What is the Product Rule?
The product rule is a fundamental differentiation technique used when you need to find the derivative of two functions multiplied together. It's one of the essential rules in calculus, alongside the chain rule and quotient rule.
Whenever you see two separate functions multiplied together—like x² · sin(x) or e^x · ln(x)—you need the product rule to differentiate them correctly.
🎯 In Simple Terms: The product rule tells you how to take the derivative when you have multiplication of two (or more) functions. You CAN'T just differentiate each function separately and multiply—that's wrong!
The Product Rule Formula
If you have two functions f(x) and g(x) multiplied together, the product rule states:
Or in Leibniz notation, if y = u · v:
Breaking Down the Formula
- f' · g - Derivative of the first function times the second function (unchanged)
- f · g' - First function (unchanged) times the derivative of the second
- Plus sign (+) - Add these two terms together
💡 Memory Trick: "First times derivative of second, PLUS second times derivative of first" — or simply remember it as: "f'g + fg'"
Some students remember: "Left d-right plus right d-left" (where "d" means derivative)
Why Can't We Just Multiply the Derivatives?
A common beginner mistake is thinking (f · g)' = f' · g'. This is WRONG! Here's why:
❌ Wrong Approach:
If f(x) = x² and g(x) = x³, then:
f'(x) = 2x and g'(x) = 3x²
Wrong answer: f'(x) · g'(x) = 2x · 3x² = 6x³
✅ Correct Approach (using product rule):
(f · g)' = f'g + fg' = (2x)(x³) + (x²)(3x²) = 2x⁴ + 3x⁴ = 5x⁴
Verification: f(x) · g(x) = x² · x³ = x⁵, so the derivative should be 5x⁴ ✅
When Do You Use the Product Rule?
Use the product rule whenever you have:
- Two (or more) functions multiplied together: x² · sin(x), e^x · cos(x)
- Polynomial times trig function: x³ · tan(x)
- Exponential times polynomial: e^x · x²
- Trig times trig: sin(x) · cos(x)
- Logarithm times polynomial: ln(x) · x²
⚠️ Don't Confuse With:
Chain Rule: One function INSIDE another → sin(x²)
Product Rule: Two functions MULTIPLIED → sin(x) · x²
Step-by-Step Process
- Identify the two functions being multiplied (call them f and g)
- Find f' - derivative of the first function
- Find g' - derivative of the second function
- Apply the formula: f'g + fg'
- Simplify your answer
Example 1: Basic Product Rule
f(x) = x²
g(x) = sin(x)
f'(x) = 2x
g'(x) = cos(x)
dy/dx = f'g + fg'
dy/dx = (2x)(sin x) + (x²)(cos x)
dy/dx = 2x sin(x) + x² cos(x)
Example 2: Product Rule with Exponential
Solution:
f(x) = e^x, so f'(x) = e^x
g(x) = x³, so g'(x) = 3x²
Apply product rule:
dy/dx = (e^x)(x³) + (e^x)(3x²)
Factor out e^x:
dy/dx = e^x(x³ + 3x²) = e^x·x²(x + 3)
Example 3: Product Rule with Logarithm
Solution:
f(x) = ln(x), so f'(x) = 1/x
g(x) = x² + 1, so g'(x) = 2x
Apply product rule:
dy/dx = (1/x)(x² + 1) + (ln x)(2x)
Simplify first term:
dy/dx = (x² + 1)/x + 2x ln(x)
dy/dx = x + 1/x + 2x ln(x)
Example 4: Three Functions
For three functions, apply the product rule twice!
Method: Group two functions first: (x² · sin x) · e^x
Step 1: Let u = x² · sin(x), find u' using product rule:
u' = 2x sin(x) + x² cos(x)
Step 2: Now use product rule on u · e^x:
dy/dx = u' · e^x + u · e^x
dy/dx = [2x sin(x) + x² cos(x)] · e^x + [x² sin(x)] · e^x
Final Answer:
dy/dx = e^x[2x sin(x) + x² cos(x) + x² sin(x)]
dy/dx = e^x[x² sin(x) + 2x sin(x) + x² cos(x)]
Product Rule Combined with Chain Rule
Often, you'll need to use BOTH the product rule AND the chain rule in the same problem:
Here, sin(3x) requires the chain rule!
f(x) = x², so f'(x) = 2x
g(x) = sin(3x), so g'(x) = cos(3x) · 3 = 3cos(3x) [chain rule!]
Apply product rule:
dy/dx = (2x)(sin 3x) + (x²)(3cos 3x)
dy/dx = 2x sin(3x) + 3x² cos(3x)
Quick Reference Table
| Function y | Derivative dy/dx | Notes |
|---|---|---|
| x · sin(x) | sin(x) + x cos(x) | Basic product rule |
| x² · e^x | 2x e^x + x² e^x = e^x(2x + x²) | Factor out e^x |
| x · ln(x) | ln(x) + x · (1/x) = ln(x) + 1 | Simplifies nicely |
| x³ · cos(x) | 3x² cos(x) - x³ sin(x) | Watch the signs! |
| (x + 1) · (x - 1) | (1)(x - 1) + (x + 1)(1) = 2x | Or expand first: x² - 1 |
| sin(x) · cos(x) | cos²(x) - sin²(x) = cos(2x) | Trig identity |
Common Mistakes to Avoid
Mistake #1: Multiplying Derivatives
❌ Wrong: (f · g)' = f' · g'
✅ Correct: (f · g)' = f' · g + f · g'
You must use the product rule formula!
Mistake #2: Forgetting One Term
❌ Wrong: d/dx[x² · sin x] = 2x sin(x) [missing second term!]
✅ Correct: d/dx[x² · sin x] = 2x sin(x) + x² cos(x)
Remember: TWO terms in the product rule!
Mistake #3: Not Applying Chain Rule When Needed
❌ Wrong: d/dx[x · sin(2x)] = sin(2x) + x cos(2x)
✅ Correct: d/dx[x · sin(2x)] = sin(2x) + x · [cos(2x) · 2] = sin(2x) + 2x cos(2x)
Don't forget: sin(2x) needs the chain rule!
Practice Problems
Test your understanding with these practice problems. Try them yourself before checking solutions!
Find the derivative of: y = x³ · cos(x)
Show Solution
f = x³, f' = 3x²
g = cos(x), g' = -sin(x)
dy/dx = f'g + fg' = (3x²)(cos x) + (x³)(-sin x)
Answer: dy/dx = 3x² cos(x) - x³ sin(x)
Find the derivative of: y = (2x + 1) · e^x
Show Solution
f = 2x + 1, f' = 2
g = e^x, g' = e^x
dy/dx = (2)(e^x) + (2x + 1)(e^x)
Factor: dy/dx = e^x[2 + 2x + 1]
Answer: dy/dx = e^x(2x + 3)
Find the derivative of: y = √x · ln(x)
Show Solution
f = √x = x^(1/2), f' = (1/2)x^(-1/2) = 1/(2√x)
g = ln(x), g' = 1/x
dy/dx = [1/(2√x)] · ln(x) + √x · (1/x)
Simplify second term: √x/x = 1/√x
Answer: dy/dx = ln(x)/(2√x) + 1/√x
Find the derivative of: y = x · sin(x) · cos(x)
Show Solution
Group as (x · sin x) · cos x, or use product rule for 3 functions
Let u = x · sin(x), then u' = sin(x) + x cos(x)
dy/dx = u' · cos(x) + u · (-sin x)
dy/dx = [sin(x) + x cos(x)] · cos(x) + [x sin(x)] · (-sin x)
dy/dx = sin(x)cos(x) + x cos²(x) - x sin²(x)
Answer: dy/dx = sin(x)cos(x) + x(cos²x - sin²x)
Or using identity: dy/dx = (1/2)sin(2x) + x cos(2x)
Tips for Mastering the Product Rule
- Always identify both functions first - Label them f and g before starting
- Write out each derivative - Calculate f' and g' separately
- Remember the pattern: "first-d-second + second-d-first"
- Watch for chain rule opportunities - Many functions need both rules
- Factor when possible - Makes answers cleaner (especially with e^x)
- Practice, practice, practice - The more you do, the more automatic it becomes
- Check your work - Use our calculator to verify your answers
Product Rule vs Other Differentiation Rules
Understanding when to use each rule is crucial:
| Situation | Rule to Use | Example |
|---|---|---|
| Two functions multiplied | Product Rule | x² · sin(x) |
| One function inside another | Chain Rule | sin(x²) |
| Two functions divided | Quotient Rule | x² / sin(x) |
| Single basic function | Basic Rules | x², sin(x), e^x |
| Two functions added | Sum Rule (differentiate each) | x² + sin(x) |
Real-World Applications
The product rule appears frequently in real-world problems:
Physics - Force and Motion
If force F(t) = m(t) · a(t) where both mass and acceleration change with time:
dF/dt = dm/dt · a + m · da/dt
Economics - Revenue
Revenue R = price × quantity. If R(x) = p(x) · q(x):
Marginal Revenue: dR/dx = p'(x) · q(x) + p(x) · q'(x)
Biology - Population Density
If population P(t) = density(t) × area(t):
Growth rate: dP/dt = d(density)/dt · area + density · d(area)/dt
Special Cases and Shortcuts
When One Function is Constant
If f(x) = c (constant), then f' = 0, so:
(c · g)' = 0 · g + c · g' = c · g'
Example: d/dx[5 · sin(x)] = 5 · cos(x)
This is just the constant multiple rule!
Product of Polynomial Terms
For (ax^m)(bx^n), you can either:
- Multiply first: abx^(m+n), then differentiate: ab(m+n)x^(m+n-1)
- Use product rule: (max^(m-1))(bx^n) + (ax^m)(nbx^(n-1))
Both give the same answer, but multiplying first is usually faster!
When Both Functions are e^x
(e^x · e^x)' can be done as:
- Product rule: e^x · e^x + e^x · e^x = 2e^(2x)
- Or simplify first: (e^x)² = e^(2x), then differentiate: 2e^(2x)
🧮 Practice with Our Product Rule Calculator
Ready to verify your solutions? Use our free product rule calculator for instant, step-by-step answers!
Try Product Rule Calculator →Advanced: Logarithmic Differentiation
For complex products of many functions, logarithmic differentiation can be easier than repeated product rule application:
Using logarithmic differentiation:
ln(y) = ln(x²) + ln(sin x) + ln(e^x) + ln(ln x)
ln(y) = 2ln(x) + ln(sin x) + x + ln(ln x)
Differentiate both sides:
(1/y) · dy/dx = 2/x + cos(x)/sin(x) + 1 + 1/(x ln x)
dy/dx = y · [2/x + cot(x) + 1 + 1/(x ln x)]
Substitute y back in for the final answer!
Summary: Key Takeaways
- ✅ Product rule formula: (f · g)' = f'g + fg'
- ✅ Memory trick: "First d-second plus second d-first"
- ✅ Use when: Two or more functions are multiplied
- ✅ Don't multiply derivatives: (fg)' ≠ f'g'
- ✅ Often combines with: Chain rule, especially for composite functions
- ✅ For three+ functions: Apply product rule multiple times
- ✅ Practice is essential: The pattern becomes automatic with repetition
What's Next?
Continue building your differentiation skills:
- 📚 Master the Chain Rule - For nested functions
- 📚 Learn the Quotient Rule - For divided functions
- 📚 Understand Implicit Differentiation - Advanced technique
- 🧮 Try All Our Calculators - 15+ free tools
Frequently Asked Questions
Q: When do I use product rule vs quotient rule?
A: Use product rule when functions are MULTIPLIED (f · g). Use quotient rule when functions are DIVIDED (f / g). However, you can rewrite division as multiplication and use product rule: f/g = f · g^(-1).
Q: Can I use product rule for more than two functions?
A: Yes! For three functions, group two together first and apply product rule twice. For f·g·h: first find (f·g)', then use product rule on (f·g)·h.
Q: What if one function is just x?
A: Still use product rule! For y = x · sin(x): dy/dx = (1)(sin x) + (x)(cos x) = sin(x) + x cos(x).
Q: Do I always need to simplify my answer?
A: While not always required, simplifying makes your answer cleaner and easier to use. Factor out common terms when possible, especially e^x.
Q: How do I check if my answer is correct?
A: Use our product rule calculator to verify! It shows step-by-step solutions so you can identify any mistakes.
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