Product and Quotient Rules

Product Rule and Quotient Rule

By the end of this lesson, you will be able to apply the product rule and quotient rule to differentiate products and quotients of functions, identify when each rule is needed, and avoid common errors.

Why We Need New Rules

When two functions are multiplied or divided, you CANNOT simply differentiate each part separately.

Common mistake: d/dx[f(x)·g(x)] ≠ f'(x)·g'(x)

Example of why this fails: Let f(x) = x² and g(x) = x³. Their product is h(x) = x⁵.

• Correct: h'(x) = 5x⁴
• Incorrect approach: f'(x)·g'(x) = 2x · 3x² = 6x³ ≠ 5x⁴

You need the product rule.

The Product Rule

If h(x) = f(x)·g(x), then:

$$h'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$$

Memory device: "First times derivative of second, plus second times derivative of first."

Or: d(uv) = u'v + uv'

Examples

Example 1: h(x) = x³ · sin(x)

• f(x) = x³, f'(x) = 3x²
• g(x) = sin(x), g'(x) = cos(x)

h'(x) = 3x² · sin(x) + x³ · cos(x)

Example 2: h(x) = (2x + 1)(x² - 3)

• f(x) = 2x + 1, f'(x) = 2
• g(x) = x² - 3, g'(x) = 2x

h'(x) = 2(x² - 3) + (2x + 1)(2x) = 2x² - 6 + 4x² + 2x = 6x² + 2x - 6

Verification: Expand first: h(x) = 2x³ - 6x + x² - 3 = 2x³ + x² - 6x - 3 h'(x) = 6x² + 2x - 6 ✓

Example 3: h(x) = e^x · (x² + 5)

h'(x) = eˣ(x² + 5) + eˣ(2x) = eˣ(x² + 2x + 5)

The Quotient Rule

If h(x) = f(x)/g(x), then:

$$h'(x) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}$$

Memory device: "Low d-High minus High d-Low, square the bottom and away we go."

Low = g(x), High = f(x), d-High = f'(x), d-Low = g'(x)

$$h'(x) = \frac{\text{Low} \cdot \text{d(High)} - \text{High} \cdot \text{d(Low)}}{(\text{Low})^2}$$

Examples

Example 1: h(x) = x²/(x + 1)

• f(x) = x², f'(x) = 2x (High)