SCSA WACE 2022-style practice: Calculator-free. (a) Differentiate y = (2x). (b) Use implicit differentiation to find dydx for the circle x^2 + y^2 = 25 at the point (3, 4).

Combines an inverse-trig derivative with implicit differentiation. (a) By the chain rule, dydx = 11 + (2x)^2 × 2 = 21 + 4x^2. (b) Differentiate both sides of x^2 + y^2 = 25 with respect to x: 2x + 2ydydx = 0, so dydx = -xy. At (3, 4) this is -34. Markers reward the chain-rule factor of 2 in (a), differentiating y^2 as 2ydydx in (b), and the evaluated gradient -34.

SCSA WACE 2023-style practice: Calculator-assumed. A spherical balloon is inflated so its volume increases at 50 cm^3 per second. Find the rate at which the radius is increasing when the radius is 5 cm. (Use V = 43π r^3.)

A related-rates problem. Differentiate V = 43π r^3 with respect to time: dVdt = 4π r^2 drdt. Substitute dVdt = 50 and r = 5: 50 = 4π (5)^2 drdt = 100π drdt. So drdt = 50100π = 12π ≈ 0.159 cm per second. Markers reward differentiating the volume formula, the chain-rule factor drdt, substituting the values last, and the final rate 12π cm per second.

§-Syllabus dot point

WASpecialist MathematicsSyllabus dot point

How do we differentiate and integrate the harder functions of Specialist?

Differentiate inverse trig functions, use implicit differentiation, and integrate rational functions, partial fractions and trig forms

WACE Specialist Unit 3 further calculus: derivatives of inverse trigonometric functions, implicit differentiation, related rates, and integration by recognition, substitution and the standard inverse-trig and logarithmic forms, with full worked 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
Derivatives of inverse trigonometric functions
Implicit differentiation
Related rates
Integration techniques

What this dot point is asking

SCSA Unit 3 extends calculus in three directions: the derivatives of the inverse circular functions, implicit differentiation (including related rates), and a wider set of integration techniques. You must move between differentiation and integration fluently, since most integration here is "differentiation run backwards".

Derivatives of inverse trigonometric functions

For a scaled argument, combine with the chain rule. For example $\dfrac{d}{dx}\arcsin\!\big(\tfrac{x}{a}\big) = \dfrac{1}{\sqrt{a^2 - x^2}}$ and $\dfrac{d}{dx}\arctan\!\big(\tfrac{x}{a}\big) = \dfrac{a}{a^2 + x^2}$ .

Implicit differentiation

When a relation links $x$ and $y$ without $y$ being isolated (for example $x^2 + y^2 = 25$ ), differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ . Every $y$ term gains a factor $\dfrac{dy}{dx}$ from the chain rule, then you solve algebraically for $\dfrac{dy}{dx}$ .

Implicit differentiation

Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$

Step 1: Differentiate term by term with respect to $x$ .

Each term is differentiated with respect to $x$ , treating $y$ as a function of $x$ . The middle term $xy$ requires the product rule, and the $y^2$ term gains a factor $\dfrac{dy}{dx}$ from the chain rule.

2x + \left(1\cdot y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0.

Step 2: Group the $\dfrac{dy}{dx}$ terms on one side.

Rearranging collects all terms containing $\dfrac{dy}{dx}$ on the left and everything else on the right. This prepares the equation for factorising out $\dfrac{dy}{dx}$ in the next step.

x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y.

Step 3: Factor out $\dfrac{dy}{dx}$ and solve.

Factorising the left side and dividing both sides by the bracket gives $\dfrac{dy}{dx}$ as a ratio of expressions in $x$ and $y$ . Substituting the given point then gives the numerical gradient.

\frac{dy}{dx} = \frac{-(2x + y)}{x + 2y}.

At the point $(1, 2)$ this gives $\dfrac{dy}{dx} = \dfrac{-(2 + 2)}{1 + 4} = -\dfrac{4}{5}$ .

Related-rates problems link two changing quantities through a known equation and use the chain rule $\dfrac{dA}{dt} = \dfrac{dA}{dx}\cdot\dfrac{dx}{dt}$ . Identify the variables, write the relation, differentiate with respect to time, then substitute the instantaneous values last.

Integration techniques

Integration in Unit 3 is built on recognition and substitution. Key standard forms (with $a > 0$ and an arbitrary constant $C$ ):

The form $\dfrac{f'(x)}{f(x)}$ is the workhorse for rational integrands: if the numerator is (a multiple of) the derivative of the denominator, the integral is a logarithm. Otherwise, use a $u$ -substitution or rearrange the numerator to manufacture $f'(x)$ .

When the integrand splits into separate fractions (for instance a numerator $x + 1$ over $x^2 + 4$ ), break it into a logarithmic piece $\dfrac{x}{x^2+4}$ and an inverse-tangent piece $\dfrac{1}{x^2+4}$ and integrate each separately.

Common traps

Confusing the inverse-trig and logarithm forms: $\dfrac{1}{a^2 + x^2}$ integrates to $\arctan$ , but $\dfrac{2x}{a^2 + x^2}$ (derivative on top) integrates to $\ln$ . Always check whether the numerator is the derivative of the denominator before choosing.
Dropping the inner factor: $\dfrac{d}{dx}\arctan(3x) = \dfrac{3}{1 + 9x^2}$ , not $\dfrac{1}{1 + 9x^2}$ . The chain rule supplies the $3$ .
Forgetting $\frac{dy}{dx}$ on $y$ terms: In implicit differentiation, $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$ , never just $2y$ .
Omitting the modulus and the constant: $\int \frac{1}{x}\,dx = \ln|x| + C$ . The $|\cdot|$ and the $+C$ both lose marks if missing.

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 20226 marksCalculator-free. (a) Differentiate

y = \arctan(2x)

. (b) Use implicit differentiation to find

\dfrac{dy}{dx}

for the circle

x^2 + y^2 = 25

at the point

(3, 4)

Show worked answer →

Combines an inverse-trig derivative with implicit differentiation.

(a) By the chain rule, $\dfrac{dy}{dx} = \dfrac{1}{1 + (2x)^2} \times 2 = \dfrac{2}{1 + 4x^2}$ .

(b) Differentiate both sides of $x^2 + y^2 = 25$ with respect to $x$ : $2x + 2y\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{x}{y}$ . At $(3, 4)$ this is $-\dfrac{3}{4}$ .

Markers reward the chain-rule factor of $2$ in (a), differentiating $y^2$ as $2y\dfrac{dy}{dx}$ in (b), and the evaluated gradient $-\dfrac{3}{4}$ .

WACE 20237 marksCalculator-assumed. A spherical balloon is inflated so its volume increases at

50

^3

per second. Find the rate at which the radius is increasing when the radius is

5

cm. (Use

V = \dfrac{4}{3}\pi r^3

Show worked answer →

A related-rates problem.

Differentiate $V = \dfrac{4}{3}\pi r^3$ with respect to time: $\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$ .

Substitute $\dfrac{dV}{dt} = 50$ and $r = 5$ : $50 = 4\pi (5)^2 \dfrac{dr}{dt} = 100\pi \dfrac{dr}{dt}$ .

So $\dfrac{dr}{dt} = \dfrac{50}{100\pi} = \dfrac{1}{2\pi} \approx 0.159$ cm per second.

Markers reward differentiating the volume formula, the chain-rule factor $\dfrac{dr}{dt}$ , substituting the values last, and the final rate $\dfrac{1}{2\pi}$ cm per second.

What this dot point is asking

Derivatives of inverse trigonometric functions

Implicit differentiation

Related rates

Integration techniques

Exam-style practice questions