3.1 The Chain Rule

The chain rule lets you differentiate composite functions, meaning a function inside another function. The core idea is to take the derivative of the outer function while keeping the inside the same, then multiply by the derivative of the inner function: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$.

Why This Matters for the AP Calculus Exam

The chain rule shows up constantly on the AP Calculus exam, both as the direct target of a question and as one step inside bigger problems. You will need it in multiple-choice and free-response questions whenever a function is built from other functions, like $\sin(3x)$, $e^{x^2}$, or $(3x+1)^5$.

It is also the foundation for later topics in this unit and beyond, including implicit differentiation, derivatives of inverse functions, and related rates. Getting comfortable with the chain rule now makes the rest of your AP Calculus work much smoother.

Key Takeaways

• A composite function is a function inside another function, written $(f \circ g)(x) = f(g(x))$, with $g(x)$ as the inner function and $f(x)$ as the outer function.
• The chain rule says $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$: differentiate the outside, then multiply by the derivative of the inside.
• In Leibniz notation, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$, where $u$ is the inner function.
• Always remember the "times the derivative of what's inside" step. Forgetting it is the most common error.
• The chain rule combines with the power, product, and quotient rules, and it can repeat when a function is nested more than once.
• Units can confirm the structure, as in $\frac{\text{psi}}{\text{min}} = \frac{\text{psi}}{\text{m}} \cdot \frac{\text{m}}{\text{min}}$.

Composite Functions

Composite functions are functions inside other functions. Given two functions $f(x)$ and $g(x)$, the composite function $(f \circ g)(x)$ is formed by applying $f$ to the output of $g$.

Written out, this is $(f \circ g)(x) = f(g(x))$. Think of $g(x)$ as the inner function and $f(x)$ as the outer function.

Composite Function Example

Take these two functions:

$\textcolor{red}{f(x)=x^2}$

$\textcolor{blue}{g(x)=3x+1}$

To work with the composite function $f(g(x))$, identify the inner and outer functions:

• $g(x)$ is the inner function. It takes $x$, multiplies by 3, and adds 1.
• $f(x)$ is the outer function. It takes the output of $g(x)$ and squares it.

So the composition looks like this:

$f(g(x))=\textcolor{red}{(}\textcolor{blue}{3x+1}\textcolor{red}{)^2}$

Recognizing the inner and outer pieces is the key skill for the chain rule.

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Definition of the Chain Rule

The chain rule is a differentiation method, just like the Power Rule, Product Rule, and Quotient Rule from Unit 2. There are two common notations.

The first is Leibniz notation:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Here is what each piece means:

• $\frac{dy}{dx}$ is the overall derivative you want to find.
• $u$ is your inner function.
• $\frac{dy}{du}$ is the derivative of the outer function with respect to the inner function.
• $\frac{du}{dx}$ is the derivative of the inner function with respect to $x$.

The second notation makes the structure clearer:

$\frac{d}{dx}\left(f(g(x))\right) = f'(g(x)) \cdot g'(x)$

To differentiate a composite function, you work from the outside in. If there is another function nested inside $g(x)$, the process repeats for that layer too.

Steps for the Chain Rule

1. Define your inner and outer functions.
2. Take the derivative of the outer function only, keeping the inside unchanged.
3. Take the derivative of the inner function with respect to $x$.
4. Multiply the two derivatives to get your final answer.

A useful habit: every time you apply the chain rule, say "times the derivative of what's inside." That keeps you from skipping the last step.

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The Chain Rule: Practice Problems

Example 1

Find the derivative of the following function with respect to $x$.

$y=(x^2+3x-1)^2$

1. Define your inner and outer functions.

The inner function is $u=x^2+3x-1$. The outer function is $y=u^2$.

2. Take the derivative of the outer function only.

$\frac{dy}{du} = 2u = 2(x^2+3x-1)$

Notice we have not touched the inner function $u$ yet.

3. Take the derivative of the inner function with respect to $x$.

$\frac{du}{dx} = 2x + 3$

4. Multiply the two derivatives.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2(x^2+3x-1) \cdot (2x+3)$

Example 2

Let $f(x)=e^x$ and $g(x)= \ln(x)$. Find the derivative of $y=f(g(x))$ with respect to $x$.

1. Define your inner and outer functions.

Here they are explicit: $f(x)$ is the outer function and $g(x)$ is the inner function.

2. Take the derivative of the outer function.

$f'(g(x))=e^{g(x)}=e^{\ln(x)}$

The derivative of $e^x$ is just $e^x$.

3. Take the derivative of the inner function with respect to $x$.

$g'(x)=\frac{1}{x}$

4. Multiply the two derivatives.

$f'(g(x))\cdot g'(x)=e^{\ln(x)}\cdot \frac{1}{x}$

Example 3

This one has more terms inside.

$y=4(5x^3+2x^2+6)^2$

1. Define your inner and outer functions.

The inner function is $u=5x^3+2x^2+6$ and the outer function is $y=4u^2$.

2. Take the derivative of the outer function.

$\frac{dy}{du}=8u=8(5x^3+2x^2+6)$

3. Take the derivative of the inner function with respect to $x$.

$\frac{du}{dx} = 15x^2+4x$

4. Multiply the two derivatives.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 8(5x^3+2x^2+6) \cdot (15x^2+4x)$

Example 4

Now try a square root.

$y=\sqrt{7x^2}$

1. Define your inner and outer functions.

The inner function is $u=7x^2$ and the outer function is $y=\sqrt{u}$.

2. Take the derivative of the outer function.

$\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2}(7x^2)^{-\frac{1}{2}}$

3. Take the derivative of the inner function with respect to $x$.

$\frac{du}{dx} = 14x$

4. Multiply the two derivatives.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2}(7x^2)^{-\frac{1}{2}} \cdot (14x)$

Example 5

Try this one yourself first.

$f(x) = \cos^2(3x)$

To help you start, the inner function is $u=\cos 3x$ and the outer function is $f(x)=u^2$. Notice this is nested twice: the $\cos$ has $3x$ inside it, so you apply the chain rule more than once.

Here is the answer:

$f'(x)=2(\cos 3x)\cdot -3\sin(3x) = -6\sin(3x)\cos(3x)$

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How to Use This on the AP Calculus Exam

MCQ

• Spot composite functions quickly. Watch for things like $\sin(3x)$, $\tan(2x-1)$, $e^{x^2}$, and $(3x+1)^5$, where the chain rule is required.
• When several rules apply at once, decide the order. For example, a product of two composite functions needs the product rule plus the chain rule on each piece.
• You may get tables or graphs of functions with names other than $f$ and $g$. Plug given values into $f'(g(x)) \cdot g'(x)$ carefully.

Free Response

• Show each layer of work: the outer derivative, the inner derivative, and the product. Clear steps make your reasoning easy to follow.
• Keep your notation consistent. Mixing up $\frac{dy}{dx}$ and $\frac{dy}{du}$ leads to mistakes.

Common Trap

• The most frequent error is forgetting to multiply by the derivative of the inside. Saying "times the derivative of what's inside" prevents it.
• Misidentifying the inner function changes the whole answer. Double check which part is truly inside.

Common Misconceptions

• Forgetting the inner derivative. Writing the derivative of $(3x+1)^5$ as $5(3x+1)^4$ is incomplete. You still multiply by the derivative of $3x+1$, which is 3, giving $15(3x+1)^4$.
• Thinking the chain rule replaces other rules. It works together with the power, product, and quotient rules, not instead of them.
• Stopping after one layer. When a function is nested more than once, like $\cos^2(3x)$, you apply the chain rule for each layer until you reach $x$.
• Differentiating the inside too early. Take the outer derivative first while keeping the inside unchanged, then handle the inside separately.
• Missing hidden compositions. Expressions like $\sin^2 x$ or $e^{x^2}$ are composite even though they look compact. If there is a function inside another function, the chain rule applies.

Related AP Calculus Guides

• 3.2 Implicit Differentiation
• 3.3 Differentiating Inverse Functions
• 3.4 Differentiating Inverse Trigonometric Functions
• 3.6 Calculating Higher-Order Derivatives
• 3.5 Selecting Procedures for Calculating Derivatives
• Unit 3 Overview: Differentiation: Composite, Implicit, and Inverse Functions