Differentiate one, keep the other — both ways, then add: To differentiate a product y = u·v: (uv)' = u'v + uv' — differentiate the first times the second, plus the first times the derivative of the second.
IB-style question — a product
Differentiate y = (2x + 1)(x² − 3).
Step by step
- Let u = 2x+1 (u' = 2), v = x²−3 (v' = 2x). Apply u'v + uv'.
- Expand and collect.
Final answer
dy/dx = 6x² + 2x − 6.
Label u, v, u', v' first: Write out u, v, u', v' before substituting — it stops you mixing up the four pieces.
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Substitute carefully, then simplify: Identify u and v, find u' and v', substitute into u'v + uv', and simplify.
Leaving the answer factorised is often neatest.
IB-style question — product with a power
Differentiate y = x²(2x − 5).
Step by step
- u = x² (u' = 2x), v = 2x − 5 (v' = 2).
- Expand and collect.
Final answer
dy/dx = 6x² − 10x.
Check by expanding first: For simple products you can expand then differentiate to check (here x²(2x−5) = 2x³−5x² → 6x²−10x).
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One factor may need the chain rule: When a factor is itself a composite (e.g. e2x or sin x), differentiate that factor with the chain rule while applying the product rule.
IB-style question — product needing the chain rule
Differentiate y = x·e2x.
Step by step
- u = x (u' = 1), v = e2x (v' = 2e2x by the chain rule).
- Factorise.
Final answer
dy/dx = e2x(1 + 2x).
Factor out the common term: Products with eˣ or a bracket usually factorise neatly — pull out the common factor for a tidy answer.