HSC Maths Advanced formula sheet: downloads and worked examples

For the 2027 HSC, use the NESA Mathematics Advanced reference sheet published with the sample examination (PDF). If you are sitting the HSC in 2026, use the sheet for the previous syllabus linked below.

A formula sheet helps you remember formulas. You still need to decide which method to use, check that it applies to the question and understand what the answer means.

Choose the sheet for your HSC year

If a download link changes, check the current Advanced exam requirements. NESA says a reference sheet will be provided. The first HSC exam for the 2024 syllabus is in 2027. Year 12 students sitting in 2026 use the 2017 syllabus.

We checked these links on 23 September 2026. The new sheet is part of the sample exam materials and could change before the exam. Our 2027 maths syllabus guide explains the other course changes.

Read the formula before substituting

The rule for differentiating a logarithm is in the Differential calculus section on page 2 of the 2027 sample sheet. Before using it, check that the expression inside the logarithm is positive.

Keep the official sheet beside you when you practise. Before using a formula:

1. Write down what the question asks you to find. A derivative, a total change and an area measure different things.
2. Write down what each symbol means. Identify which values can change, which are constant and whether one function is inside another.
3. Check which values you can use. Logarithms, denominators and square roots can limit the values for which an expression is defined.
4. Decide how to check your answer. You could substitute it back into the question, sketch a graph or differentiate an antiderivative.

We wrote these exercises to explain standard methods. They do not cover every formula on the sheet, give an official marking scheme or predict exam questions.

Worked example: the chain rule and a logarithm's domain

Find the derivative of $f(x)=\ln(3x-2)$ and state its real domain.

Start with the domain, which is the set of values you can put into the function. The expression inside the logarithm must be positive:

$$3x-2>0\quad\Longrightarrow\quad x>\frac{2}{3}.$$

For a positive differentiable function $u(x)$, the chain rule gives

$$\frac{d}{dx}\ln u(x)=\frac{u'(x)}{u(x)}.$$

Here $u(x)=3x-2$ and $u'(x)=3$, so

$$f'(x)=\frac{3}{3x-2},\qquad x>\frac{2}{3}.$$

The numerator comes from differentiating the expression inside the logarithm. Writing $1/(3x-2)$ leaves out that step. Writing $x\ne2/3$ as the domain only rules out the value that makes the denominator zero. You can put values below $2/3$ into the fraction, but not into the original logarithm.

At $x=1$, the derivative is $3$. To check whether this is reasonable, calculate how quickly the original function changes between two nearby values:

$$\frac{f(1.001)-f(1)}{0.001} =\frac{\ln(1.003)}{0.001}\approx2.996.$$

The result is close to $3$, which agrees with our derivative. This checks the answer, while the chain rule explains how to find it.

Try a similar question

For $g(x)=\ln(5-2x)$, find the derivative and domain.

The derivative of the inside function is now negative. The answer is $g'(x)=-2/(5-2x)$ with $x<5/2$. The method is the same, but both the sign and the domain change. Find the inside function and differentiate it each time so you do not copy the positive numerator from the first example.

Worked example: an integral is not always the total area

For $y=x-1$ on $0\le x\le3$, find both the definite integral and the total area between the graph and the $x$-axis.

An antiderivative is $x^2/2-x$. Therefore

$$\int_0^3(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_0^3 =\frac{3}{2}.$$

The graph crosses the axis at $x=1$. Between $0$ and $1$, it is below the axis, so the integral over that part is negative. To find the total area, count the area on both sides of the axis as positive:

$$A=-\int_0^1(x-1)\,dx+\int_1^3(x-1)\,dx =\frac12+2=\frac52.$$

Check by sketching the graph. The first triangle has base $1$ and height $1$, so its area is $1/2$. The second has base $2$ and height $2$, so its area is $2$. Together, they have an area of $5/2$ square units.

The question asks for two different things. The definite integral allows the negative and positive parts to cancel, while the total area counts both as positive. The answer $3/2$ is correct for the integral but wrong for the total area.

Check the antiderivative too

Differentiate $x^2/2-x$ and check that you get $x-1$. This checks the antiderivative. To check the area calculation too, split the area at $x=1$ and compare it with the two triangles.

What to practise after a mistake

Try a new question on the same idea without looking at the worked answer. Keep your first attempt so you can compare the two and explain what you changed. Our Maths Advanced past-paper guide has official questions you can practise next.

If you are considering Maths Advanced tutoring, bring a question and your working so we can help you choose a method. You can find more subject guides in the HSC resource hub.