2.6 – The Chain Rule

Learning Objectives

By the end of this section, you should be able to:

  • State and apply the Chain Rule using the notation: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)
  • Identify composite functions and their inner/outer components
  • Differentiate composite functions using the Chain Rule
  • Apply the Chain Rule in combination with other differentiation rules
  • Solve real-world problems involving rates of change of composite functions
Section Progress

Notation Guide

For the Chain Rule, we'll use this notation:

  • Let \( y = f(u) \) where \( u = g(x) \)
  • \( \frac{dy}{du} \) = derivative of outer function with respect to u
  • \( \frac{du}{dx} \) = derivative of inner function with respect to x
  • The Chain Rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

The Chain Rule

If \( y = f(u) \) and \( u = g(x) \), then:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

In words: The derivative of a composite function equals the derivative of the outer function (with respect to the inner function) times the derivative of the inner function (with respect to x).

🔗 Easy to Remember! 🔗

"Derivative of the outside, leave the inside alone, times derivative of the inside"

Or simply: "Outside derivative times inside derivative"

Understanding Function Composition

The Chain Rule is used for composite functions - functions within functions.

Example: \( y = (3x^2 + 2)^5 \)

  • Inner function: \( u = 3x^2 + 2 \) (the "inside" part)
  • Outer function: \( y = u^5 \) (what's happening to the inside)

The Chain Rule tells us how to differentiate this "function within a function."

Why We Need the Chain Rule

We cannot simply apply basic rules to composite functions. For example:

If \( y = (x^2 + 1)^3 \), we cannot just say:

\[ \frac{dy}{dx} = 3(x^2 + 1)^2 \] ❌ (This is incomplete!)

The correct application of Chain Rule gives:

\[ \frac{dy}{dx} = 3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2 \] ✓

The Chain Rule accounts for how the inner function changes with respect to x.

Chain Rule Calculator

Enter outer function f(u) and inner function u(x):

Step-by-step solution:

Worked Examples

Example 1: Basic Chain Rule Application

Power Rule with Chain Rule

Problem: Find the derivative of \( y = (3x^2 + 2)^5 \)

1

Identify inner and outer functions:

Let \( u = 3x^2 + 2 \) (inner function)

Then \( y = u^5 \) (outer function)

2

Find \( \frac{dy}{du} \) and \( \frac{du}{dx} \):

\[ \frac{dy}{du} = 5u^4 \]

\[ \frac{du}{dx} = 6x \]

3

Apply the Chain Rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

\[ \frac{dy}{dx} = 5u^4 \cdot 6x \]

4

Substitute back for u:

\[ \frac{dy}{dx} = 5(3x^2 + 2)^4 \cdot 6x \]

\[ \frac{dy}{dx} = 30x(3x^2 + 2)^4 \]

Example 2: Chain Rule with Trigonometric Functions

Trigonometry Application

Problem: Find the derivative of \( y = \sin(2x^3) \)

1

Identify inner and outer functions:

Let \( u = 2x^3 \) (inner function)

Then \( y = \sin(u) \) (outer function)

2

Find \( \frac{dy}{du} \) and \( \frac{du}{dx} \):

\[ \frac{dy}{du} = \cos(u) \]

\[ \frac{du}{dx} = 6x^2 \]

3

Apply the Chain Rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

\[ \frac{dy}{dx} = \cos(u) \cdot 6x^2 \]

4

Substitute back for u:

\[ \frac{dy}{dx} = \cos(2x^3) \cdot 6x^2 \]

\[ \frac{dy}{dx} = 6x^2 \cos(2x^3) \]

Example 3: Multiple Chain Rule Applications

Triple Composition

Problem: Find the derivative of \( y = \sqrt{\sin(3x^2)} \)

1

Identify all layers:

Let \( w = 3x^2 \) (innermost)

Let \( u = \sin(w) \) (middle)

Then \( y = \sqrt{u} = u^{1/2} \) (outermost)

2

Apply Chain Rule multiple times:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dw} \cdot \frac{dw}{dx} \]

3

Find each derivative:

\[ \frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \]

\[ \frac{du}{dw} = \cos(w) \]

\[ \frac{dw}{dx} = 6x \]

4

Combine and substitute back:

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \cos(w) \cdot 6x \]

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin(w)}} \cdot \cos(w) \cdot 6x \]

\[ \frac{dy}{dx} = \frac{3x \cos(3x^2)}{\sqrt{\sin(3x^2)}} \]

When to Use the Chain Rule

Use the Chain Rule when you have:

  • A function inside another function: \( f(g(x)) \)
  • Any composite function
  • Functions with parentheses containing more than just x
  • Trigonometric functions of polynomials: \( \sin(x^2), \cos(3x) \), etc.
  • Exponential functions of polynomials: \( e^{x^2}, 2^{3x} \), etc.

Common patterns:

  • \( (expression)^n \) → Power Rule + Chain Rule
  • \( \sin(expression) \) → Trig derivative + Chain Rule
  • \( e^{expression} \) → Exponential derivative + Chain Rule

Chain Rule Practice

Test your understanding with these practice problems:

Common Mistakes to Avoid

Mistake Why It's Wrong Correct Approach
Forgetting the inner derivative Only takes derivative of outer function Multiply by derivative of inner function
\( \frac{d}{dx}[f(g(x))] = f'(g(x)) \) Missing multiplication by g'(x) \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
Not identifying all layers Misses intermediate functions in complex compositions Break down into all component functions
Forgetting to substitute back Leaves answer in terms of intermediate variables Always express final answer in terms of original variable

Quick Concept Check

Question 1: If \( y = f(u) \) and \( u = g(x) \), what is \( \frac{dy}{dx} \)?

\( \frac{dy}{du} \cdot \frac{du}{dx} \)
\( \frac{dy}{du} + \frac{du}{dx} \)
\( \frac{dy}{du} \div \frac{du}{dx} \)
\( \frac{dy}{du} \)
✓ Correct! This is the Chain Rule formula.
✗ Remember the correct formula: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

Question 2: Find \( \frac{dy}{dx} \) for \( y = (x^2 + 1)^3 \)

\( 6x(x^2 + 1)^2 \)
\( 3(x^2 + 1)^2 \)
\( 6x^2(x^2 + 1)^2 \)
\( 2x(x^2 + 1)^3 \)
✓ Correct! You applied both the power rule and chain rule properly.
✗ Remember: After taking the outer derivative, multiply by the derivative of the inside

Problem-Solving Strategy

  1. Identify if you have a composite function
  2. Label the inner function as u and outer function as f(u)
  3. Find \( \frac{dy}{du} \) (derivative of outer function)
  4. Find \( \frac{du}{dx} \) (derivative of inner function)
  5. Apply the Chain Rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)
  6. Substitute back to express in terms of x
  7. Simplify your final answer

AP Exam Tip

On the AP Calculus exam, Chain Rule problems are extremely common. Remember to:

  • Always check if a function is composite before differentiating
  • Show the u-substitution step in free-response questions
  • Watch for multiple layers of composition
  • Combine Chain Rule with other rules (Product, Quotient) when needed
  • Practice recognizing common composite patterns quickly

Practice Problems

Problem 1: Basic Chain Rule

Find \( \frac{dy}{dx} \) for \( y = (2x - 5)^4 \)

Let \( u = 2x - 5 \), then \( y = u^4 \)

\[ \frac{dy}{du} = 4u^3 \], \( \frac{du}{dx} = 2 \)

\[ \frac{dy}{dx} = 4u^3 \cdot 2 = 8u^3 \]

\[ \frac{dy}{dx} = 8(2x - 5)^3 \]

Problem 2: Chain Rule with Trig

Differentiate \( y = \cos(4x^3) \)

Let \( u = 4x^3 \), then \( y = \cos(u) \)

\[ \frac{dy}{du} = -\sin(u) \], \( \frac{du}{dx} = 12x^2 \)

\[ \frac{dy}{dx} = -\sin(u) \cdot 12x^2 \]

\[ \frac{dy}{dx} = -12x^2 \sin(4x^3) \]

Problem 3: Multiple Rules Combined

Find \( \frac{dy}{dx} \) for \( y = (x^2 + 1)^3 \cdot \sin(2x) \)

Use Product Rule + Chain Rule:

Let \( u = (x^2 + 1)^3 \), \( v = \sin(2x) \)

\[ \frac{dy}{dx} = u'v + uv' \]

Find u' using Chain Rule:

Let \( w = x^2 + 1 \), then \( u = w^3 \)

\[ u' = 3w^2 \cdot 2x = 6x(x^2 + 1)^2 \]

Find v' using Chain Rule:

Let \( z = 2x \), then \( v = \sin(z) \)

\[ v' = \cos(z) \cdot 2 = 2\cos(2x) \]

Combine:

\[ \frac{dy}{dx} = 6x(x^2 + 1)^2 \cdot \sin(2x) + (x^2 + 1)^3 \cdot 2\cos(2x) \]

\[ \frac{dy}{dx} = 2(x^2 + 1)^2[3x\sin(2x) + (x^2 + 1)\cos(2x)] \]

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