When two functions of x are multiplied, differentiate using the product rule: if y = uv, then dy/dx = u′v + uv′. You differentiate one factor at a time and add the results. In some questions, expanding the brackets first is quicker and gives the same answer.
This lesson follows differentiating powers and feeds into the chain rule. Both are needed as soon as a factor is a bracket raised to a power.
How does the product rule work, step by step?
- Name the factors: write u = … and v = ….
- Differentiate both separately: write u′ and v′ beside them.
- Substitute into u′v + uv′. Keep brackets around each part.
- Simplify, for example by factorising a common bracket.
Writing u, v, u′ and v′ in a small block keeps the structure clear, and stops you mixing which factor is which.
Worked example
Differentiate y = (2x + 1)(x² − 3x).
Method A, product rule. Let u = 2x + 1 and v = x² − 3x. Then u′ = 2 and v′ = 2x − 3.
dy/dx = u′v + uv′ = 2(x² − 3x) + (2x + 1)(2x − 3).
The first part is 2x² − 6x. The second part is 4x² − 6x + 2x − 3 = 4x² − 4x − 3.
Adding gives dy/dx = 6x² − 10x − 3.
Method B, expand first. y = 2x³ − 6x² + x² − 3x = 2x³ − 5x² − 3x.
Then dy/dx = 6x² − 10x − 3. Both methods agree, so the answer is confirmed.
For simple polynomial brackets, Method B is shorter. The product rule earns its place when expansion is awkward.
The mistake to watch for
The pull of “just differentiate each part and multiply” is strong, because it looks tidy.
Mistaken working: dy/dx = u′ × v′ = 2 × (2x − 3) = 4x − 6
The student multiplied the two derivatives.
Compare with the answer above: 4x − 6 has no x² term, but the original expression becomes a cubic when expanded, so its derivative must contain x². That quick degree check catches the error. The correct structure is u′v + uv′, two terms added.
Check yourself
1. Differentiate y = (x − 2)(x² + 5), then check by expanding.
Show answer
u = x − 2, v = x² + 5, so u′ = 1 and v′ = 2x.
dy/dx = (x² + 5) + 2x(x − 2) = x² + 5 + 2x² − 4x = 3x² − 4x + 5.
Check: expanding gives y = x³ − 2x² + 5x − 10, so dy/dx = 3x² − 4x + 5. ✓
2. Differentiate y = (3x + 1)√x and simplify to a single fraction.
Show answer
u = 3x + 1, v = x1/2, so u′ = 3 and v′ = ½x−1/2.
dy/dx = 3x1/2 + (3x + 1) × 1/(2√x) = 3√x + (3x + 1)/(2√x).
Put over 2√x: (6x + 3x + 1)/(2√x) = (9x + 1)/(2√x).
Check: y = 3x3/2 + x1/2 gives (9/2)x1/2 + (1/2)x−1/2, which is the same. ✓
3. Find the gradient of y = (x + 1)(x² − 4) at x = 2.
Show answer
u = x + 1, v = x² − 4, u′ = 1, v′ = 2x.
dy/dx = (x² − 4) + (x + 1)(2x). At x = 2: 0 + 3 × 4 = 12.
Check: y = x³ + x² − 4x − 4, so dy/dx = 3x² + 2x − 4 = 12 + 4 − 4 = 12. ✓
Where this leads next
The next step is the chain rule for composite expressions, which lets you deal with factors like (2x − 1)³. The calculus shape and rate explorer and the non-calculator working trainer are useful for practice.
Choosing between expanding and using a rule is a judgement skill, and it is one we work on directly in online one-to-one Additional Mathematics tuition.