2.4 – The Product Rule

Learning Objectives

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

  • State and apply the Product Rule using the notation: \( \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} \)
  • Differentiate products of functions using the u-v notation
  • Recognize when to use the Product Rule vs. other rules
  • Simplify derivatives obtained using the Product Rule
  • Apply the Product Rule to solve real-world problems
Section Progress

Notation Guide

In this section, we'll use the following notation:

  • Let \( y = u \cdot v \) where u and v are functions of x
  • \( \frac{du}{dx} \) = derivative of u with respect to x
  • \( \frac{dv}{dx} \) = derivative of v with respect to x
  • The Product Rule: \( \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} \)

The Product Rule

If \( y = u \cdot v \), then:

\[ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} \]

In words: The derivative of a product equals the second function times the derivative of the first, plus the first function times the derivative of the second.

Why We Need the Product Rule

The derivative of a product is NOT the product of the derivatives. For example:

If \( u = x^2 \) and \( v = x^3 \), then:

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

But the actual derivative of \( y = u \cdot v = x^5 \) is:

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

These are different! This shows why we need a special rule for products.

Product Rule Calculator

Enter u and v functions to find the derivative of their product:

Step-by-step solution:

Worked Examples

Example 1: Basic Product Rule Application

Fundamental Application

Problem: Find the derivative of \( y = (x^2 + 3x)(4x - 1) \)

1

Identify u and v:

Let \( u = x^2 + 3x \) and \( v = 4x - 1 \)

2

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

\[ \frac{du}{dx} = 2x + 3 \]

\[ \frac{dv}{dx} = 4 \]

3

Apply the Product Rule:

\[ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} \]

\[ \frac{dy}{dx} = (4x - 1)(2x + 3) + (x^2 + 3x)(4) \]

4

Expand and simplify:

\[ \frac{dy}{dx} = (8x^2 + 12x - 2x - 3) + (4x^2 + 12x) \]

\[ \frac{dy}{dx} = 8x^2 + 10x - 3 + 4x^2 + 12x \]

\[ \frac{dy}{dx} = 12x^2 + 22x - 3 \]

Example 2: Product with Radical Functions

Advanced Application

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

1

Rewrite using exponents and identify u and v:

\[ y = x^3 \cdot x^{1/2} = x^{7/2} \]

Let \( u = x^3 \) and \( v = x^{1/2} \)

2

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

\[ \frac{du}{dx} = 3x^2 \]

\[ \frac{dv}{dx} = \frac{1}{2}x^{-1/2} \]

3

Apply the Product Rule:

\[ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} \]

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

4

Simplify:

\[ \frac{dy}{dx} = 3x^{5/2} + \frac{1}{2}x^{5/2} = \frac{7}{2}x^{5/2} \]

5

Verify with power rule:

Since \( y = x^{7/2} \), using power rule:

\[ \frac{dy}{dx} = \frac{7}{2}x^{5/2} \] ✓

Example 3: Real-World Application

Economics Context

Problem: A company's revenue is \( R = x \cdot p(x) \), where \( x \) is quantity sold and \( p(x) = 100 - 0.5x \) is the price function. Find the marginal revenue.

1

Identify u and v:

Let \( u = x \) and \( v = 100 - 0.5x \)

So \( R = u \cdot v \)

2

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

\[ \frac{du}{dx} = 1 \]

\[ \frac{dv}{dx} = -0.5 \]

3

Apply the Product Rule:

\[ \frac{dR}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} \]

\[ \frac{dR}{dx} = (100 - 0.5x)(1) + (x)(-0.5) \]

4

Simplify:

\[ \frac{dR}{dx} = 100 - 0.5x - 0.5x = 100 - x \]

5

Interpretation:

The marginal revenue is \( \frac{dR}{dx} = 100 - x \). This means selling one additional unit increases revenue by \( 100 - x \) dollars.

When to Use the Product Rule

Use the Product Rule when you have:

  • A product of two different functions: \( y = u \cdot v \)
  • A product where one function is not a constant multiple of the other
  • Functions that cannot be easily multiplied together before differentiating

When NOT to use the Product Rule:

  • When one function is a constant (use Constant Multiple Rule)
  • When you can easily multiply the functions first
  • For sums or differences of functions (use Sum/Difference Rule)

Product Rule Practice

Test your understanding with these practice problems:

Common Mistakes to Avoid

Mistake Why It's Wrong Correct Approach
\( \frac{d}{dx}[uv] = \frac{du}{dx} \cdot \frac{dv}{dx} \) Derivative of product ≠ Product of derivatives Use Product Rule: \( v\frac{du}{dx} + u\frac{dv}{dx} \)
Forgetting to differentiate both factors Product Rule requires derivatives of both u and v Always include both \( v\frac{du}{dx} \) and \( u\frac{dv}{dx} \)
Not simplifying the result Leaves answer in unnecessarily complex form Expand and combine like terms
Using Product Rule for sums Product Rule is only for products Use Sum Rule for sums: \( \frac{d}{dx}[u+v] = \frac{du}{dx} + \frac{dv}{dx} \)

Quick Concept Check

Question 1: If \( y = u \cdot v \), what is \( \frac{dy}{dx} \)?

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

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

\( 9x^2 - 4x + 3 \)
\( 6x^2 - 4x + 3 \)
\( (2x)(3) = 6x \)
\( 3x^2 + 2x - 2 \)
✓ Correct! You applied the Product Rule properly.
✗ Remember: \( \frac{d}{dx}[uv] = v\frac{du}{dx} + u\frac{dv}{dx} \), not \( \frac{du}{dx} \cdot \frac{dv}{dx} \)

Problem-Solving Strategy

  1. Identify if you have a product of two functions: \( y = u \cdot v \)
  2. Label the functions as u and v
  3. Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) separately
  4. Apply the Product Rule: \( v\frac{du}{dx} + u\frac{dv}{dx} \)
  5. Simplify the result by expanding and combining like terms
  6. Verify your answer makes sense

AP Exam Tip

On the AP Calculus exam, Product Rule problems often appear in both multiple-choice and free-response sections. Remember to:

  • Show all steps when using the Product Rule
  • Clearly identify u and v in free-response questions
  • Simplify your final answer
  • Check your work by multiplying first (when possible) and differentiating
  • Watch for opportunities to simplify before differentiating

Practice Problems

Problem 1: Basic Product

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

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

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

\[ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} = (2x^2 - 1)(3x^2) + (x^3)(4x) \]

\[ \frac{dy}{dx} = 6x^4 - 3x^2 + 4x^4 = 10x^4 - 3x^2 \]

Problem 2: With Constants

Differentiate \( y = (3x^2 + 2)(4x^3 - 5x) \)

Let \( u = 3x^2 + 2 \), \( v = 4x^3 - 5x \)

\[ \frac{du}{dx} = 6x \], \( \frac{dv}{dx} = 12x^2 - 5 \)

\[ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} = (4x^3 - 5x)(6x) + (3x^2 + 2)(12x^2 - 5) \]

\[ \frac{dy}{dx} = 24x^4 - 30x^2 + 36x^4 - 15x^2 + 24x^2 - 10 \]

\[ \frac{dy}{dx} = 60x^4 - 21x^2 - 10 \]

Problem 3: Multiple Factors

Find \( \frac{dy}{dx} \) for \( y = (x^2 + 1)(x^3 - 2x)(x + 4) \)

Method 1: Multiply first, then differentiate

First multiply: \( (x^2 + 1)(x^3 - 2x) = x^5 - 2x^3 + x^3 - 2x = x^5 - x^3 - 2x \)

Then multiply by \( (x + 4) \): \( (x^5 - x^3 - 2x)(x + 4) = x^6 + 4x^5 - x^4 - 4x^3 - 2x^2 - 8x \)

Differentiate: \( \frac{dy}{dx} = 6x^5 + 20x^4 - 4x^3 - 12x^2 - 4x - 8 \)

Method 2: Extended Product Rule

Let \( u = x^2 + 1 \), \( v = x^3 - 2x \), \( w = x + 4 \)

\[ \frac{d}{dx}[uvw] = vw\frac{du}{dx} + uw\frac{dv}{dx} + uv\frac{dw}{dx} \]

\[ = (x^3 - 2x)(x + 4)(2x) + (x^2 + 1)(x + 4)(3x^2 - 2) + (x^2 + 1)(x^3 - 2x)(1) \]

(This expands to the same result as Method 1)

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