The product rule differentiates two functions multiplied together. If y = uv, then dy/dx = u′v + uv′: differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second. Writing out u, v, u′ and v′ before you start keeps the two terms straight, and factorising the answer makes later steps much easier.
The product rule
- If y = uv, then dy/dx = u′v + uv′
- Use the chain rule to find u′ or v′ if either is a composite
- √x = x^(1/2) and √(ax + b) = (ax + b)^(1/2)
Factorise the answer by taking out the lowest power of any repeated bracket.
What is the product rule?
When a function is a product of two parts, such as x²(2x − 1) or x(2x + 1)⁵, the product rule differentiates it without expanding. Label u and v, find u′ and v′, then write u′v + uv′. Setting out the four pieces in a small table first stops you mixing them up.
The product rule with the chain rule
When one factor is itself a composite, use the chain rule to find its derivative before substituting. For example, v = (2x + 1)⁵ gives v′ = 5(2x + 1)⁴ × 2 = 10(2x + 1)⁴.
Then tidy the answer by factorising out the common bracket. Take out the lowest power of the repeated bracket; exam marks reward the factorised form.
Products with roots
Write any root as a power before differentiating: √x = x^(1/2), and √(2x + 1) = (2x + 1)^(1/2). After using the product rule, combine the two terms over a common denominator to give a single fraction.
Stationary points of a product
With f′(x) found, the usual follow-up tasks apply. Solving f′(x) = 0 is much easier when the derivative is factorised, because each factor gives a stationary point directly.
Products with eˣ, ln x and trig
The product rule works with any two functions. Many questions multiply a polynomial by eˣ, ln x or a trig function. You still need the product rule: you cannot differentiate the two parts separately and multiply the results.
Worked examples
Example 1. Differentiate y = x²(2x − 1) using the product rule.
- u = x², v = 2x − 1, u′ = 2x, v′ = 2.
- u′v + uv′ = 2x(2x − 1) + 2x².
- Simplify: 4x² − 2x + 2x² = 6x² − 2x.
Answer: dy/dx = 6x² − 2x.
Marks: M1 for u′v + uv′, A1 for the answer.
Example 2. Differentiate y = x(2x + 1)⁵, giving a factorised answer.
- u = x, v = (2x + 1)⁵, u′ = 1, v′ = 10(2x + 1)⁴.
- u′v + uv′ = (2x + 1)⁵ + 10x(2x + 1)⁴.
- Take out (2x + 1)⁴: (2x + 1)⁴[(2x + 1) + 10x].
Answer: dy/dx = (2x + 1)⁴(12x + 1).
Marks: M1 for the chain rule on v, M1 for factorising, A1 for (2x + 1)⁴(12x + 1).
Example 3. Differentiate y = x√(2x + 1), giving a single fraction.
- u = x, v = (2x + 1)^(1/2), u′ = 1, v′ = (2x + 1)^(−1/2).
- u′v + uv′ = (2x + 1)^(1/2) + x(2x + 1)^(−1/2).
- Combine over (2x + 1)^(1/2): ((2x + 1) + x) / √(2x + 1).
Answer: dy/dx = (3x + 1) / √(2x + 1).
Marks: M1 for the product rule with the correct v′, A1 for combining into one fraction.
Example 4. Find the x-coordinates of the stationary points of y = x(x − 6)².
- u = x, v = (x − 6)², u′ = 1, v′ = 2(x − 6).
- f′(x) = (x − 6)² + 2x(x − 6) = (x − 6)(3x − 6) = 3(x − 6)(x − 2).
- Set f′(x) = 0.
Answer: x = 2 and x = 6.
Marks: M1 for the product rule, M1 for factorising, A1 for both values.
Example 5. Differentiate y = x³(x + 2)², factorising fully.
- u′ = 3x², v′ = 2(x + 2).
- 3x²(x + 2)² + 2x³(x + 2).
- Take out x²(x + 2): x²(x + 2)[3(x + 2) + 2x].
Answer: dy/dx = x²(x + 2)(5x + 6).
Marks: M1 for u′v + uv′, M1 for the common factor, A1 for the answer.
Common mistakes
- Multiplying the derivatives. The derivative of uv is u′v + uv′, not u′v′.
- Forgetting the chain rule inside. If v = (2x + 1)⁵, then v′ = 10(2x + 1)⁴, not 5(2x + 1)⁴.
- Expanding instead of factorising. Take out the common bracket; it makes f′(x) = 0 easy to solve.
- Taking out the wrong power. Factor out the lowest power of the shared bracket.
- Leaving roots as roots. Rewrite √(2x + 1) as (2x + 1)^(1/2) before differentiating.
How the product rule is examined
The product rule is tested on the Pure 3 paper (WMA13) of Edexcel International A-Level, and in second-year Pure for Edexcel A-Level (9MA0). A typical question asks you to:
- show that the derivative of a product can be written in a given factorised form
- find the coordinates of the stationary points
- find a tangent or normal at a given point
- combine the product rule with eˣ, ln x or trig functions
Practise with the Pure 3 differentiation exam questions, which have full mark schemes, and the Pure 3 differentiation worksheet. The differentiation revision engine creates new questions on demand.
The product rule relies on the chain rule; its partner for fractions is the quotient rule, and it is reversed in integration by parts.
Frequently asked questions
What is the product rule?
If y = uv, then dy/dx = u′v + uv′. Differentiate each part in turn, keeping the other fixed, and add.
When should I use the product rule?
When two functions of x are multiplied together and expanding is impractical, such as x(2x + 1)⁵ or x²eˣ.
Does the order matter in the product rule?
No. u′v + uv′ is the same as uv′ + u′v, because addition is commutative. Order does matter in the quotient rule.