Chain Rule, Product Rule and Quotient Rule: An A Level Guide
· Webrich Software · 6 min read
Differentiating xⁿ, eˣ, ln x, sin x and cos x on their own is the easy part of A level calculus. The marks are in differentiating combinations of them — a function inside another function, two functions multiplied together, or one divided by another. That’s what the chain rule, product rule and quotient rule are for.
This guide shows how to recognise which rule a question needs, works through examples step by step, and lists the errors examiners see most often. All three rules are Year 2 (A level) content on every board.
Step one: recognise the structure
Before you differentiate anything, ask: what is the last thing I’d do if I were evaluating this function for a number?
| If the function is… | Example | Use |
|---|---|---|
| a function of a function | (3x − 1)⁵, e^(sin x), ln(x² + 1) | Chain rule |
| two functions multiplied | x² e^(3x), x sin x | Product rule |
| one function divided by another | (2x + 1)/(x − 3), sin x / x | Quotient rule |
Many exam questions need more than one rule. In x(2x + 1)⁴, the last operation is a multiplication (product rule), but differentiating (2x + 1)⁴ itself needs the chain rule.
The chain rule
If y is a function of u, and u is a function of x, then
dy/dx = (dy/du) × (du/dx)
In words: differentiate the outside function (leaving the inside alone), then multiply by the derivative of the inside.
Example 1: a power of a bracket
Differentiate y = (3x − 1)⁵.
Let u = 3x − 1, so y = u⁵.
- dy/du = 5u⁴
- du/dx = 3
So dy/dx = 5u⁴ × 3 = 15(3x − 1)⁴.
The most common wrong answer is 5(3x − 1)⁴ — forgetting to multiply by the derivative of the inside.
Example 2: a square root
Differentiate y = √(1 + x²).
Write it as a power: y = (1 + x²)^½. Let u = 1 + x².
- dy/du = ½u^(−½)
- du/dx = 2x
So dy/dx = ½(1 + x²)^(−½) × 2x = x / √(1 + x²).
Example 3: exponentials and logarithms
For y = e^(sin x), the outside is e^u and the inside is sin x:
dy/dx = e^(sin x) × cos x = cos x e^(sin x).
For y = ln(kx), where k is a positive constant:
dy/dx = (1/(kx)) × k = 1/x.
The k cancels. That’s not a coincidence: ln(kx) = ln k + ln x, and ln k is a constant, so its derivative is zero.
Shortcut worth knowing: for any function f, the derivative of f(ax + b) is a f′(ax + b). So d/dx sin(4x) = 4 cos(4x) and d/dx e^(2x + 5) = 2e^(2x + 5).
Practise the chain rule in our chain rule subtopic, and try the free questions on differentiating exponentials and logarithms, which use it throughout.
The product rule
If y = uv, where u and v are both functions of x, then
dy/dx = u (dv/dx) + v (du/dx)
Example 4: polynomial times exponential
Differentiate y = x² e^(3x).
- u = x², so du/dx = 2x
- v = e^(3x), so dv/dx = 3e^(3x) (chain rule)
dy/dx = x² × 3e^(3x) + e^(3x) × 2x = 3x² e^(3x) + 2x e^(3x)
Now factorise: dy/dx = x e^(3x)(3x + 2).
Factorising isn’t decoration. If the question goes on to ask for stationary points, the factorised form tells you immediately that dy/dx = 0 when x = 0 or x = −2/3 (e^(3x) is never zero).
Example 5: finding a stationary point
Find the stationary point of y = x e^(−x).
- u = x, du/dx = 1
- v = e^(−x), dv/dx = −e^(−x)
dy/dx = x × (−e^(−x)) + e^(−x) × 1 = e^(−x)(1 − x).
Setting dy/dx = 0: e^(−x) > 0 for all x, so 1 − x = 0 and x = 1. Then y = 1 × e^(−1) = 1/e. The stationary point is (1, 1/e).
To classify it, look at the sign of dy/dx: for x slightly less than 1 it’s positive, and for x slightly more than 1 it’s negative, so the point is a maximum.
Practise in our product rule subtopic.
The quotient rule
If y = u/v, then
dy/dx = (v (du/dx) − u (dv/dx)) / v²
This one is printed in every board’s formula booklet — but the order of the terms in the numerator matters, so it’s worth knowing well enough to use without hesitating.
Example 6: a rational function
Differentiate y = (2x + 1)/(x − 3).
- u = 2x + 1, du/dx = 2
- v = x − 3, dv/dx = 1
dy/dx = ((x − 3) × 2 − (2x + 1) × 1) / (x − 3)² = (2x − 6 − 2x − 1) / (x − 3)² = −7 / (x − 3)²
Because (x − 3)² is positive for every x ≠ 3, the gradient is always negative: the function is decreasing on each side of its asymptote.
Example 7: proving the derivative of tan x
Write tan x = sin x / cos x.
- u = sin x, du/dx = cos x
- v = cos x, dv/dx = −sin x
dy/dx = (cos x × cos x − sin x × (−sin x)) / cos²x = (cos²x + sin²x) / cos²x = 1/cos²x = sec²x.
This is a favourite “show that” question. Proofs of the derivatives of sec x, cosec x and cot x work in exactly the same way.
Practise in our quotient rule subtopic.
Putting the rules together
Example 8: product rule and chain rule
Differentiate y = x(2x + 1)⁴, and find the equation of the tangent where x = 0.
- u = x, du/dx = 1
- v = (2x + 1)⁴, dv/dx = 4(2x + 1)³ × 2 = 8(2x + 1)³
dy/dx = x × 8(2x + 1)³ + (2x + 1)⁴ × 1
Take out the common factor (2x + 1)³:
dy/dx = (2x + 1)³(8x + 2x + 1) = (2x + 1)³(10x + 1)
At x = 0: y = 0 × 1⁴ = 0, and dy/dx = 1³ × 1 = 1. The tangent is y = x.
Mistakes that cost marks
- Forgetting the inside derivative in the chain rule — the single most common error.
- Swapping the order in the quotient rule. v du/dx comes first; if you write u dv/dx − v du/dx your answer has the wrong sign.
- Not simplifying. “Show that” questions need the exact form given, and stationary point questions are much easier from a factorised derivative.
- Treating a product as a chain, e.g. differentiating x sin x as cos x. The correct answer is sin x + x cos x.
- Losing brackets with negative terms — especially −u dv/dx when dv/dx is itself negative, as in Example 7.
Exam tip: write down u, v, du/dx and dv/dx explicitly before you substitute into the rule. It takes ten seconds, earns method marks, and makes your own errors easy to spot when you check.
Where this leads
These three rules unlock the rest of Year 2 calculus: implicit differentiation (the chain rule applied to y), parametric differentiation, connected rates of change, and — run in reverse — integration by substitution and integration by parts.
If you’re still on Year 1 content, make sure differentiation from first principles is solid first; it’s free to practise here. Then work through the Further differentiation notes. The A Level Maths app has 15 exam-style questions on each of the chain, product and quotient rules, every one with a worked explanation of the right answer and of each wrong one.
Frequently asked questions
Which of these rules is in the formula booklet?
Only the quotient rule, and it is printed by all four boards. The chain rule and the product rule are not printed anywhere, so learn them by heart.
Can I use the product rule instead of the quotient rule?
Yes. Write u/v as u × v⁻¹ and use the product rule with the chain rule. You get the same answer, though it often needs more simplifying.
