TASMath MethodsSyllabus dot point

How do we differentiate products, quotients and composites, and use the result to optimise and sketch?

Further differentiation and applications: product, quotient and chain rules, curve sketching, optimisation and rates of change

Product, quotient and chain rules, second-derivative curve sketching, optimisation and rates of change for TCE Mathematics Methods Unit 3.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

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What this dot point is asking

Unit 3 of TCE Mathematics Methods builds the differentiation you met earlier into a full toolkit. By the end of this dot point you should be able to differentiate any combination of polynomial, exponential, logarithmic and trigonometric pieces, then turn the derivative into information about a graph or a real situation.

The three rules

The chain rule differentiates a composite $y = f(g(x))$ :

\frac{dy}{dx} = f'(g(x)) \cdot g'(x).

The product rule differentiates $y = u(x)\,v(x)$ :

\frac{dy}{dx} = u'v + uv'.

The quotient rule differentiates $y = \dfrac{u(x)}{v(x)}$ :

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.

These combine. A term like $x^2 e^{3x}$ needs the product rule, and the $e^{3x}$ factor itself needs the chain rule to give $3e^{3x}$ .

Curve sketching with derivatives

Stationary points occur where $f'(x) = 0$ . To classify them, use the second derivative: if $f''(x) > 0$ the point is a local minimum, if $f''(x) < 0$ it is a local maximum. Where $f''(x) = 0$ and the concavity changes you have a point of inflection. The first-derivative sign test is an alternative when $f''$ is awkward.

Optimisation and rates of change

Optimisation asks for the largest or smallest value of a quantity. The recipe: write the quantity as a function of one variable (use a constraint to eliminate any second variable), differentiate, set the derivative to zero, solve, confirm the nature of the stationary point, and check the endpoints of the domain.

Rates of change use $\dfrac{dy}{dx}$ directly as "rate of change of $y$ per unit $x$ ". If $x$ is time, $\dfrac{dy}{dt}$ is a rate per second (or per hour). Related rates link two rates through the chain rule.

Optimising a fenced enclosure

Problem

A farmer has $120$ m of fencing for a rectangular paddock against a straight river (no fence needed on the river side). Maximise the area.

Solution

Let the two sides perpendicular to the river be $x$ and the side parallel be $y$ . The fencing used is $2x + y = 120$ , so $y = 120 - 2x$ .

Area:

A = xy = x(120 - 2x) = 120x - 2x^2.

Differentiate and set to zero:

\frac{dA}{dx} = 120 - 4x = 0 \quad\Rightarrow\quad x = 30.

Since $\dfrac{d^2A}{dx^2} = -4 < 0$ , this is a maximum. Then $y = 120 - 2(30) = 60$ , and

A = 30 \times 60 = 1800 \text{ m}^2.

The largest area is $1800$ square metres, achieved with sides $30$ m and $60$ m.

Putting it together

A typical extended question gives a modelling function (a profit curve, a concentration over time, a height profile), asks for the rate of change at an instant, then asks for the maximum or minimum and the value of the variable that produces it. Always state units and confirm the nature of any stationary point before claiming it is the answer.

Exam-style practice questions

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

2024 TASC3 marksFind and classify the stationary point(s) of the function f(x) = x e^(2x). Show all algebraic working.

Show worked answer →

Differentiate using the product rule with u = x and v = e^(2x), so u' = 1 and v' = 2 e^(2x):
f'(x) = (1) e^(2x) + x (2 e^(2x)) = e^(2x)(1 + 2x).

Set f'(x) = 0. Since e^(2x) is never zero, we need 1 + 2x = 0, so x = -1/2.
f(-1/2) = (-1/2) e^(-1) = -1/(2e). The stationary point is (-1/2, -1/(2e)).

Classify with the second derivative: f''(x) = 2 e^(2x)(1 + 2x) + e^(2x)(2) = e^(2x)(4x + 4).
At x = -1/2: f''(-1/2) = e^(-1)(-2 + 4) = 2/e > 0. Since f'' is positive, the point is a local minimum. One mark for f'(x), one for the coordinates, one for the classification.

2023 TASC3 marksTwo positive numbers, x and y, are such that xy = 16. Find the minimum value of x + y. You are not required to justify the minimum value.

Show worked answer →

Use the constraint to write the sum as a function of one variable. From xy = 16, y = 16/x. Let S = x + y = x + 16/x for x > 0.

Differentiate: dS/dx = 1 - 16/x^2.
Set to zero: 1 - 16/x^2 = 0, so x^2 = 16, giving x = 4 (taking the positive root since x > 0).
Then y = 16/4 = 4.

Minimum value of x + y = 4 + 4 = 8. Marks are for setting up S in one variable, differentiating and solving, and stating the minimum value of 8.

2023 TASC4 marksa) Find the derivative of f(x) = (3x^2 - 2)(4x + 1) by expanding the brackets first. b) Show how the same answer can be obtained using the product rule for derivatives.

Show worked answer →

a) Expand: f(x) = 12x^3 + 3x^2 - 8x - 2. Differentiate term by term: f'(x) = 36x^2 + 6x - 8.

b) Product rule with u = 3x^2 - 2 (so u' = 6x) and v = 4x + 1 (so v' = 4):
f'(x) = u'v + uv' = (6x)(4x + 1) + (3x^2 - 2)(4)
= 24x^2 + 6x + 12x^2 - 8 = 36x^2 + 6x - 8.

The two methods agree, which is the point of the question. Two marks for each method, with the product-rule part requiring the correct u'v + uv' structure before simplifying.