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Reading Time: 5 min
Last Updated: March 4, 2026
Main Ideas: 4
Reading Time: 5 min
Last Updated: March 4, 2026
Main Ideas: 4

Topic 3.1 Notes – The Chain Rule

Verified for 2027 AP® Calculus AB Exam
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The rule that lets you differentiate composite functions. When one function is plugged into another, their rates of change multiply together. This idea shows up constantly for the rest of the course, so getting comfortable spotting and applying it matters.

1. Composite Functions and Why the Chain Rule Exists

A composite function happens when one function is inside another:

(f∘g)(x)=f(g(x)) (f \circ g)(x) = f(g(x))

  • Inner function: g(x) g(x)
  • Outer function: f(x) f(x)

Example structures:

  • (4x−5)3 (4x - 5)^3 → outer: cube; inner: 4x−54x - 5
  • 2x2+7 \sqrt{2x^2 + 7} → outer: square root; inner: 2x2+72x^2 + 7
  • sin⁡(5x2) \sin(5x^2) → outer: sine; inner: 5x25x^2

Here’s the key idea:
Derivatives measure rates of change. If one quantity depends on another, which depends on xx, the total rate is the product of those rates.

Think of it like this:

  • Let u=g(x)u = g(x)
  • Let y=f(u)y = f(u)

Then the total rate of change is:

dydx=dydu⋅dudx \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

That’s the Chain Rule.

The Chain Rule Formula

If y=f(g(x)) y = f(g(x)) , then

dydx=f′(g(x))⋅g′(x) \frac{dy}{dx} = f'(g(x)) \cdot g'(x)

In words:
Take the derivative of the outside, keep the inside the same, then multiply by the derivative of the inside.

If there are more layers, you keep repeating this idea.

2. How to Apply the Chain Rule

When you’re staring at a function, the first job is identifying layers.

Step-by-step process

  1. Identify inner and outer functions
    • Parentheses, radicals, trig arguments, exponents are clues.
  2. Differentiate the outer function
    • Treat the inside as one single variable.
    • Do not touch the inside yet.
  3. Multiply by the derivative of the inner function
  4. Simplify if helpful

Example

Find ddx((2x3−x)5) \frac{d}{dx} \left( (2x^3 - x)^5 \right)

  • Outer: something to the 5th power
  • Inner: 2x3−x2x^3 - x

Derivative:

5(2x3−x)4⋅(6x2−1) 5(2x^3 - x)^4 \cdot (6x^2 - 1)

Power rule on the outside, derivative of the inside multiplied on.

On quizzes, the most common mistake is stopping after the power rule and forgetting the inner derivative.

3. Common Composite Structures You Must Know

These patterns appear constantly in MCQs and FRQs.

a. Power of a Function

(ax2+bx+c)n (ax^2 + bx + c)^n

Derivative pattern:

n(ax2+bx+c)n−1⋅(2ax+b) n(ax^2 + bx + c)^{n-1} \cdot (2ax + b)

b. Trig Functions with Inner Expressions

sin⁡(u),cos⁡(u),tan⁡(u) \sin(u), \quad \cos(u), \quad \tan(u)

Derivatives:

  • sin⁡(u)→cos⁡(u)⋅u′ \sin(u) \to \cos(u)\cdot u'
  • cos⁡(u)→−sin⁡(u)⋅u′ \cos(u) \to -\sin(u)\cdot u'
  • tan⁡(u)→sec⁡2(u)⋅u′ \tan(u) \to \sec^2(u)\cdot u'

If the angle isn’t just xx, you need the Chain Rule.

c. Exponentials

eg(x) e^{g(x)}

Derivative:

eg(x)⋅g′(x) e^{g(x)} \cdot g'(x)

The exponential stays exactly the same. Multiply by the inside derivative.

d. Logarithms

ln⁡(g(x)) \ln(g(x))

Derivative:

1g(x)⋅g′(x) \frac{1}{g(x)} \cdot g'(x)

Students often forget the inner derivative here.

e. Multiple Layers

Example:
sin⁡3(4x) \sin^3(4x)

Layers:

  1. Cube
  2. Sine
  3. 4x4x

Derivative:

3sin⁡2(4x)⋅cos⁡(4x)⋅4 3\sin^2(4x) \cdot \cos(4x) \cdot 4

You move from outside inward, multiplying each derivative.

4. When to Use It and Where Students Mess Up

When You Need the Chain Rule

  • A function raised to a power
  • A trig function with something inside
  • An exponential like e3xe^{3x}
  • A logarithm like ln⁡(7x2+1)\ln(7x^2+1)
  • A radical rewritten as a power

If you can clearly point to an “inside expression,” you likely need it.

Common Mistakes

  • Forgetting the inner derivative
  • Losing parentheses, especially with trig
  • Sign errors with cosine
  • Not recognizing hidden powers, like x2+4=(x2+4)1/2 \sqrt{x^2+4} = (x^2+4)^{1/2}

On no-calculator multiple choice, small sign mistakes are how they trap you. On FRQs, forgetting the inner derivative usually costs the majority of the point.

Key Takeaways

The Chain Rule says ddxf(g(x))=f′(g(x))⋅g′(x) \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) .
Always identify layers before differentiating.
The derivative of eg(x) e^{g(x)} is eg(x)⋅g′(x) e^{g(x)} \cdot g'(x) .
The derivative of ln⁡(g(x)) \ln(g(x)) is 1g(x)⋅g′(x) \frac{1}{g(x)} \cdot g'(x) .
If you see a function inside another function, multiply by the derivative of the inside.

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