What is product rule?

If y = u v then dydx = u'v + uv'.

What is quotient rule?

If y = uv then dydx = u'v - uv'v^2.

What is quotient rule numerator order?

It is u'v - uv', not uv' - u'v. The wrong order flips the sign of the whole derivative.

What is 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.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do the product, quotient and chain rules let us differentiate and apply more complex functions?

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

WACE Year 12 Mathematics Methods Unit 3 further differentiation: the product, quotient and chain rules combined, optimisation, rates of change and the second derivative for curve sketching, with worked SCSA-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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 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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksCalculator-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 20246 marksCalculator-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.