Differentiate products of functions
Apply the product rule to two differentiable functions and simplify without differentiating each factor in isolation.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If y = uv, what remains unchanged in the first product-rule term?
What you will learn
- Use d(uv)/dx = u′v + uv′
- Choose and label the two factors before differentiating
- Combine the product rule with standard derivative and chain rules
Differentiate one factor at a time
For y = u(x)v(x), the product rule is y′ = u′v + uv′. Differentiate the first factor and leave the second unchanged, then leave the first unchanged and differentiate the second.
This rule is not u′v′. Labeling u and v before substituting helps preserve the two terms and their signs.
Use the rule even when one factor is a composite
For y = x² sin x, take u = x² and v = sin x. Then y′ = 2x sin x + x² cos x.
If one factor contains ax + b, apply the chain rule inside that derivative. Simplify only after both product-rule terms are written.
Pause and explain
Differentiate x sin x.
Worked example
Differentiate y = (x² + 1)eˣ.
Show the worked solution
- Let u = x² + 1 and v = eˣ, so u′ = 2x and v′ = eˣ.
- Apply y′ = u′v + uv′: 2xeˣ + (x² + 1)eˣ.
- Factor out eˣ if desired.
Answer dy/dx = eˣ(x² + 2x + 1) = eˣ(x + 1)².
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Differentiate y = x³ cos x.
Give me a hint
Set u = x³ and v = cos x.
Compare my reasoning
- u′ = 3x² and v′ = −sin x.
- Apply the rule: y′ = 3x² cos x + x³(−sin x).
- Keep the negative sign in the second term.
dy/dx = 3x² cos x − x³ sin x.
Look for these in your work
- Factors labelled
- Both terms included
- Cosine sign retained
Differentiate y = (2x + 1)ln x for x > 0.
Give me a hint
Use the product rule, then differentiate each factor.
Compare my reasoning
- u′ = 2 and v′ = 1/x.
- y′ = 2 ln x + (2x + 1)/x.
- Combine terms if desired.
dy/dx = 2 ln x + 2 + 1/x, for x > 0.
Look for these in your work
- Product-rule terms correct
- Logarithm domain stated
- Algebra simplified
A model is R(x) = x e^(−x). Find R′(x) and the stationary input where R′(x) = 0.
Give me a hint
Use the product rule and factor the result.
Compare my reasoning
- R′ = 1·e^(−x) + x·(−e^(−x)).
- Factor e^(−x): R′ = e^(−x)(1 − x).
- Since e^(−x) is never zero, the stationary input is x = 1.
R′(x) = e^(−x)(1 − x); stationary input x = 1.
Look for these in your work
- Product and chain rules combined
- Common factor identified
- Nonzero exponential used to solve
Common mistakes
- Multiplying the two derivatives u′v′
- Forgetting one of the two product-rule terms
- Differentiating a factor twice
- Dropping the chain factor inside a composite factor
What can you explain now?
What is the product rule for u(x)v(x)?
Compare with the explanation
u′v + uv′
Differentiate each factor once while keeping the other factor unchanged.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How can you remember the two terms in the product rule, and what does each term differentiate?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.