Product Rule: Complete Tutorial

📅 Updated: November 28, 2024 ⏱️ 22 min read 🎯 Essential Technique

Learn how to differentiate products of functions with detailed examples and common mistakes to avoid. Master the product rule with our comprehensive guide featuring intuitive explanations, 20+ worked examples, and proven strategies for calculus success.

🔥 Why the Product Rule Matters

The Product Rule is essential whenever you need to differentiate two functions multiplied together. It appears constantly in physics (force × distance), economics (price × quantity), and engineering. Master this rule to unlock advanced calculus!

🎯 What is the Product Rule?

The Product Rule allows us to differentiate products of two (or more) functions multiplied together.

When to Use the Product Rule

Use the Product Rule whenever you have two functions multiplied together:

💡 Examples Requiring Product Rule

x² · sin(x) → polynomial times trig function
e^x · ln(x) → exponential times logarithm
(x + 1) · (x² - 3) → polynomial times polynomial
x · cos(x) → simple product

❌ Common Misconception

d/dx[x² · sin(x)] = 2x · cos(x) ← WRONG!

You CANNOT just differentiate each function separately. You MUST use the Product Rule!

💡 Quick Recognition Test

Ask yourself: "Are two functions being multiplied together?" If yes, use the Product Rule!

📐 The Formula & Intuition

The Product Rule Formula

d/dx[u · v] = u' · v + u · v'

In Words: "First times derivative of second, PLUS second times derivative of first"

Alternative Notation

(fg)' = f'g + fg'

The 3-Step Process

🎯 Product Rule in 3 Steps

1 Identify the two functions

Label the first function as u (or f) and the second as v (or g)

2 Find the derivative of each

Calculate u' and v' separately

3 Apply the formula: u'v + uv'

First derivative times second original, PLUS first original times second derivative

💡 Memory Aid: "Derivative Dance"

First derivative, second stays → PLUS → First stays, second derivative

Think: "Derivative dances with original, then original dances with derivative!"

🌱 Basic Product Rule Examples

Let's start with simple examples to build confidence.

📝 Example 1: Polynomial × Polynomial EASY

Find: d/dx[x² · (x + 1)]

1 Identify:

u = x²
v = x + 1

2 Find derivatives:

u' = 2x
v' = 1

3 Apply formula: u'v + uv'

= 2x(x + 1) + x²(1)
= 2x² + 2x + x²
= 3x² + 2x

📝 Example 2: Polynomial × Trig EASY

Find: d/dx[x · sin(x)]

1 Identify: u = x, v = sin(x)

2 Find derivatives: u' = 1, v' = cos(x)

3 Apply formula:

= (1)·sin(x) + x·cos(x)
= sin(x) + x·cos(x)

📝 Example 3: Polynomial × Exponential EASY

Find: d/dx[x² · e^x]

Solution: u = x², v = e^x → u' = 2x, v' = e^x

Answer: 2x·e^x + x²·e^x = e^x(2x + x²)

📝 Example 4: Trig × Trig EASY

Find: d/dx[sin(x) · cos(x)]

Solution: u = sin(x), v = cos(x) → u' = cos(x), v' = -sin(x)

Answer: cos²(x) - sin²(x)

⚡ Intermediate Product Rule Examples

📝 Example 5: Product + Chain Rule MEDIUM

Find: d/dx[x² · sin(3x)]

Note: sin(3x) requires Chain Rule!

Solution:

u = x², u' = 2x
v = sin(3x), v' = 3cos(3x) (Chain Rule!)
= 2x·sin(3x) + x²·3cos(3x)
= 2x·sin(3x) + 3x²·cos(3x)

📝 Example 6: e^x × Polynomial MEDIUM

Find: d/dx[e^x · (x³ - 2x + 5)]

Solution:

u = e^x, u' = e^x
v = x³ - 2x + 5, v' = 3x² - 2
= e^x(x³ - 2x + 5) + e^x(3x² - 2)
= e^x(x³ + 3x² - 2x + 3)

📝 Example 7: ln(x) × Polynomial MEDIUM

Find: d/dx[x² · ln(x)]

Solution: 2x·ln(x) + x²·(1/x) = 2x·ln(x) + x

📝 Example 8: Three Functions Product MEDIUM

Find: d/dx[x · sin(x) · e^x]

For three functions: Apply Product Rule twice!

Solution: (x·sin(x))'·e^x + (x·sin(x))·e^x

= [sin(x) + x·cos(x)]·e^x + x·sin(x)·e^x

= e^x[sin(x) + x·cos(x) + x·sin(x)]

🚀 Advanced Product Rule Applications

📝 Example 9: Product + Quotient Rule HARD

Find: d/dx[x² · (sin(x)/x)]

Simplify first: x² · sin(x)/x = x · sin(x)

Answer: sin(x) + x·cos(x)

📝 Example 10: Nested Product & Chain HARD

Find: d/dx[e^(x²) · cos(2x)]

Solution:

u = e^(x²), u' = 2x·e^(x²)
v = cos(2x), v' = -2sin(2x)
= 2x·e^(x²)·cos(2x) - 2e^(x²)·sin(2x)

📝 Example 11: Implicit Differentiation HARD

Find dy/dx: xy + y² = 5

Differentiate both sides (use Product Rule on xy):

(1)y + x(dy/dx) + 2y(dy/dx) = 0

Answer: dy/dx = -y/(x + 2y)

⚠️ Common Mistakes to Avoid

❌ Mistake #1: Differentiating Each Function Separately

Wrong: d/dx[x² · sin(x)] = 2x · cos(x)

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

Why: You MUST use the formula u'v + uv', not just u'· v'!

❌ Mistake #2: Forgetting the Second Term

Wrong: d/dx[x · e^x] = e^x (forgot the u'v term)

Right: d/dx[x · e^x] = (1)·e^x + x·e^x = e^x(1 + x)

Why: Product Rule has TWO terms: u'v + uv'

❌ Mistake #3: Wrong Order/Sign

Wrong: d/dx[x · sin(x)] = x·cos(x) + sin(x) (reversed order)

Order doesn't matter for addition, but consistency helps! Stick to u'v + uv'

❌ Mistake #4: Forgetting Chain Rule Within Product Rule

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

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

Why: sin(2x) derivative is 2cos(2x) (Chain Rule!)

🎯 Practice Problems

Practice Set 1: Basic EASY

d/dx[x · cos(x)]
d/dx[x³ · e^x]
d/dx[(x + 1) · (x - 2)]
d/dx[x² · ln(x)]

✅ Solutions

cos(x) - x·sin(x)
3x²·e^x + x³·e^x = e^x(x³ + 3x²)
2x - 1
2x·ln(x) + x

Practice Set 2: Intermediate MEDIUM

d/dx[x² · sin(2x)]
d/dx[e^x · cos(x)]
d/dx[(x² + 1) · ln(x)]
d/dx[x · tan(x)]

✅ Solutions

2x·sin(2x) + 2x²·cos(2x)
e^x·cos(x) - e^x·sin(x) = e^x(cos(x) - sin(x))
2x·ln(x) + (x² + 1)/x
tan(x) + x·sec²(x)

Practice Set 3: Advanced HARD

d/dx[x·e^(x²)]
d/dx[sin(x)·cos(2x)]
d/dx[x²·ln(x²)]
d/dx[e^x·sin(x)·cos(x)]

✅ Solutions

e^(x²) + 2x²·e^(x²) = e^(x²)(1 + 2x²)
cos(x)·cos(2x) - 2sin(x)·sin(2x)
2x·ln(x²) + 2x
e^x·sin(x)·cos(x) + e^x·cos²(x) - e^x·sin²(x)

🚀 Practice with Our Calculator

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📚 Key Takeaways

Master the Product Rule!

Formula: (uv)' = u'v + uv'
Memory: "First derivative times second, PLUS first times second derivative"
Practice: Do 20-30 problems to build fluency
Combine: Often used WITH Chain Rule - watch for composite functions!