Chain rule — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Basic chain rule formula for single-variable composite functions, application of the rule to power, trigonometric, and exponential functions, chain rule for tangent line problems, and recognition of common student mistakes.

You should already know: How to decompose composite functions, derivative rules for basic functions, the point-slope form of a line.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Chain rule?

The chain rule is the core differentiation rule for composite functions, which take the form $y=f(g(x))$, where an inner function $u=g(x)$ is plugged into an outer function $f(u)$. Per the College Board AP Calculus AB Course and Exam Description (CED), this unit makes up 9-13% of total exam weight, with the chain rule accounting for roughly 4-6% of total exam points. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and it is almost always combined with other differentiation rules, so mastery is non-negotiable for a high score.

Synonyms for the chain rule include the "composite function differentiation rule" or "outer-inner rule". The core intuition is that the rate of change of a composite function equals the product of the rate of change of the outer function (evaluated at the inner function) and the rate of change of the inner function. Most functions you encounter on the AP exam are composite, so you cannot rely on basic differentiation rules alone to find their derivatives. For example, $y=(x^2+3x{)}^5$ is not a basic power function—its derivative must account for how both the outer power function and inner quadratic change with respect to $x$.

2. Basic Chain Rule: Decomposition and Core Formula

To apply the chain rule correctly, you first need to correctly decompose a composite function into its outer and inner components. A simple rule for decomposition: the inner function is what you calculate first when plugging in a value of $x$, and the outer function is what you calculate last. For example, for $y=e^{x^2}$, when you plug in $x=2$, you first calculate $x^2=4$, then calculate $e^4$, so inner function is $u=x^2$, outer function is $y=e^u$.

The core chain rule formulas are: $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$ in Leibniz notation, and: $$\frac{d}{dx}\left[f(g(x))\right]=f^{\prime }(g(x))\cdot g^{\prime }(x)$$ in prime notation. The intuition for the product form is simple: if $y$ changes twice as fast as $u$, and $u$ changes three times as fast as $x$, then $y$ changes $2\times 3=6$ times as fast as $x$, so the rates multiply. The only time you do not need chain rule is when the inner function is just $y=x$, so its derivative is 1, and the product simplifies to just the outer derivative.

Worked Example

Find the derivative of $y=(2x^3-5x+1{)}^4$.

1. Decompose the function into outer and inner parts: $y=u^4$ (outer), $u=2x^3-5x+1$ (inner).
2. Differentiate the outer function with respect to $u$: $\frac{dy}{du}=4u^3$.
3. Differentiate the inner function with respect to $x$: $\frac{du}{dx}=6x^2-5$.
4. Multiply the derivatives and substitute back $u=2x^3-5x+1$: $$\frac{dy}{dx}=4u^3(6x^2-5)=4(2x^3-5x+1{)}^3(6x^2-5)$$
5. Simplify (optional for AP, but tidied): $\frac{dy}{dx}=(24x^2-20)(2x^3-5x+1{)}^3$.

Exam tip: Always explicitly label your inner and outer functions on scratch work—if you cannot name both clearly, you are almost guaranteed to misapply the rule.

3. Combining Chain Rule with Other Differentiation Rules

The chain rule is almost never tested in isolation on the AP exam. You will almost always need to combine it with product rule, quotient rule, or derivatives of non-power basic functions. The most common generalized forms of the chain rule for common function types are:

• Generalized power rule: $\frac{d}{dx}[u^n]=n{u}^{n-1}{u}^{\prime }$
• Generalized trigonometric: $\frac{d}{dx}[\sin u]=\cos u\cdot u^{\prime }$, $\frac{d}{dx}[\cos u]=-\sin u\cdot u^{\prime }$, etc.
• Generalized exponential (base $e$): $\frac{d}{dx}[e^u]=e^u\cdot u^{\prime }$

The key rule when combining rules is: apply the rule for the outermost operation first, then work inward to apply chain rule to any composite inner parts. For example, if the outermost operation is a product of two functions, apply product rule first, then use chain rule to differentiate any composite factors inside the product rule.

Worked Example

Find the derivative of $f(x)=e^{3x}\cos (2x)$.

1. The outermost operation is multiplication of two functions, so apply product rule first: $$f^{\prime }(x)=\frac{d}{dx}[e^{3x}]\cdot \cos (2x)+e^{3x}\cdot \frac{d}{dx}[\cos (2x)]$$
2. Differentiate $e^{3x}$ with chain rule: outer $e^u$, inner $u=3x$, so $\frac{d}{dx}[e^{3x}]=e^{3x}\cdot 3=3e^{3x}$.
3. Differentiate $\cos (2x)$ with chain rule: outer $\cos u$, inner $u=2x$, so $\frac{d}{dx}[\cos (2x)]=-\sin (2x)\cdot 2=-2\sin (2x)$.
4. Substitute back into the product rule and simplify: $$f^{\prime }(x)=3e^{3x}\cos (2x)+e^{3x}(-2\sin (2x))=e^{3x}\left(3\cos (2x)-2\sin (2x)\right)$$

Exam tip: When combining multiple rules, write out each step one at a time instead of trying to write the final derivative in one step—partial credit is almost always available for correct intermediate steps on FRQ.

4. Chain Rule for Tangent and Normal Line Problems

A very common AP exam question asks you to find the equation of the tangent line (or slope of the normal line) to a composite function at a given point. To solve these problems, you need the chain rule to calculate the slope of the tangent line at the given point, since the function is composite. Recall that the tangent line at $x=a$ follows point-slope form: $y-f(a)=f^{\prime }(a)(x-a)$, where $f(a)$ is the $y$-coordinate of the point, and $f^{\prime }(a)$ is the slope of the tangent, which requires chain rule to compute. For a normal line, the slope is the negative reciprocal of the tangent slope: $m_{\text{normal}}=-\frac{1}{f^{\prime }(a)}$.

Worked Example

Find the equation of the tangent line to $y=\sqrt{4x^2+9}$ at $x=2$.

1. First find the $y$-coordinate of the point at $x=2$: $y=\sqrt{4(2{)}^2+9}=\sqrt{25}=5$, so the point is $(2,5)$.
2. Rewrite the function to make differentiation easier: $y=(4x^2+9{)}^{1/2}$. Decompose into outer $y=u^{1/2}$, inner $u=4x^2+9$.
3. Apply the generalized power rule (chain rule for powers): $$y^{\prime }=\frac{1}{2}{u}^{-1/2}\cdot \frac{du}{dx}=\frac{1}{2}(4x^2+9{)}^{-1/2}\cdot 8x=\frac{4x}{\sqrt{4x^2+9}}$$
4. Evaluate the slope at $x=2$: $y^{\prime }(2)=\frac{4(2)}{\sqrt{16+9}}=\frac{8}{5}$.
5. Write the tangent line in point-slope form: $y-5=\frac{8}{5}(x-2)$, or in slope-intercept form: $y=\frac{8}{5}x+\frac{9}{5}$.

Exam tip: Always calculate $f(a)$ (the point) before calculating $f^{\prime }(a)$—this is an easy 1 point on FRQ that you can get even if you mess up the derivative calculation.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to multiply by $g^{\prime }(x)$, leaving off the inner derivative. For example, differentiating $(x^2+1{)}^3$ to get $3(x^2+1{)}^2$ instead of $3(x^2+1{)}^2(2x)$. Why: Students stop after differentiating the outer function, forgetting the inner function changes with $x$. Correct move: After differentiating the outer function, always ask: "Is the inner function anything other than just $x$?" If not, multiply by its derivative before moving on.
• Wrong move: Reversing outer/inner differentiation, getting $g^{\prime }(x)f(x)$ instead of $f^{\prime }(g(x))g^{\prime }(x)$. For example, differentiating $\sin (5x)$ to get $5\sin (x)$. Why: Confusion about which function to differentiate first. Correct move: Always explicitly label outer and inner on scratch paper, differentiate the outer function first, then multiply by the inner derivative.
• Wrong move: Using chain rule for a product of two functions instead of product rule. For example, differentiating $x^2\sin x$ to get $2x\cos x$. Why: Confusing composition $f(g(x))$ with multiplication $f(x)g(x)$. Correct move: Before differentiating, confirm if the function is one function plugged into another, or two separate functions multiplied—only use chain rule for composition.
• Wrong move: Substituting $x=a$ into the inner function before differentiating for tangent line problems. For example, for $y=(x^2+1{)}^3$ at $x=2$, substituting $x=2$ first to get $y=125$, then differentiating to get slope 0. Why: Confusing when to substitute values. Correct move: Differentiate the entire function first with chain rule, then substitute $x=a$ into the derivative to get the slope.
• Wrong move: Applying chain rule before product/quotient rule when the outermost operation is a product/quotient. For example, differentiating $(x+1)(x^2+2{)}^3$ by applying chain rule to the entire expression first. Why: Forgetting that outermost operation determines rule order. Correct move: Always apply the rule for the outermost operation first, then work inward to apply chain rule to composite inner parts.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is equal to $\frac{d}{dx}\left[\tan (x^3-2x)\right]$? A) ${\sec }^2(3x^2-2)$ B) $(3x^2-2){\sec }^2(x^3-2x)$ C) ${\sec }^2(x^3-2x)$ D) $(3x^2-2)\tan (x^3-2x)$

Worked Solution: We first recognize $\tan (x^3-2x)$ is a composite function with outer $f(u)=\tan u$ and inner $u=x^3-2x$. By the chain rule, the derivative is $f^{\prime }(u)\cdot u^{\prime }$. The derivative of $\tan u$ is ${\sec }^{2}u$, and the derivative of $x^3-2x$ is $3x^2-2$. Substituting back $u=x^3-2x$, we get the derivative $(3x^2-2){\sec }^2(x^3-2x)$. Correct answer: B.

Question 2 (Free Response)

Let $g(x)=(3x^2-12)e^{-x}$. (a) Find $g^{\prime }(x)$. (b) Find all values of $x$ where the graph of $g(x)$ has a horizontal tangent line. Justify your answer. (c) Find the slope of the normal line to $g(x)$ at $x=0$.

Worked Solution: (a) $g(x)$ is a product of two functions, so we use product rule, with chain rule for $e^{-x}$: $$g^{\prime }(x)=\frac{d}{dx}[3x^2-12]\cdot e^{-x}+(3x^2-12)\cdot \frac{d}{dx}[e^{-x}]=6x{e}^{-x}+(3x^2-12)(-e^{-x})$$ Simplify: $g^{\prime }(x)=e^{-x}(-3x^2+6x+12)=-3e^{-x}(x^2-2x-4)$.

(b) Horizontal tangents occur where $g^{\prime }(x)=0$. $e^{-x}$ is never zero for real $x$, so we solve $x^2-2x-4=0$. By quadratic formula: $$x=\frac{2\pm \sqrt{4+16}}{2}=1\pm \sqrt{5}$$ These are the only values where $g^{\prime }(x)=0$, so they are the locations of all horizontal tangent lines.

(c) Evaluate $g^{\prime }(0)=-3e^0(0-0-4)=12$. The slope of the normal line is the negative reciprocal: $m_{\text{normal}}=-\frac{1}{12}$.

Question 3 (Application / Real-World Style)

A population of bacteria in a controlled experiment grows according to the function $P(t)=1000e^{0.1t^2}$, where $P(t)$ is the number of bacteria after $t$ hours, $t\ge 0$. Find the rate of change of the bacterial population at $t=3$ hours, and include units in your answer.

Worked Solution: The rate of change of the population is given by $P^{\prime }(t)$, found via chain rule. Let $u=0.1t^2$, so $P(t)=1000e^u$. By chain rule: $$P^{\prime }(t)=1000e^u\cdot \frac{du}{dt}=1000e^{0.1t^2}(0.2t)=200t{e}^{0.1t^2}$$ Evaluate at $t=3$: $P^{\prime }(3)=200(3)e^{0.1(9)}=600e^{0.9}\approx 600(2.46)\approx 1476$.

Interpretation: After 3 hours, the bacterial population is increasing at a rate of approximately 1476 bacteria per hour.

7. Quick Reference Cheatsheet

8. What's Next

The chain rule is the foundational prerequisite for all remaining topics in Unit 3, and it is required for nearly all integration and application topics later in the AP Calculus AB course. Next, you will apply the chain rule to implicit differentiation, where you differentiate implicit functions by applying the chain rule to terms with $y$ as a function of $x$. Without correctly identifying when to multiply by $\frac{dy}{dx}$ for implicit terms, implicit differentiation will be impossible. The chain rule is also critical for related rates problems, a heavily tested FRQ topic on the AP exam. Later, when you learn u-substitution for integration, you will reverse the chain rule, so mastering it now makes integration much more intuitive.

Implicit differentiation Derivatives of inverse functions Related rates U-substitution