Product Rule (Edexcel A Level Maths: Pure): Revision Note
Exam code: 9MA0
Product rule
What is the product rule?
The product rule is a formula that allows you to differentiate a product of two functions
If where u and v are functions of x then the product rule is:
In function notation, if then the product rule can be written as:
The easiest way to remember the product rule is, for where u and v are functions of x:
Examiner Tips and Tricks
The product rule formulae are NOT in the formulae booklet – you need to know them.
Don't confuse the product of two functions with a composite function:
The product of two functions is two functions multiplied together
A composite function is a function of a function
To differentiate composite functions you need to use the chain rule
Worked Example
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Questions students ask about Product Rule
Real questions from students revising this topic on Save My Exams
What is the difference between the product rule and the chain rule?
Use the product rule when two functions are multiplied together, and the chain rule when one function sits inside another.
- A product is two functions multiplied: with f(x) = 3x² and g(x) = x − 3, the product is 3x²(x − 3)
- Differentiate a product with y = uv ⇒ dy/dx = u(dv/dx) + v(du/dx), or y′ = uv′ + vu′
- A composite function is a function of a function: the same two functions give fg(x) = f(g(x)) = 3(x − 3)²
- Differentiate a composite with the chain rule
Quick test: can you split the expression into two chunks joined by a multiplication sign? Product rule. Is there a bracket, power, eˣ, ln or trig function with something more complicated than x inside it? Chain rule.
You often need both. In this page's worked example y = (5 sin x − 7 cos x)e³ˣ⁺² is a product, but differentiating v = e³ˣ⁺² needs the chain rule, giving dv/dx = 3e³ˣ⁺².
When do you need to use the product rule?
Use the product rule whenever y is two functions of x multiplied together, especially when you cannot easily expand them.
Typical products that need it:
- A power of x times an exponential: y = xeˣ gives dy/dx = xeˣ + eˣ
- A power of x times a log: y = 5x ln 5x gives dy/dx = 5 + 5 ln 5x
- A trig expression times an exponential: y = (5 sin x − 7 cos x)e³ˣ⁺²
How to use it:
- Split y into u and v
- Differentiate each to get du/dx and dv/dx
- Substitute into dy/dx = u(dv/dx) + v(du/dx)
Both parts must contain x. You do not need it for something like y = 5eˣ, where one factor is a constant, or for y = e⁵ˣ, which is composite and needs the chain rule.
If both parts are polynomials you can expand instead: f(x) = x³(x² + 7) gives 5x⁴ + 21x² either way.
How do you factorise your answer when using the product rule?
Look for what both terms have in common — usually an exponential, a number or a bracket — take it outside a bracket, then collect like terms inside.
- Write out u(dv/dx) + v(du/dx) in full
- Spot the common factor: eˣ, e³ˣ⁺², a number, a power of x or a repeated bracket
- Take it outside and simplify what is left
Worked example from this page: y = (5 sin x − 7 cos x)e³ˣ⁺²
- dy/dx = (5 sin x − 7 cos x)(3e³ˣ⁺²) + (e³ˣ⁺²)(5 cos x + 7 sin x)
- Both terms contain e³ˣ⁺², so take it out: e³ˣ⁺²[3(5 sin x − 7 cos x) + 5 cos x + 7 sin x]
- Expand and collect: 15 sin x + 7 sin x = 22 sin x, and −21 cos x + 5 cos x = −16 cos x
- dy/dx = (22 sin x − 16 cos x)e³ˣ⁺²
Simpler ones factorise the same way: xeˣ + eˣ = eˣ(x + 1).
