Further differentiation and applications: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Apply the product, quotient and chain rules to differentiate functions, and use the derivative in optimisation, rates of change and curve sketching”

9 min readUpdated 2 Jun 2026

Quick answer

Combine the product, quotient and chain rules to differentiate composite and combined functions, then use the first derivative for rates of change and optimisation and the second derivative for concavity and curve sketching.

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What this dot point is asking
The three combining rules
Combining the rules
Rates of change
Optimisation
The second derivative and curve sketching
Exam-style questions

What this dot point is asking

SCSA Unit 3 extends your differentiation toolkit so you can handle functions built from several pieces multiplied, divided or nested inside one another, and then apply the derivative to real contexts: maxima and minima, rates of change, and sketching. This dot point is examined in both the calculator-free and calculator-assumed sections of the WACE written examination.

The three combining rules

The differentiation rules

Chain rule: If $y = f(g(x))$ then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$ .
Product rule: If $y = u v$ then $\dfrac{dy}{dx} = u'v + uv'$ .
Quotient rule: If $y = \dfrac{u}{v}$ then $\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2}$ .

The skill SCSA tests is recognising which rule (often several) a function needs and applying them in the correct order. Always differentiate the outermost structure first, then work inwards.

Standard derivatives you must know

$\dfrac{d}{dx}(x^n) = n x^{n-1}$
$\dfrac{d}{dx}(\sin x) = \cos x$ , $\dfrac{d}{dx}(\cos x) = -\sin x$
$\dfrac{d}{dx}(e^{x}) = e^{x}$ , $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$

Combining the rules

A single function can require all three rules. The strategy is to label the structure clearly before differentiating.

Differentiating a combined function

Step 1: Identify the structure and label the factors

The function $y = x^2 \cos(3x)$ is a product of two factors, so the product rule applies. Label $u = x^2$ and $v = \cos(3x)$ . The factor $v$ is itself a composite function, so the chain rule will also be needed when differentiating it.

Step 2: Differentiate each factor

Differentiating $u$ is straightforward: $u' = 2x$ . For $v = \cos(3x)$ , the chain rule gives an inner derivative of $3$ , so $v' = -\sin(3x) \cdot 3 = -3\sin(3x)$ .

Step 3: Apply the product rule $y' = u'v + uv'$

Substituting into $\dfrac{dy}{dx} = u'v + uv'$ and simplifying:

\frac{dy}{dx} = 2x \cos(3x) + x^2 \cdot (-3\sin(3x))

\frac{dy}{dx} = 2x\cos(3x) - 3x^2 \sin(3x)

Final answer: $\dfrac{dy}{dx} = 2x\cos(3x) - 3x^2\sin(3x)$

Rates of change

The derivative $\dfrac{dy}{dx}$ is the instantaneous rate of change of $y$ with respect to $x$ . If a quantity $Q$ depends on time $t$ , then $\dfrac{dQ}{dt}$ is the rate of change of $Q$ . For example, if the volume of water in a tank is $V(t) = 50t - t^2$ litres, the inflow rate at time $t$ is $V'(t) = 50 - 2t$ litres per unit time, which is zero at $t = 25$ (the tank stops filling).

Optimisation

To find a maximum or minimum of a quantity in context:

Write the quantity to be optimised as a function of one variable, using any constraint to eliminate other variables.
Differentiate and solve $f'(x) = 0$ for the stationary points.
Classify each stationary point (sign of $f'$ either side, or the sign of $f''$ ).
Check the endpoints of the domain and interpret the answer in context.

Optimisation in context

Step 1: Introduce variables and use the constraint

A farmer has $40$ m of fencing for three sides of a rectangular pen against a wall (the wall forms the fourth side). Let the two sides perpendicular to the wall each have length $x$ and the side parallel to the wall have length $y$ . The fencing constraint gives $2x + y = 40$ , so $y = 40 - 2x$ , reducing the problem to one variable.

Step 2: Write the area as a function of one variable

Substituting the constraint into $A = xy$ gives the area function to maximise:

A(x) = x(40 - 2x) = 40x - 2x^2.

Step 3: Differentiate and solve $A'(x) = 0$

Differentiating gives $A'(x) = 40 - 4x$ . Setting this to zero: $40 - 4x = 0$ , so $x = 10$ .

Step 4: Confirm a maximum and find all dimensions

Since $A''(x) = -4 < 0$ , the stationary point is a local maximum. Then $y = 40 - 2(10) = 20$ , giving $A = 10 \times 20 = 200$ m $^2$ .

Final answer: Maximum area is $200$ m $^2$ with sides $10$ m and $20$ m.

The second derivative and curve sketching

The second derivative $f''(x)$ measures concavity:

$f''(x) > 0$ : the curve is concave up (holds water).
$f''(x) < 0$ : the curve is concave down.
A point of inflection occurs where $f''$ changes sign.

The second derivative test classifies a stationary point $x = a$ where $f'(a) = 0$ : if $f''(a) > 0$ it is a local minimum, and if $f''(a) < 0$ it is a local maximum. When $f''(a) = 0$ the test is inconclusive and you must use a sign table for $f'$ .

A complete curve sketch combines intercepts, stationary points, the nature of each stationary point, points of inflection, and end behaviour.

Common errors

Forgetting the inner derivative in the chain rule: $\dfrac{d}{dx}\cos(3x)$ is $-3\sin(3x)$ , not $-\sin(3x)$ . The factor of $3$ is essential.
Quotient rule numerator order: It is $u'v - uv'$ , not $uv' - u'v$ . The wrong order flips the sign of the whole derivative.
Not classifying the stationary point: Solving $f'(x) = 0$ only locates candidates. You must confirm whether each is a maximum, minimum or inflection before answering an optimisation question.
Ignoring the domain endpoints: In a restricted-domain optimisation problem the true maximum can occur at an endpoint, not at a stationary point.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2022-style7 marks

Calculator-free. Consider

y=\dfrac{x^{2}}{e^{x}}

. (a) Find

\dfrac{dy}{dx}

. (b) Find the coordinates of the stationary points. (c) Classify each stationary point using the second derivative or a sign table.

Show worked answer →

A combined-rules question leading to stationary points.

(a) Write $y=x^{2}e^{-x}$ and use the product rule: $\dfrac{dy}{dx}=2xe^{-x}+x^{2}(-e^{-x})=e^{-x}(2x-x^{2})=xe^{-x}(2-x)$ .

(b) $\dfrac{dy}{dx}=0$ when $x=0$ or $x=2$ . Then $y(0)=0$ and $y(2)=\dfrac{4}{e^{2}}\approx 0.541$ . Stationary points $(0,0)$ and $\left(2,\dfrac{4}{e^{2}}\right)$ .

(c) Using a sign table for $\dfrac{dy}{dx}=xe^{-x}(2-x)$ : at $x=0$ the sign changes $-$ to $+$ , a local minimum; at $x=2$ it changes $+$ to $-$ , a local maximum.

Markers reward the product rule with the internal chain rule, both stationary points, and a valid classification.

WACE 2024-style6 marks

Calculator-assumed. An open-top box is made from a square sheet of side

24

cm by cutting equal squares of side

x

from each corner and folding up the sides. (a) Show that the volume is

V(x)=x(24-2x)^{2}

. (b) Find the value of

x

that maximises the volume.

Show worked answer →

A classic optimisation problem.

(a) The base is a square of side $24-2x$ and the height is $x$ , so $V(x)=x(24-2x)^{2}$ for $0<x<12$ .

(b) Expand: $V(x)=x(576-96x+4x^{2})=576x-96x^{2}+4x^{3}$ . Then $V'(x)=576-192x+12x^{2}=12(x^{2}-16x+48)=12(x-4)(x-12)$ . Setting $V'(x)=0$ gives $x=4$ or $x=12$ . Since $x=12$ gives zero volume, $x=4$ is the maximum (confirmed by $V''(4)=-192+24(4)<0$ ). The maximum volume is $V(4)=4\times 16^{2}=1024$ cm $^{3}$ .

Markers reward the volume model, the factored derivative, rejecting $x=12$ , and confirming a maximum.

Further differentiation and applications: WACE Year 12 Mathematics Methods Unit 3

What this dot point is asking

The three combining rules

Standard derivatives you must know

Combining the rules

Step 1: Identify the structure and label the factors

Step 2: Differentiate each factor

Step 3: Apply the product rule $y' = u'v + uv'$

Rates of change

Optimisation

Step 1: Introduce variables and use the constraint

Step 2: Write the area as a function of one variable

Step 3: Differentiate and solve $A'(x) = 0$

Step 4: Confirm a maximum and find all dimensions

The second derivative and curve sketching

Exam-style questions

Practise this

Sources & how we know this

What this dot point is asking

The three combining rules

Standard derivatives you must know

Combining the rules

Step 1: Identify the structure and label the factors

Step 2: Differentiate each factor

Step 3: Apply the product rule y′=u′v+uv′y' = u'v + uv'y′=u′v+uv′

Rates of change

Optimisation

Step 1: Introduce variables and use the constraint

Step 2: Write the area as a function of one variable

Step 3: Differentiate and solve A′(x)=0A'(x) = 0A′(x)=0

Step 4: Confirm a maximum and find all dimensions

The second derivative and curve sketching

Exam-style questions

Practise this

Sources & how we know this

Step 3: Apply the product rule $y' = u'v + uv'$

Step 3: Differentiate and solve $A'(x) = 0$