Further differentiation and applications (TCE Mathematics Methods, Tasmania)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

You must differentiate products, quotients and composite functions fluently, then apply the derivative to sketch curves (stationary points and their nature) and to solve optimisation and rates-of-change problems.

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What this dot point is asking
The three combining rules
Curve sketching with derivatives
Optimisation
Rates of change and related rates
Exam-style questions

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. TASC examines this material in both the calculator-free and calculator-assumed sections of the external Level 3/4 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 $\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^{2}}$ .

These rules combine. A term like $x^{2}e^{3x}$ needs the product rule, and the $e^{3x}$ factor itself needs the chain rule to give $3e^{3x}$ . The examiner's skill is recognising which rule (often several) a function needs and applying them in the correct order: differentiate the outermost structure first, then work inwards.

Standard derivatives you must know

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

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 the alternative when $f''$ is awkward to evaluate.

Stationary point nature

At a stationary point $x = a$ : $f''(a) > 0 \Rightarrow$ minimum, $f''(a) < 0 \Rightarrow$ maximum. If $f''(a) = 0$ you must test the sign of $f'$ on each side.

A complete sketch combines intercepts, stationary points, the nature of each stationary point, any points of inflection, and the end behaviour as $x \to \pm\infty$ .

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 the 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, with units.

Optimising a fenced enclosure

Problem

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

Set up one variable

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

Build the function

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

Differentiate and solve

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

Confirm and interpret

Since $\dfrac{d^{2}A}{dx^{2}} = -4 < 0$ , 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.

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 its rate of change. For example, if the volume of water in a tank is $V(t) = 50t - t^{2}$ litres, the inflow rate is $V'(t) = 50 - 2t$ litres per unit time, which is zero at $t = 25$ .

Related rates link two rates through the chain rule: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$ . You differentiate the relationship between the quantities, then substitute the known rate to find the unknown one.

Common errors

Forgetting the inner derivative in the chain rule: When differentiating $e^{3x}$ , $\sin(2x)$ or $\ln(5x-1)$ , students drop the factor from the inside. The correct results are $3e^{3x}$ , $2\cos(2x)$ and $\dfrac{5}{5x-1}$ .
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.
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 TASC 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.

Practice3 marks

Calculator-free. Find and classify the stationary point(s) of

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 $v' = 2e^{2x}$ :

f'(x) = (1)e^{2x} + x(2e^{2x}) = e^{2x}(1 + 2x).

Set $f'(x) = 0$ . Since $e^{2x}$ is never zero, we need $1 + 2x = 0$ , so $x = -\tfrac{1}{2}$ .
Then $f\left(-\tfrac{1}{2}\right) = -\tfrac{1}{2}e^{-1} = -\dfrac{1}{2e}$ . The stationary point is $\left(-\tfrac{1}{2}, -\dfrac{1}{2e}\right)$ .

Classify with the second derivative: $f''(x) = 2e^{2x}(1 + 2x) + e^{2x}(2) = e^{2x}(4x + 4)$ .
At $x = -\tfrac{1}{2}$ : $f''\left(-\tfrac{1}{2}\right) = e^{-1}(2) = \dfrac{2}{e} > 0$ , so the point is a local minimum. One mark for $f'(x)$ , one for the coordinates, one for the classification.

Practice6 marks

Calculator-assumed. An open rectangular tank with a square base of side

x

metres is to hold

32

^3

of water. (a) Show that the surface area of the base and four sides is

S(x) = x^{2} + \dfrac{128}{x}

. (b) Find the value of

x

that minimises the material used, and the minimum surface area.

Show worked answer →

(a) The volume is $x^{2}h = 32$ , so $h = \dfrac{32}{x^{2}}$ . The open tank has a base $x^{2}$ and four sides each $xh$ , so

S = x^{2} + 4xh = x^{2} + 4x\cdot\frac{32}{x^{2}} = x^{2} + \frac{128}{x}.

(b) Differentiate: $S'(x) = 2x - \dfrac{128}{x^{2}}$ . Set $S'(x) = 0$ : $2x = \dfrac{128}{x^{2}}$ , so $2x^{3} = 128$ , giving $x^{3} = 64$ and $x = 4$ .
Check it is a minimum: $S''(x) = 2 + \dfrac{256}{x^{3}} > 0$ for $x > 0$ , confirming a minimum.
Then $S(4) = 16 + \dfrac{128}{4} = 16 + 32 = 48$ m $^2$ .

Markers reward the correct surface-area model, the factored derivative, $x = 4$ , and confirming a minimum.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC

What this dot point is asking

The three combining rules

Standard derivatives you must know

Curve sketching with derivatives

Optimisation

Problem

Set up one variable

Build the function

Differentiate and solve

Confirm and interpret

Rates of change and related rates

Exam-style questions

Practise this

Sources & how we know this