Chain, product and quotient rules for polynomial and power functions (QCE Math Methods Unit 2)

Syllabus dot point

“Apply the chain rule, the product rule and the quotient rule to differentiate functions built from polynomial and power functions, and use the results to find gradients and tangents”

8 min readUpdated 7 Oct 2026

Quick answer

Use the chain rule for a function of a function, $\dfrac{d}{dx}f(g(x))=f'(g(x))\,g'(x)$ . Use the product rule for $u\,v$ , which gives $u'v+uv'$ . Use the quotient rule for $\dfrac{u}{v}$ , which gives $\dfrac{u'v-uv'}{v^2}$ . Factorise the answer so that it is easy to use.

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What this dot point is asking
The three rules
Choosing the rule
Stationary points from a product
Practice questions

What this dot point is asking

In the 2025 QCAA Mathematical Methods syllabus, Unit 2 Topic 5 is titled Further differentiation, and it introduces the rules for differentiating products, quotients and composite functions. In Unit 3 the same rules are extended to exponential, logarithmic and trigonometric functions, which is covered on the Unit 3 product, quotient and chain rules page. The official syllabus is at qcaa.qld.edu.au.

The three rules

Chain rule: Differentiate the outer function, keep the inner function, then multiply by the derivative of the inner function.
Product rule: For $y=uv$ , $\dfrac{dy}{dx}=u'v+uv'$ .
Quotient rule: For $y=\dfrac{u}{v}$ , $\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}$ .

Chain rule:

y=(3x^2-1)^4

The outer function is the fourth power and the inner function is $3x^2-1$ .

\frac{dy}{dx}=4(3x^2-1)^3\cdot 6x=24x(3x^2-1)^3.

At $x=1$ the gradient is $24(1)(2)^3=192$ .

Product rule:

y=x^2(2x+1)^3

Let $u=x^2$ and $v=(2x+1)^3$ . Then $u'=2x$ and, by the chain rule, $v'=3(2x+1)^2\cdot 2=6(2x+1)^2$ .

\frac{dy}{dx}=2x(2x+1)^3+6x^2(2x+1)^2.

Factorise by taking out $2x(2x+1)^2$ :

\frac{dy}{dx}=2x(2x+1)^2\big[(2x+1)+3x\big]=2x(2x+1)^2(5x+1).

At $x=1$ the gradient is $2(1)(3)^2(6)=108$ .

Quotient rule:

y=\dfrac{x^2+1}{x-1}

Let $u=x^2+1$ and $v=x-1$ , so $u'=2x$ and $v'=1$ .

\frac{dy}{dx}=\frac{2x(x-1)-(x^2+1)(1)}{(x-1)^2}=\frac{x^2-2x-1}{(x-1)^2}.

At $x=2$ the gradient is $\dfrac{4-4-1}{1}=-1$ .

Choosing the rule

A bracket raised to a power, such as $(ax+b)^n$ , needs the chain rule.
Two factors multiplied, such as $x^2(2x+1)^3$ , need the product rule, and the second factor needs the chain rule as well.
A fraction with $x$ in the denominator needs the quotient rule, unless you can rewrite it as a product with a negative power.

Stationary points from a product

Factorised answers make stationary points easy. For $y=(x+1)^2(x-3)^3$ :

\frac{dy}{dx}=2(x+1)(x-3)^3+3(x+1)^2(x-3)^2=(x+1)(x-3)^2(5x-3).

Setting $\dfrac{dy}{dx}=0$ gives $x=-1$ , $x=3$ and $x=\tfrac35$ .

Common trap

Four common errors. Forgetting to multiply by the derivative of the inner function in the chain rule. Writing the quotient rule numerator as $uv'-u'v$ , which reverses the sign. Expanding a product before differentiating when it is easier to factorise afterwards. Dropping the brackets when substituting a negative value into the answer.

Practice questions

1. Differentiate $y=(2x+5)^3$ .

Model answer: By the chain rule, $\dfrac{dy}{dx}=3(2x+5)^2\cdot 2=6(2x+5)^2$ .

2. Differentiate $y=x(x-4)^2$ and simplify.

Model answer: With $u=x$ and $v=(x-4)^2$ , $\dfrac{dy}{dx}=(x-4)^2+x\cdot 2(x-4)=(x-4)(x-4+2x)=(x-4)(3x-4)$ .

3. Find the gradient of $y=\dfrac{2x+1}{x^2+1}$ at $x=0$ .

Model answer: The quotient rule gives $\dfrac{dy}{dx}=\dfrac{2(x^2+1)-(2x+1)(2x)}{(x^2+1)^2}$ . At $x=0$ this is $\dfrac{2-0}{1}=2$ .

4. Find the $x$ -values where $y=(x+1)^2(x-3)^3$ has a horizontal tangent.

Model answer: The derivative is $(x+1)(x-3)^2(5x-3)$ . Setting it to zero gives $x=-1$ , $x=3$ and $x=\tfrac35$ .

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA