Further calculus (WACE Year 12 Mathematics Specialist Unit 3)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

Differentiate inverse trigonometric functions using $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$ , $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$ , apply implicit differentiation by differentiating both sides with respect to $x$ and using the chain rule on $y$ terms, and integrate by reversing these results, by substitution, and by recognising the standard forms that give $\arctan$ and $\ln$ .

Jump to a section

What this dot point is asking
Derivatives of inverse trigonometric functions
Implicit differentiation
Related rates
Integration techniques
Exam-style questions

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

Inverse trig derivatives

\frac{d}{dx}\big(\arcsin x\big) = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\big(\arccos x\big) = \frac{-1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\big(\arctan x\big) = \frac{1}{1+x^2}.

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 $C$ ):

Standard integrals

\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\!\Big(\frac{x}{a}\Big) + C, \qquad \int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\!\Big(\frac{x}{a}\Big) + C,

\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + 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 $f'(x)$ .

Integration by recognising a logarithm

Evaluate $\displaystyle\int \frac{2x}{x^2 + 4}\,dx$

The denominator is $f(x) = x^2 + 4$ , whose derivative is $f'(x) = 2x$ , exactly the numerator.

\int \frac{2x}{x^2 + 4}\,dx = \ln\big(x^2 + 4\big) + C.

No modulus signs are needed because $x^2 + 4 > 0$ for all real $x$ . Contrast this with $\displaystyle\int \frac{1}{x^2 + 4}\,dx = \frac{1}{2}\arctan\!\Big(\frac{x}{2}\Big) + C$ , where the numerator is constant, not the derivative of the denominator: that is a standard inverse-tangent form, not a logarithm.

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 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-style6 marks

Calculator-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 2023-style7 marks

Calculator-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.

Practise this

Sources & how we know this

Mathematics Specialist ATAR course: syllabus and support materials — SCSA
Past ATAR course examinations — SCSA

What this dot point is asking

Derivatives of inverse trigonometric functions

Implicit differentiation

Find dydx\frac{dy}{dx}dxdy​ for x2+xy+y2=7x^2 + xy + y^2 = 7x2+xy+y2=7

Step 1: Differentiate term by term with respect to xxx.

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

Step 3: Factor out dydx\dfrac{dy}{dx}dxdy​ and solve.

Related rates

Integration techniques

Evaluate ∫2xx2+4 dx\displaystyle\int \frac{2x}{x^2 + 4}\,dx∫x2+42x​dx

Exam-style questions

Practise this

Sources & how we know this

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

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

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

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

Evaluate $\displaystyle\int \frac{2x}{x^2 + 4}\,dx$