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Lesson 1: Product and Quotient Rules

Estimated time: 35-45 minutes

Learning Objectives

By the end of this lesson, you will be able to:

State and apply the Product Rule
State and apply the Quotient Rule
Recognize when each rule is needed
Combine these rules with the basic differentiation rules from Module 2

Why We Need New Rules

In Module 2 you learned the power rule, constant multiple rule, and sum/difference rule. These let you differentiate polynomials easily. But what about a function like f(x) = x² sin x? This is a product of two functions, and the derivative of a product is not simply the product of the derivatives.

Common Mistake

d/dx [f(x) · g(x)] ≠ f'(x) · g'(x). For example, d/dx [x · x] = d/dx [x²] = 2x, but (1)(1) = 1 ≠ 2x. We need the Product Rule.

The Product Rule

Product Rule

If f and g are differentiable, then:

d/dx [f(x) · g(x)] = f'(x) · g(x) + f(x) · g'(x)

In words: "derivative of the first times the second, plus the first times the derivative of the second."

Example 1: Basic Product Rule

Find d/dx [x² · sin x].

Step 1: Identify f(x) = x² and g(x) = sin x.

Step 2: Compute f'(x) = 2x and g'(x) = cos x.

Step 3: Apply the Product Rule:

d/dx [x² sin x] = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x

Example 2: Product of Three Factors

Find d/dx [x · e^x · cos x].

Strategy: Group as (x · e^x) · cos x. Apply the Product Rule twice, or use the extended formula:

d/dx [f · g · h] = f' g h + f g' h + f g h'

Solution:

= (1)(e^x)(cos x) + (x)(e^x)(cos x) + (x)(e^x)(−sin x)

= e^x cos x + x e^x cos x − x e^x sin x

= e^x(cos x + x cos x − x sin x)

The Quotient Rule

Quotient Rule

If f and g are differentiable and g(x) ≠ 0, then:

d/dx [f(x)/g(x)] = [f'(x) · g(x) − f(x) · g'(x)] / [g(x)]²

Memory aid: "Low d-high minus high d-low, all over low squared." (Here "high" = numerator, "low" = denominator.)

Example 3: Basic Quotient Rule

Find d/dx [(3x + 1)/(x² + 4)].

Step 1: f(x) = 3x + 1, g(x) = x² + 4.

Step 2: f'(x) = 3, g'(x) = 2x.

Step 3: Apply the Quotient Rule:

= [3(x² + 4) − (3x + 1)(2x)] / (x² + 4)²

= [3x² + 12 − 6x² − 2x] / (x² + 4)²

= (−3x² − 2x + 12) / (x² + 4)²

Example 4: Deriving the Tangent Derivative

Use the Quotient Rule to find d/dx [tan x].

Step 1: Write tan x = sin x / cos x.

Step 2: f(x) = sin x, g(x) = cos x, so f'(x) = cos x, g'(x) = −sin x.

Step 3:

= [(cos x)(cos x) − (sin x)(−sin x)] / cos² x

= [cos² x + sin² x] / cos² x

= 1 / cos² x = sec² x

When to Use Which Rule

Before reaching for the Quotient Rule, check if you can simplify first:

Constant in the denominator: d/dx [f(x)/5] = f'(x)/5. No Quotient Rule needed.
Rewrite as a product: d/dx [f(x)/x²] can be written as d/dx [f(x) · x⁻²] and handled with the Product Rule and power rule.
True quotient of two non-constant functions: Use the Quotient Rule.

Example 5: Simplify Before Differentiating

Find d/dx [(x³ + 2x)/x].

Simplify first: (x³ + 2x)/x = x² + 2.

Differentiate: d/dx [x² + 2] = 2x.

Much easier than using the Quotient Rule on the original form!

Example 6: Combining Product and Quotient Rules

Find d/dx [x² e^x / (x + 1)].

Strategy: Let the numerator be u = x² e^x and denominator v = x + 1.

First find u' using the Product Rule: u' = 2x e^x + x² e^x = e^x(2x + x²).

Then apply Quotient Rule:

= [e^x(2x + x²)(x + 1) − x² e^x(1)] / (x + 1)²

= e^x[x(2 + x)(x + 1) − x²] / (x + 1)²

Factor out e^x and simplify further: = e^x x(x² + 3x + 2 − x) / (x + 1)² = e^x x(x² + 2x + 2) / (x + 1)²

Key Takeaways

Product Rule: (fg)' = f'g + fg'. The derivative of a product is NOT the product of the derivatives.
Quotient Rule: (f/g)' = (f'g − fg') / g². Remember "low d-high minus high d-low."
Always check if you can simplify before applying a rule.
When multiple rules are needed, work inside out: find derivatives of components first, then combine.

Check Your Understanding

1. Find d/dx [x³ · ln x].

Answer: Using the Product Rule with f = x³, g = ln x: f' = 3x², g' = 1/x. So d/dx [x³ ln x] = 3x² ln x + x³(1/x) = 3x² ln x + x² = x²(3 ln x + 1).

2. Find d/dx [(2x − 5)/(x² + 1)].

Answer: f = 2x − 5, g = x² + 1, f' = 2, g' = 2x. Quotient Rule: [2(x² + 1) − (2x − 5)(2x)] / (x² + 1)² = [2x² + 2 − 4x² + 10x] / (x² + 1)² = (−2x² + 10x + 2) / (x² + 1)².

3. True or false: d/dx [f(x) · g(x)] = f'(x) · g'(x).

Answer: False. The correct formula is d/dx [f · g] = f' g + f g'. The product of derivatives is almost never equal to the derivative of the product.

4. Find d/dx [e^x / x²].

Answer: Quotient Rule: [e^x · x² − e^x · 2x] / x⁴ = e^x(x² − 2x) / x⁴ = e^x(x − 2) / x³.

Ready for More?

Next Lesson

In Lesson 2, you will learn the Chain Rule, the most widely used differentiation technique for composite functions.

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Module Progress

You have completed Lesson 1! Keep going to master all differentiation rules.

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