2.7 – Derivatives of Composite Functions

Learning Objectives

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

  • Apply the Chain Rule to complex composite functions with multiple layers
  • Combine the Chain Rule with Product and Quotient Rules
  • Differentiate functions involving trigonometric, exponential, and logarithmic compositions
  • Recognize and apply derivative patterns for common composite function types
  • Solve applied problems involving rates of change of composite quantities
Section Progress

Building on the Chain Rule

In Section 2.6, we learned the Chain Rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

Now we'll apply this to more complex composite functions and combine it with other differentiation rules.

Function Decomposition Strategy

When facing complex composite functions, use this systematic approach:

  1. Identify the outermost operation (what happens last in order of operations)
  2. Work inward, identifying each layer of composition
  3. Apply the Chain Rule for each layer, working from outside in
  4. Combine with other rules (Product, Quotient) as needed

Common Composite Function Patterns

Power of a Function

\( y = [f(x)]^n \)

\[ \frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x) \]

Trigonometric Composition

\( y = \sin(f(x)) \)

\[ \frac{dy}{dx} = \cos(f(x)) \cdot f'(x) \]

Exponential Composition

\( y = e^{f(x)} \)

\[ \frac{dy}{dx} = e^{f(x)} \cdot f'(x) \]

Logarithmic Composition

\( y = \ln(f(x)) \)

\[ \frac{dy}{dx} = \frac{1}{f(x)} \cdot f'(x) \]

Composite Function Analyzer

Enter a composite function to see its decomposition:

Worked Examples

Example 1: Triple Composition

Multiple Chain Rule Applications

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

1

Decompose the function:

Let \( u = 3x^2 + 2 \) (innermost)

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

Then \( y = \ln(v) \) (outermost)

2

Find each derivative:

\[ \frac{dy}{dv} = \frac{1}{v} \]

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

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

3

Apply Chain Rule:

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

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

4

Substitute back:

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

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

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

Example 2: Chain Rule + Product Rule

Multiple Rules Combined

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

1

Identify the structure:

This is a product: \( u \cdot v \) where:

\( u = x^2 \) and \( v = e^{3x} \)

2

Apply Product Rule:

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

3

Find u' and v':

\[ u' = 2x \]

For \( v = e^{3x} \), use Chain Rule:

Let \( w = 3x \), then \( v = e^w \)

\[ \frac{dv}{dw} = e^w \], \( \frac{dw}{dx} = 3 \)

\[ v' = e^w \cdot 3 = 3e^{3x} \]

4

Combine results:

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

\[ \frac{dy}{dx} = 2xe^{3x} + 3x^2e^{3x} \]

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

Example 3: Chain Rule + Quotient Rule

Complex Rational Composite

Problem: Find the derivative of \( y = \frac{\sin(2x)}{x^2 + 1} \)

1

Identify the structure:

This is a quotient: \( \frac{u}{v} \) where:

\( u = \sin(2x) \) and \( v = x^2 + 1 \)

2

Apply Quotient Rule:

\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \]

3

Find u' and v':

\[ v' = 2x \]

For \( u = \sin(2x) \), use Chain Rule:

Let \( w = 2x \), then \( u = \sin(w) \)

\[ \frac{du}{dw} = \cos(w) \], \( \frac{dw}{dx} = 2 \)

\[ u' = \cos(w) \cdot 2 = 2\cos(2x) \]

4

Combine results:

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

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

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

Strategy for Complex Derivatives

When facing a complex derivative problem:

  1. Identify the main structure - Is it a product? Quotient? Composition?
  2. Apply the appropriate primary rule (Product, Quotient, or Chain Rule)
  3. Use Chain Rule for any composite parts
  4. Work systematically from outside in
  5. Simplify your final answer

Composite Function Practice

Test your skills with these challenging problems:

Common Mistakes to Avoid

Mistake Why It's Wrong Correct Approach
Forgetting inner derivatives in multiple compositions Only applies Chain Rule to some layers Apply Chain Rule for every layer of composition
Mixing up order of operations Applies rules in wrong sequence Always work from outside in
Not recognizing when to combine rules Uses only one rule when multiple are needed Identify the overall structure first
Algebraic errors in simplification Correct derivative but messy final answer Factor and simplify carefully

Quick Concept Check

Question 1: What is the derivative of \( y = e^{2x^2} \)?

\( 4xe^{2x^2} \)
\( e^{2x^2} \)
\( 2e^{2x^2} \)
\( 4x^2e^{2x^2} \)
✓ Correct! Chain Rule: derivative of e^u is e^u times derivative of u.
✗ Remember: For e^(f(x)), derivative is e^(f(x)) · f'(x)

Question 2: Which rule would you use first for \( y = x^2 \cdot \sin(3x) \)?

Product Rule
Chain Rule
Quotient Rule
Power Rule
✓ Correct! The main structure is a product, so start with Product Rule.
✗ Look at the overall structure first - this is a product of two functions.

Problem-Solving Strategy for Composite Functions

  1. Analyze the structure - Identify the main operation
  2. Choose the primary rule - Product, Quotient, or Chain Rule
  3. Decompose composite parts - Identify inner functions
  4. Apply Chain Rule systematically - Work from outside in
  5. Combine results - Use appropriate combination of rules
  6. Simplify - Factor and reduce the final expression

AP Exam Tip

On the AP Calculus exam, complex derivatives often combine multiple rules. Remember to:

  • Show your work clearly, especially the decomposition steps
  • Label which rule you're using at each step
  • Check that you've applied Chain Rule to every composite part
  • Simplify your final answer - this can earn you points
  • Practice recognizing common patterns quickly

Practice Problems

Problem 1: Multiple Composition

Find \( \frac{dy}{dx} \) for \( y = \cos^3(2x) \)

Rewrite: \( y = [\cos(2x)]^3 \)

Let \( u = \cos(2x) \), then \( y = u^3 \)

\[ \frac{dy}{du} = 3u^2 \]

For \( u = \cos(2x) \), let \( v = 2x \), then \( u = \cos(v) \)

\[ \frac{du}{dv} = -\sin(v) \], \( \frac{dv}{dx} = 2 \)

\[ \frac{du}{dx} = -\sin(v) \cdot 2 = -2\sin(2x) \]

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

Problem 2: Product + Chain Rule

Differentiate \( y = x \cdot e^{x^2} \)

Use Product Rule:

Let \( u = x \), \( v = e^{x^2} \)

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

\[ u' = 1 \]

For \( v = e^{x^2} \), use Chain Rule:

Let \( w = x^2 \), then \( v = e^w \)

\[ \frac{dv}{dw} = e^w \], \( \frac{dw}{dx} = 2x \)

\[ v' = e^w \cdot 2x = 2xe^{x^2} \]

\[ \frac{dy}{dx} = (1)(e^{x^2}) + (x)(2xe^{x^2}) \]

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

Problem 3: Quotient + Chain Rule

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

Use Quotient Rule:

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

\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \]

\[ v' = 2x \]

For \( u = \ln(3x) \), use Chain Rule:

Let \( w = 3x \), then \( u = \ln(w) \)

\[ \frac{du}{dw} = \frac{1}{w} \], \( \frac{dw}{dx} = 3 \)

\[ u' = \frac{1}{w} \cdot 3 = \frac{3}{3x} = \frac{1}{x} \]

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

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

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

Back ⬅ Section 2.6 Next ➡ Section 2.8