Maths › Differentiation › The product, quotient and chain rules
The product, quotient and chain rules
Year 12 differentiated by rewriting; that only stretches so far. Three rules finish the job properly: the chain rule for functions inside functions, the product rule for multiplied pairs, and the quotient rule for fractions, and between them they differentiate everything this course can write down.
Builds on Differentiating trig, exponentials and logs.
IN THIS TOPIC
- Differentiate composite functions with the chain rule.
- Differentiate products and quotients with their rules, laid out cleanly.
- Use the booklet derivatives of sec x, cosec x and cot x.
- Combine the rules on nested expressions like cos2 x, tan2 2x and e3x/x.
COMMON MISCONCEPTION
The derivative of a quotient is the quotient of the derivatives.
The chain rule
A function inside a function is a chain of two stages, and rates through a chain multiply.
Here u names the inner function. In practice the rule reads as an instruction. Differentiate the outside, leaving the inside untouched, then multiply by the inside's derivative.
WORKED EXAMPLE
Outside first, then inside
Differentiate y = (3x2 + 1)5.
Set u = 3x2 + 1, so y = u5, giving dy/du = 5u4 and du/dx = 6x.
Multiply: dy/dx = 5u4 × 6x = 30x(3x2 + 1)4.
With practice the u stays silent. On a show-your-working question, naming it anyway is the safest mark in calculus.
Products and quotients
When two functions multiply, the product rule shares the differentiation between them, (uv)' = u'v + uv'. When they divide, the quotient rule keeps the order strict.
That minus sign in the numerator is essential. Differentiate top and bottom separately and you get a different answer, and one substitution on any example will show you how different.
WORKED EXAMPLE
A spec-fluency pair
Differentiate y = 2x4 sin x, and then y = e3x/x.
Product rule: u = 2x4, v = sin x gives dy/dx = 8x3 sin x + 2x4 cos x.
Quotient rule: u = e3x, v = x gives (3e3x × x − e3x × 1)/x2 = e3x(3x − 1)/x2.
At x = 1 the quotient's true gradient is 2e3 ≈ 40.2, while the naive top-over-bottom fake gives 3e3 ≈ 60.3. One number separates them for good.
Factorise the quotient answer with e3x out front. That is the presentation the mark scheme prints.
GUIDED PRACTICE
A square of a function
Differentiate y = cos2 x, giving the answer as a single trig term, before opening the working.
Show the working
Chain rule with u = cos x: y = u2, so dy/dx = 2u × u' = 2 cos x × (−sin x).
The double angle formula folds it: dy/dx = −sin 2x.
Squared trig functions are chains with the square on the outside, and their derivatives almost always tidy through a double angle.
The rest of the booklet, and layered rules
Three more derivatives sit in the formulae booklet and the specification expects you to use them. sec x differentiates to sec x tan x, cosec x to −cosec x cot x, and cot x to −cosec2 x. Look them up. They are not on the memorise list, and hunting for them under exam pressure costs less time than getting a sign wrong from memory.
Layered expressions are where marks quietly leak away. Work outside in, one rule per line, and carry every k as you go. Trying to do two layers in a single step is how the fours and twos get lost.
INDEPENDENT PRACTICE
Rules within rules
Differentiate y = tan2 2x.
Show the working
Two chains deep: y = u2 with u = tan 2x, and u' = 2 sec2 2x from the shelf.
dy/dx = 2u × u' = 2 tan 2x × 2 sec2 2x = 4 tan 2x sec2 2x.
One layer per line. The 2 from the square and the 2 from the inner angle are separate animals, and combining them in your head is how one of them goes missing.
ASSESSMENT FOCUS
- Chain rule: outside differentiated with the inside untouched, times the inside's derivative. Name u when working is being marked.
- Product rule is symmetric, quotient rule is ordered: u'v − uv', minus in the middle, all over v squared.
- Choose the rule from the shape of the expression, composed or multiplied or divided, and say which one you are using.
- Factorise your answer. Equivalent algebra earns the same accuracy marks, but e3x(3x − 1)/x2 is the form the next part of the question will want: stationary points drop straight out of a factorised derivative and hide in an expanded one.
- sec, cosec and cot derivatives are booklet entries. Open the booklet instead of trusting a half-remembered sign.
- For nested expressions apply one rule per line, outside in, and carry every k.
CHECK YOURSELF
Differentiate y = x2 ln x, and find the gradient where x = e.
Show a hint
Product rule, and ln e is 1.
Show the answer
Product rule: dy/dx = 2x ln x + x2 × (1/x) = 2x ln x + x.
At x = e: 2e × 1 + e = 3e ≈ 8.15.
That x2/x collapsing to x is the tidy-up the question exists to test.
Chains multiply rates, products share the differentiation, quotients keep strict order over v squared.
Pick the rule from the shape, name your u, and tidy the answer by factorising.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the the product, quotient and chain rules questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Differentiate composite functions with the chain rule.
- Differentiate products and quotients with their rules, laid out cleanly.
- Use the booklet derivatives of sec x, cosec x and cot x.
- Combine the rules on nested expressions like cos2 x, tan2 2x and e3x/x.
Open the full revision checklist to see every objective in the course in one place.