The product and quotient rules (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“The product and quotient rules differentiate functions formed by multiplying or dividing two differentiable functions.”

8 min readUpdated 2 Jun 2026

Quick answer

To differentiate a product $f(x)=u(x)v(x)$ use the product rule $f'=u'v+uv'$ . To differentiate a quotient $f(x)=\dfrac{u(x)}{v(x)}$ use the quotient rule $f'=\dfrac{u'v-uv'}{v^2}$ . Identify $u$ and $v$ , differentiate each, then substitute.

Jump to a section

What this dot point is asking
The product rule
The quotient rule
Choosing the right rule
Products and quotients with a chain inside
Common errors
Bringing it together
Exam-style questions

What this dot point is asking

You need to recognise when a function is a product of two functions or a quotient of two functions, choose the correct rule, and apply it carefully. These rules build on the basic derivatives of polynomials, exponentials, logarithms and trigonometric functions, so accuracy in those is assumed.

The product rule

If $f(x)=u(x)\,v(x)$ where $u$ and $v$ are both differentiable, then

f'(x)=u'(x)\,v(x)+u(x)\,v'(x).

A useful shorthand is "first times derivative of second, plus second times derivative of first" - but written as $u'v+uv'$ the order does not matter because addition is commutative.

Product rule

\frac{d}{dx}\big[u\,v\big]=u'v+uv'

When to use it

Use the product rule whenever the function is a multiplication of two non-trivial expressions, for example $x^2 e^x$ , $x\sin x$ , or $(2x+1)\ln x$ . If one factor is a constant you do not need the product rule - just keep the constant.

Differentiating

x^2 e^x

Find the derivative of $f(x)=x^2 e^x$ .

Step 1: Recognise the structure and identify $u$ and $v$

The function is a product of two non-constant expressions, so the product rule applies. Label the two factors:

u=x^2 \qquad \text{and} \qquad v=e^x.

Step 2: Differentiate each factor separately

Differentiate $u$ using the power rule, and $v$ using the fact that the exponential function is its own derivative:

u'=2x \qquad \text{and} \qquad v'=e^x.

Step 3: Apply the product rule

The product rule states $f'(x)=u'v+uv'$ . Substituting the four expressions found above:

f'(x)=2x\,e^x+x^2 e^x.

Step 4: Factorise the result

Both terms share the factors $x$ and $e^x$ . Factorising makes the answer more compact and is useful if you later need to solve $f'(x)=0$ :

f'(x)=x e^x(2+x).

Final answer: $f'(x)=xe^x(2+x)$ .

The quotient rule

If $f(x)=\dfrac{u(x)}{v(x)}$ where $u$ and $v$ are differentiable and $v(x)\neq 0$ , then

f'(x)=\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\big[v(x)\big]^2}.

The order in the numerator does matter here because of the subtraction: it is always (derivative of top)(bottom) minus (top)(derivative of bottom), all over (bottom) squared.

Quotient rule

\frac{d}{dx}\!\left[\frac{u}{v}\right]=\frac{u'v-uv'}{v^2}

Differentiating

\dfrac{x}{x^2+1}

Find the derivative of $f(x)=\dfrac{x}{x^2+1}$ .

Step 1: Recognise the structure and identify $u$ and $v$

The function is a fraction with two distinct expressions, so the quotient rule applies. The numerator is $u$ and the denominator is $v$ :

u=x \qquad \text{and} \qquad v=x^2+1.

Step 2: Differentiate each part

The numerator is a simple linear expression and the denominator is a polynomial, so both derivatives follow the power rule:

u'=1 \qquad \text{and} \qquad v'=2x.

Step 3: Apply the quotient rule

The quotient rule states $f'(x)=\dfrac{u'v-uv'}{v^2}$ . It is important to keep the subtraction in the numerator in the correct order, "derivative of top times bottom" minus "top times derivative of bottom":

f'(x)=\frac{(1)(x^2+1)-(x)(2x)}{(x^2+1)^2}.

Step 4: Simplify the numerator

Expand the numerator and collect like terms to obtain a clean expression:

f'(x)=\frac{x^2+1-2x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}.

The denominator $(x^2+1)^2$ cannot be simplified further, so this is the final form.

Final answer: $f'(x)=\dfrac{1-x^2}{(x^2+1)^2}$ .

Choosing the right rule

A product (two things multiplied): product rule.
A quotient (one thing divided by another): quotient rule, or rewrite as a product with a negative power and use the product and chain rules.
A sum or difference: differentiate term by term - no special rule needed.

Sometimes a quotient is easier as a product. For example $\dfrac{e^x}{x^2}=e^x x^{-2}$ can be differentiated with the product rule, which some students find less error-prone than the quotient rule. Both approaches give the same answer; choose whichever you can execute cleanly under exam pressure.

Simplify before you differentiate

If a quotient simplifies algebraically - for instance $\dfrac{x^3+x}{x}=x^2+1$ - simplify first. Differentiating $x^2+1$ directly is far quicker and safer than applying the quotient rule to the original.

Products and quotients with a chain inside

The real exam difficulty is that the inner factors often need the chain rule themselves. For $y=(2x+1)^3 e^{x}$ , treat it as a product with $u=(2x+1)^3$ and $v=e^x$ . Here $u'=3(2x+1)^2(2)=6(2x+1)^2$ uses the chain rule, and $v'=e^x$ , so

\frac{dy}{dx}=6(2x+1)^2 e^x+(2x+1)^3 e^x=(2x+1)^2 e^x\big(6+(2x+1)\big)=(2x+1)^2 e^x(2x+7).

Differentiate each factor fully (chain rule included) before substituting into the product or quotient rule, then factorise.

Common errors

Forgetting the subtraction order in the quotient rule

The numerator of the quotient rule is $u'v-uv'$ , not $uv'-u'v$ . Swapping the two terms flips the sign of the whole derivative. A reliable check: differentiate $\dfrac{1}{x}=x^{-1}$ using the quotient rule (with $u=1$ , $v=x$ ) and confirm you get $-\dfrac{1}{x^2}$ , which matches the power rule. If you get $+\dfrac{1}{x^2}$ , you have the order wrong.

Bringing it together

The product and quotient rules are the workhorses of Stage 2 differentiation. They combine constantly with the chain rule - for instance $\dfrac{d}{dx}\big[x e^{2x}\big]$ needs the product rule on the outside and the chain rule to differentiate $e^{2x}$ . Master identifying $u$ and $v$ cleanly, write the rule before substituting, and factorise your answer where possible so it is ready for the curve-sketching and optimisation applications that follow.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style2 marks

Calculator-assumed. Find

\dfrac{dy}{dx}

for

y = \dfrac{\ln(5x^2)}{x^2}

. There is no need to simplify your answer.

Show worked answer →

This is a quotient, so use the quotient rule with $u = \ln(5x^2)$ and $v = x^2$ .

First the derivatives of the parts. By the chain rule, $u' = \dfrac{10x}{5x^2} = \dfrac{2}{x}$ . The denominator gives $v' = 2x$ .

Apply the quotient rule $\dfrac{u'v - uv'}{v^2}$ :

\frac{dy}{dx} = \frac{\left(\frac{2}{x}\right)(x^2) - \ln(5x^2)(2x)}{(x^2)^2} = \frac{2x - 2x\ln(5x^2)}{x^4}.

The question says there is no need to simplify, so full marks go to a correct unsimplified quotient-rule expression. One mark is for differentiating $\ln(5x^2)$ to $\dfrac{2}{x}$ , and one mark for the correct quotient-rule structure.

SACE 2022-style3 marks

Calculator-assumed. Determine

\dfrac{dy}{dx}

for

y = \dfrac{e^x + e^{-x}}{x}

. You do not need to simplify your answer.

Show worked answer →

Use the quotient rule with $u = e^x + e^{-x}$ and $v = x$ .

Differentiate the parts: $u' = e^x - e^{-x}$ (the chain rule sends $e^{-x}$ to $-e^{-x}$ ), and $v' = 1$ .

Quotient rule $\dfrac{u'v - uv'}{v^2}$ :

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

Marks: one for $u'$ with the correct sign on $e^{-x}$ , one for $v' = 1$ with the quotient-rule set-up, and one for a correctly assembled final expression. No simplification is required.

SACE 2021-style2 marks

Calculator-free. Differentiate

y = x^2\sin x

with respect to

x

Show worked answer →

This is a product, so use the product rule with $u = x^2$ and $v = \sin x$ .

The derivatives of the parts are $u' = 2x$ and $v' = \cos x$ .

Product rule $u'v + uv'$ :

\frac{dy}{dx} = 2x\sin x + x^2\cos x.

Marks: one for the product-rule structure and one for both derivatives correct ( $u' = 2x$ and $v' = \cos x$ ). Writing the derivative of $\sin x$ as $-\cos x$ is the usual slip.

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

The product rule

When to use it

Step 1: Recognise the structure and identify uuu and vvv

Step 2: Differentiate each factor separately

Step 3: Apply the product rule

Step 4: Factorise the result

The quotient rule

Step 1: Recognise the structure and identify uuu and vvv

Step 2: Differentiate each part

Step 3: Apply the quotient rule

Step 4: Simplify the numerator

Choosing the right rule

Products and quotients with a chain inside

Common errors

Bringing it together

Exam-style questions

Practise this

Sources & how we know this

Step 1: Recognise the structure and identify $u$ and $v$

Step 1: Recognise the structure and identify $u$ and $v$