. A non-linear inside contributes a variable factor, and then the omission is visible:

$\frac{d}{dx}\left(x^2+1\right)^3 = 3(x^2+1)^2 \cdot 2x = 6x(x^2+1)^2$

Dropping the $2x$ here does not shift the answer by a constant; it changes the degree of the result. On a multiple-choice item that is easy to spot. On a free-response item it shows up as a wrong derivative at the substitution step, several lines later.

Compositions Inside Compositions

For $y = \sqrt{\sin(2x)}$ the layers are: square root, sine, and $2x$ . Peeling gives

$\frac{dy}{dx} = \frac{1}{2\sqrt{\sin(2x)}} \cdot \cos(2x) \cdot 2 = \frac{\cos(2x)}{\sqrt{\sin(2x)}}$

Three layers, three factors, and the last one is a constant. The habit that prevents errors is counting: name three layers, write three factors, then simplify.

Common Misconception Check

Myth 1: "Once the outer derivative is written, the problem is done." Truth: The outer derivative is the shape of the answer, not the answer. The size comes from the inner rate.
Myth 2: "A constant inner rate can be ignored because it is just a number." Truth: That constant is the difference between $\sin(3x)$ and $3\sin(3x)$ as derivatives. It is exactly the factor the exam tests.
Myth 3: "The chain rule only applies to powers." Truth: It applies to every composition, including the ones that look like plain functions because the argument hides a coefficient.

Topic-Specific Study Method

The Bracket Habit

Borrowed idea: Put the inside in brackets before touching the derivative. Why does that help? Because the bracket turns an invisible layer into a visible object you can point at. Differentiate the outside, copy the bracket back unchanged, then multiply by the derivative of what is inside. How many factors should the result have? Exactly as many as there are layers, and no more. If the answer looks short, a factor was merged away before you could check it.

Given	Bracketed inside	Derivative
$\sin(x^2)$	$x^2$	$2x\cos(x^2)$
$(5-2x)^7$	$5-2x$	$-14(5-2x)^6$
$e^{\sqrt{x}}$	$\sqrt{x}$	$\frac{e^{\sqrt{x}}}{2\sqrt{x}}$
$\ln(3x+4)$	$3x+4$	$\frac{3}{3x+4}$

① Differentiate five composites and, for each, underline the trailing factor before simplifying. The underlining is the whole exercise.

② Take one composite and differentiate it with the inside named $u$ and again without naming it. Both routes should end at the same expression.

③ Write a composite whose inner rate is $-1$ and differentiate it. The sign is the factor students drop most often when the inside is something like $1-x$ .

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $y = \sqrt{4x+9}$ .

Standard Execution:

Rewrite. $y = (4x+9)^{1/2}$ .

Outside. $\frac{1}{2}(4x+9)^{-1/2}$ .

Inside rate. Multiply by 4.

Simplify. $\frac{2}{\sqrt{4x+9}}$ .

Exam Variation (Context / Parameter Shift): The temperature of a sample is modeled by $T(t) = 80 - 60e^{-0.05t}$ degrees Celsius, with $t$ in minutes. Find $T'(t)$ and interpret the sign at $t = 0$ .

Solution: The outer derivative of $-60e^{u}$ is $-60e^{u}$ and the inner rate of $-0.05t$ is $-0.05$ , so $T'(t) = 3e^{-0.05t}$ . At $t = 0$ this is $3$ degrees per minute, positive because the sample is warming, and the rate decays as the sample approaches room temperature.

Common Traps & Failure Signals:

The frozen inside (Signal: the bracket was differentiated but never multiplied): the outer derivative was treated as the whole answer.
The vanished negative (Signal: a plus where the inside decreases): the inner rate of a decreasing argument was taken as positive.
The merged factor (Signal: a number that absorbed the coefficient): simplification ran ahead of checking, so the chain factor cannot be inspected.

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

Factor Dropped: You wrote the outer derivative and closed the line.
Bracket Lost $\rightarrow$ Rewrite the inside in brackets before differentiating anything.
Evaluation Confused: You read $f'(g(x))$ as the product of two derivatives rather than the outer derivative evaluated at the inner function, which reverses the structure of the answer even when the pieces are right.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; A released part asks you to differentiate an explicitly written composite in five of the nine sets, and the inner function is rarely linear
10-Year Total Count: 5 of 9 Exams · 6 FRQ Parts; Counted where the composite is written out and its derivative is scored, not where a composition only appears inside a supplied rate
FRQ Years Appeared: 2018 (Q2a), 2021 (Q2a), 2022 (Q4d), 2023 (Q1d, Q2a), 2026 (Q4A)

Practice Examples

MCQ 1

A Square Root

If $y = \sqrt{6x-5}$ , what is $\frac{dy}{dx}$ ?

MCQ 2

A Negative Inner Rate

If $g(x) = \cos(2-5x)$ , what is $g'(x)$ ?

MCQ 3

A Log of a Power

If $h(x) = \ln(x^2+4)$ , what is $h'(x)$ ?

FRQ 1 • A Composed Model

The volume of a balloon is modeled by $V(t) = \frac{4}{3}\pi(2+0.5t)^3$ cubic centimeters, where $t$ is measured in seconds and $0 \le t \le 6$ .

(a) Find $V'(t)$ , leaving the bracket intact.

(b) Find the rate of change of the volume at $t = 4$ seconds, with units.

(c) Explain why $V'(t)$ is increasing on the interval.

(d) A classmate writes $V'(t) = 4\pi(2+0.5t)^2$ and calls the missing factor a rounding detail. Explain what that omission does to the reported rate at $t = 4$ .

Cheatsheet

Differentiating $f(g(x))$

	Move	Watch for
Linear inside	multiply by the slope	the constant factor
Power inside	multiply by the power's derivative	degree of the answer
Negative inside	multiply by the negative	sign of the result
Nested layers	one factor per layer	factor count
Trig or log inside	keep the argument intact	argument never changes

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 3.3 Implicit Differentiation Technique

Unit 3Differentiation of

f(g(x))

Concept 4.184Built for May 2027 AP Exam

Differentiation of $f(g(x))$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A composite written in a form you did not choose: a root of a quadratic, a trig function of a product, an exponential of a quotient, a log of a power.
What you’ll do: Bracket the inside, differentiate outside, close the bracket, then multiply by the inside rate.
Memory hook: Bracket it, peel it, multiply by it. The three verbs match the three moves.
Use it on the exam: For MCQs: Check the exponent and the coefficient separately. Most wrong options get one of them right and the other wrong, so reading only the shape of the answer is usually enough. For FRQs: Write the inner rate on its own line before simplifying. Graders award the chain point on the presence of that factor, and a derivative that collapses it into a single number can cost the point even when the value is right.

Concept Review

Three Moves, Always in This Order

Figure 2 — Three numbered steps, then three compositions carried through them. The middle column keeps the inside exactly as it was; the right-hand column is the factor that most answers drop.

Three numbered steps for the chain rule and a table of three compositions showing the outer derivative, the untouched inside, the inside rate and the product

The Four Shapes That Cover Most Parts

Shape	Example	Outer derivative	Inner rate
Power	$(2x-7)^4$	$4(2x-7)^3$	$2$
Root	$\sqrt{x^2+9}$	$\frac{1}{2\sqrt{x^2+9}}$	$2x$
Trigonometric	$\cos(1-3x)$	$-\sin(1-3x)$	$-3$
Exponential or log	$e^{5x}$ , $\ln(x^2+1)$	$e^{5x}$ , $\frac{1}{x^2+1}$	$5$ , $2x$

Every row is the same sentence with different nouns. If you can fill this table from memory, the rest of the section is bookkeeping.

When the Inside Is Not Linear

A linear inside contributes a constant, which is why so many pupils believe the chain rule is optional. A non-linear inside contributes a variable factor, and then the omission is visible:

$\frac{d}{dx}\left(x^2+1\right)^3 = 3(x^2+1)^2 \cdot 2x = 6x(x^2+1)^2$

Compositions Inside Compositions

For $y = \sqrt{\sin(2x)}$ the layers are: square root, sine, and $2x$ . Peeling gives

$\frac{dy}{dx} = \frac{1}{2\sqrt{\sin(2x)}} \cdot \cos(2x) \cdot 2 = \frac{\cos(2x)}{\sqrt{\sin(2x)}}$

Three layers, three factors, and the last one is a constant. The habit that prevents errors is counting: name three layers, write three factors, then simplify.

Common Misconception Check

Myth 1: "Once the outer derivative is written, the problem is done." Truth: The outer derivative is the shape of the answer, not the answer. The size comes from the inner rate.
Myth 2: "A constant inner rate can be ignored because it is just a number." Truth: That constant is the difference between $\sin(3x)$ and $3\sin(3x)$ as derivatives. It is exactly the factor the exam tests.
Myth 3: "The chain rule only applies to powers." Truth: It applies to every composition, including the ones that look like plain functions because the argument hides a coefficient.

Topic-Specific Study Method

The Bracket Habit

Given	Bracketed inside	Derivative
$\sin(x^2)$	$x^2$	$2x\cos(x^2)$
$(5-2x)^7$	$5-2x$	$-14(5-2x)^6$
$e^{\sqrt{x}}$	$\sqrt{x}$	$\frac{e^{\sqrt{x}}}{2\sqrt{x}}$
$\ln(3x+4)$	$3x+4$	$\frac{3}{3x+4}$

① Differentiate five composites and, for each, underline the trailing factor before simplifying. The underlining is the whole exercise.

② Take one composite and differentiate it with the inside named $u$ and again without naming it. Both routes should end at the same expression.

③ Write a composite whose inner rate is $-1$ and differentiate it. The sign is the factor students drop most often when the inside is something like $1-x$ .

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $y = \sqrt{4x+9}$ .

Standard Execution:

Rewrite. $y = (4x+9)^{1/2}$ .

Outside. $\frac{1}{2}(4x+9)^{-1/2}$ .

Inside rate. Multiply by 4.

Simplify. $\frac{2}{\sqrt{4x+9}}$ .

Solution: The outer derivative of $-60e^{u}$ is $-60e^{u}$ and the inner rate of $-0.05t$ is $-0.05$ , so $T'(t) = 3e^{-0.05t}$ . At $t = 0$ this is $3$ degrees per minute, positive because the sample is warming, and the rate decays as the sample approaches room temperature.

Common Traps & Failure Signals:

The frozen inside (Signal: the bracket was differentiated but never multiplied): the outer derivative was treated as the whole answer.
The vanished negative (Signal: a plus where the inside decreases): the inner rate of a decreasing argument was taken as positive.
The merged factor (Signal: a number that absorbed the coefficient): simplification ran ahead of checking, so the chain factor cannot be inspected.

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

Factor Dropped: You wrote the outer derivative and closed the line.
Bracket Lost $\rightarrow$ Rewrite the inside in brackets before differentiating anything.
Evaluation Confused: You read $f'(g(x))$ as the product of two derivatives rather than the outer derivative evaluated at the inner function, which reverses the structure of the answer even when the pieces are right.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; A released part asks you to differentiate an explicitly written composite in five of the nine sets, and the inner function is rarely linear
10-Year Total Count: 5 of 9 Exams · 6 FRQ Parts; Counted where the composite is written out and its derivative is scored, not where a composition only appears inside a supplied rate
FRQ Years Appeared: 2018 (Q2a), 2021 (Q2a), 2022 (Q4d), 2023 (Q1d, Q2a), 2026 (Q4A)

Practice Examples

MCQ 1

A Square Root

If $y = \sqrt{6x-5}$ , what is $\frac{dy}{dx}$ ?

MCQ 2

A Negative Inner Rate

If $g(x) = \cos(2-5x)$ , what is $g'(x)$ ?

MCQ 3

A Log of a Power

If $h(x) = \ln(x^2+4)$ , what is $h'(x)$ ?

FRQ 1 • A Composed Model

The volume of a balloon is modeled by $V(t) = \frac{4}{3}\pi(2+0.5t)^3$ cubic centimeters, where $t$ is measured in seconds and $0 \le t \le 6$ .

(a) Find $V'(t)$ , leaving the bracket intact.

(b) Find the rate of change of the volume at $t = 4$ seconds, with units.

(c) Explain why $V'(t)$ is increasing on the interval.

(d) A classmate writes $V'(t) = 4\pi(2+0.5t)^2$ and calls the missing factor a rounding detail. Explain what that omission does to the reported rate at $t = 4$ .

Cheatsheet

Differentiating $f(g(x))$

	Move	Watch for
Linear inside	multiply by the slope	the constant factor
Power inside	multiply by the power's derivative	degree of the answer
Negative inside	multiply by the negative	sign of the result
Nested layers	one factor per layer	factor count
Trig or log inside	keep the argument intact	argument never changes

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 3.3 Implicit Differentiation Technique

Unit 3Differentiation of

f(g(x))

Concept 4.184Built for May 2027 AP Exam

Differentiation of $f(g(x))$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A composite written in a form you did not choose: a root of a quadratic, a trig function of a product, an exponential of a quotient, a log of a power.
What you’ll do: Bracket the inside, differentiate outside, close the bracket, then multiply by the inside rate.
Memory hook: Bracket it, peel it, multiply by it. The three verbs match the three moves.
Use it on the exam: For MCQs: Check the exponent and the coefficient separately. Most wrong options get one of them right and the other wrong, so reading only the shape of the answer is usually enough. For FRQs: Write the inner rate on its own line before simplifying. Graders award the chain point on the presence of that factor, and a derivative that collapses it into a single number can cost the point even when the value is right.

Concept Review

Three Moves, Always in This Order

Figure 2 — Three numbered steps, then three compositions carried through them. The middle column keeps the inside exactly as it was; the right-hand column is the factor that most answers drop.

Three numbered steps for the chain rule and a table of three compositions showing the outer derivative, the untouched inside, the inside rate and the product

The Four Shapes That Cover Most Parts

Shape	Example	Outer derivative	Inner rate
Power	$(2x-7)^4$	$4(2x-7)^3$	$2$
Root	$\sqrt{x^2+9}$	$\frac{1}{2\sqrt{x^2+9}}$	$2x$
Trigonometric	$\cos(1-3x)$	$-\sin(1-3x)$	$-3$
Exponential or log	$e^{5x}$ , $\ln(x^2+1)$	$e^{5x}$ , $\frac{1}{x^2+1}$	$5$ , $2x$

Every row is the same sentence with different nouns. If you can fill this table from memory, the rest of the section is bookkeeping.

When the Inside Is Not Linear

A linear inside contributes a constant, which is why so many pupils believe the chain rule is optional. A non-linear inside contributes a variable factor, and then the omission is visible:

$\frac{d}{dx}\left(x^2+1\right)^3 = 3(x^2+1)^2 \cdot 2x = 6x(x^2+1)^2$

Compositions Inside Compositions

For $y = \sqrt{\sin(2x)}$ the layers are: square root, sine, and $2x$ . Peeling gives

$\frac{dy}{dx} = \frac{1}{2\sqrt{\sin(2x)}} \cdot \cos(2x) \cdot 2 = \frac{\cos(2x)}{\sqrt{\sin(2x)}}$

Three layers, three factors, and the last one is a constant. The habit that prevents errors is counting: name three layers, write three factors, then simplify.

Common Misconception Check

Myth 1: "Once the outer derivative is written, the problem is done." Truth: The outer derivative is the shape of the answer, not the answer. The size comes from the inner rate.
Myth 2: "A constant inner rate can be ignored because it is just a number." Truth: That constant is the difference between $\sin(3x)$ and $3\sin(3x)$ as derivatives. It is exactly the factor the exam tests.
Myth 3: "The chain rule only applies to powers." Truth: It applies to every composition, including the ones that look like plain functions because the argument hides a coefficient.

Topic-Specific Study Method

The Bracket Habit

Given	Bracketed inside	Derivative
$\sin(x^2)$	$x^2$	$2x\cos(x^2)$
$(5-2x)^7$	$5-2x$	$-14(5-2x)^6$
$e^{\sqrt{x}}$	$\sqrt{x}$	$\frac{e^{\sqrt{x}}}{2\sqrt{x}}$
$\ln(3x+4)$	$3x+4$	$\frac{3}{3x+4}$

① Differentiate five composites and, for each, underline the trailing factor before simplifying. The underlining is the whole exercise.

② Take one composite and differentiate it with the inside named $u$ and again without naming it. Both routes should end at the same expression.

③ Write a composite whose inner rate is $-1$ and differentiate it. The sign is the factor students drop most often when the inside is something like $1-x$ .

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $y = \sqrt{4x+9}$ .

Standard Execution:

Rewrite. $y = (4x+9)^{1/2}$ .

Outside. $\frac{1}{2}(4x+9)^{-1/2}$ .

Inside rate. Multiply by 4.

Simplify. $\frac{2}{\sqrt{4x+9}}$ .

Solution: The outer derivative of $-60e^{u}$ is $-60e^{u}$ and the inner rate of $-0.05t$ is $-0.05$ , so $T'(t) = 3e^{-0.05t}$ . At $t = 0$ this is $3$ degrees per minute, positive because the sample is warming, and the rate decays as the sample approaches room temperature.

Common Traps & Failure Signals:

The frozen inside (Signal: the bracket was differentiated but never multiplied): the outer derivative was treated as the whole answer.
The vanished negative (Signal: a plus where the inside decreases): the inner rate of a decreasing argument was taken as positive.
The merged factor (Signal: a number that absorbed the coefficient): simplification ran ahead of checking, so the chain factor cannot be inspected.

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

Factor Dropped: You wrote the outer derivative and closed the line.
Bracket Lost $\rightarrow$ Rewrite the inside in brackets before differentiating anything.
Evaluation Confused: You read $f'(g(x))$ as the product of two derivatives rather than the outer derivative evaluated at the inner function, which reverses the structure of the answer even when the pieces are right.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; A released part asks you to differentiate an explicitly written composite in five of the nine sets, and the inner function is rarely linear
10-Year Total Count: 5 of 9 Exams · 6 FRQ Parts; Counted where the composite is written out and its derivative is scored, not where a composition only appears inside a supplied rate
FRQ Years Appeared: 2018 (Q2a), 2021 (Q2a), 2022 (Q4d), 2023 (Q1d, Q2a), 2026 (Q4A)

Practice Examples

MCQ 1

A Square Root

If $y = \sqrt{6x-5}$ , what is $\frac{dy}{dx}$ ?

MCQ 2

A Negative Inner Rate

If $g(x) = \cos(2-5x)$ , what is $g'(x)$ ?

MCQ 3

A Log of a Power

If $h(x) = \ln(x^2+4)$ , what is $h'(x)$ ?

FRQ 1 • A Composed Model

The volume of a balloon is modeled by $V(t) = \frac{4}{3}\pi(2+0.5t)^3$ cubic centimeters, where $t$ is measured in seconds and $0 \le t \le 6$ .

(a) Find $V'(t)$ , leaving the bracket intact.

(b) Find the rate of change of the volume at $t = 4$ seconds, with units.

(c) Explain why $V'(t)$ is increasing on the interval.

(d) A classmate writes $V'(t) = 4\pi(2+0.5t)^2$ and calls the missing factor a rounding detail. Explain what that omission does to the reported rate at $t = 4$ .

Cheatsheet

Differentiating $f(g(x))$

	Move	Watch for
Linear inside	multiply by the slope	the constant factor
Power inside	multiply by the power's derivative	degree of the answer
Negative inside	multiply by the negative	sign of the result
Nested layers	one factor per layer	factor count
Trig or log inside	keep the argument intact	argument never changes

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 3.3 Implicit Differentiation Technique

On the AP Exam

Concept Review

Three Moves, Always in This Order

The Four Shapes That Cover Most Parts

When the Inside Is Not Linear

Compositions Inside Compositions

Common Misconception Check

Topic-Specific Study Method

The Bracket Habit

Training & Error Diagnosis

Past 10-Year AP Exam Appearance

Practice Examples

A Square Root

A Negative Inner Rate

A Log of a Power

FRQ 1 • A Composed Model

Cheatsheet

Differentiating f(g(x))f(g(x))f(g(x))

Ready to check your understanding?

On the AP Exam

Concept Review

Three Moves, Always in This Order

The Four Shapes That Cover Most Parts

When the Inside Is Not Linear

Compositions Inside Compositions

Common Misconception Check

Topic-Specific Study Method

The Bracket Habit

Training & Error Diagnosis

Past 10-Year AP Exam Appearance

Practice Examples

A Square Root

A Negative Inner Rate

A Log of a Power

FRQ 1 • A Composed Model

Cheatsheet

Differentiating f(g(x))f(g(x))f(g(x))

Ready to check your understanding?

On the AP Exam

Concept Review

Three Moves, Always in This Order

The Four Shapes That Cover Most Parts

When the Inside Is Not Linear

Compositions Inside Compositions

Common Misconception Check

Topic-Specific Study Method

The Bracket Habit

Training & Error Diagnosis

Past 10-Year AP Exam Appearance

Practice Examples

A Square Root

A Negative Inner Rate

A Log of a Power

FRQ 1 • A Composed Model

Cheatsheet

Differentiating f(g(x))f(g(x))f(g(x))

Ready to check your understanding?

Differentiating $f(g(x))$

Differentiating $f(g(x))$

Differentiating $f(g(x))$