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.

Year 12-13EDEXCEL 9MA0 7.4

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.
  • Combine the rules on expressions like cos2 x and e3x/x.

WHAT YOU PROBABLY THINK

The derivative of a quotient is the quotient of the derivatives.

The chain rule

A function inside a function is a chain of machines, and rates through a chain multiply,

dydx = dydu × dudxNOT IN THE BOOKLET — LEARN IT

with u naming the inner function. In practice the rule reads: differentiate the outside leaving the inside alone, then multiply by the inside's derivative.

The chain rule as linked machines: x feeds the inner function u equals 3 x squared plus 1, whose output feeds the outer function y equals u to the fifth, and the overall rate is the product of the two link ratesxu = 3x² + 1y = u⁵du/dx = 6xdy/du = 5u⁴two links, two ratesdy/dx = dy/du × du/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴rates through a chain multiply
FIG. 1The chain rule as linked machines: x into u = 3x² + 1, u into y = u⁵. Two link rates, multiplied, give the whole chain's rate.

WORKED EXAMPLE

Outside first, then inside

Differentiate y = (3x2 + 1)5.

Set u = 3x2 + 1, so y = u5: dy/du = 5u4 and du/dx = 6x.

Multiply: dy/dx = 5u4 × 6x = 30x(3x2 + 1)4.

With practice the u stays silent, but on show-your-working questions naming it 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,

(uv)' = u'v − uv'v2IN THE FORMULAE BOOKLET

and the minus sign in its numerator is why the opening lie fails: differentiating top and bottom separately gives a different, wrong answer, as one substitution shows on any example.

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, lie dead.

Factorising the quotient answer, e3x out front, is the presentation the mark scheme prints.

The curve of cos squared x above its derivative minus sine 2x: the squared curve bounces between 0 and 1, and its gradient, found by the chain rule, is a single wave of amplitude 1y = cos² xdy/dx = −sin 2xflat wherever the wave below crosses zero
FIG. 2cos² x over its derivative −sin 2x: the chain rule's product 2 cos x × (−sin x), folded by the double angle formula into one wave.

YOUR TURN

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 outside, and their derivatives almost always tidy through a double angle.

TRY IT UNSEEN

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.

Work outside-in, one layer per line, and the nesting never tangles; trying to do both layers at once is where the fours and twos get lost.

THE EXAM BIT

  • Chain rule: outside differentiated with the inside untouched, times the inside's derivative; name u when marks are shown.
  • Product rule symmetric, quotient rule ordered: u'v − uv', minus in the middle, all over v squared.
  • Choose the rule from the expression's shape: composed, multiplied or divided, and say which you are using.
  • Factorise derivative answers; e3x(3x − 1)/x2 earns what its expanded twin forfeits.
  • 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.

The x2/x collapse to x is the tidy-up this 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.

CHECK YOUR PROGRESS

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  • Differentiate composite functions with the chain rule.
  • Differentiate products and quotients with their rules, laid out cleanly.
  • Combine the rules on expressions like cos2 x and e3x/x.

No animated video for this topic yet; these notes stand alone.