HSC Maths Differentiation: Complete Guide with Examples

Last updated: 20 July 2026

Differentiation is the heart of HSC Maths Advanced. It's worth 20-25% of the exam on its own, and it underpins integration, optimisation, and rates of change. If your differentiation is weak, nothing else in calculus works.

This guide covers every rule, when to use each one, worked examples, and the errors that cost marks.

What is differentiation?

Differentiation finds the rate of change of a function at any point. If you have a function f(x), its derivative f'(x) tells you how fast f(x) is changing at each value of x.

Geometrically, the derivative at a point gives the gradient of the tangent line at that point. This is why differentiation shows up in optimisation (finding maximum and minimum values) and in questions about rates of change.

The four rules

Power rule

ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}

The most basic rule. Bring the power down, then subtract 1 from the exponent.

Example: Differentiate f(x) = 5x³

f(x)=5×3x2=15x2f'(x) = 5 \times 3x^{2} = 15x^{2}

Example: Differentiate g(x) = 1/x⁴

Rewrite as x⁻⁴, then: g'(x) = -4x⁻⁵ = -4/x⁵

Product rule

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Use this when two functions are multiplied together. Differentiate the first, keep the second. Then keep the first, differentiate the second. Add the results.

Example: Differentiate f(x) = x² × sin(x)

Let u = x², v = sin(x) u' = 2x, v' = cos(x)

f(x)=2xsin(x)+x2cos(x)f'(x) = 2x·sin(x) + x^{2}·cos(x)

Quotient rule

ddx(uv)=vdudxudvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

Use this when one function is divided by another. The mnemonic "low D-high minus high D-low, over low squared" helps remember the order.

Example: Differentiate f(x) = (3x + 1) / (x² + 2)

Let u = 3x + 1, v = x² + 2 u' = 3, v' = 2x

f(x)=[(x2+2)(3)(3x+1)(2x)]/(x2+2)2f'(x) = [(x^{2} + 2)(3) - (3x + 1)(2x)] / (x^{2} + 2)^{2} f(x)=[3x2+66x22x]/(x2+2)2f'(x) = [3x^{2} + 6 - 6x^{2} - 2x] / (x^{2} + 2)^{2} f(x)=[3x22x+6]/(x2+2)2f'(x) = [-3x^{2} - 2x + 6] / (x^{2} + 2)^{2}

Chain rule

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

The chain rule applies when you have a function inside another function. Differentiate the outside, keep the inside, then multiply by the derivative of the inside.

This is the most important rule in HSC Maths and the one students forget most often.

Example: Differentiate f(x) = (3x² + 1)⁵

Outside function: (something)⁵ → derivative is 5(something)⁴ Inside function: 3x² + 1 → derivative is 6x

f(x)=5(3x2+1)4×6x=30x(3x2+1)4f'(x) = 5(3x^{2} + 1)^{4} \times 6x = 30x(3x^{2} + 1)^{4}

Example: Differentiate g(x) = sin(4x³)

Outside: sin(something) → cos(something) Inside: 4x³ → 12x²

g(x)=cos(4x3)×12x2=12x2cos(4x3)g'(x) = cos(4x^{3}) \times 12x^{2} = 12x^{2}·cos(4x^{3})

Derivatives of specific functions

These need to be memorised. They're partly on the formula sheet but you should know them cold.

Function Derivative
ln(x) 1/x
sin(x) cos(x)
cos(x) -sin(x)
tan(x) sec²(x)

For Extension 1, you also need:

Function Derivative
sin⁻¹(x) 1/√(1-x²)
cos⁻¹(x) -1/√(1-x²)
tan⁻¹(x) 1/(1+x²)

Applications

Finding tangent lines

The derivative gives the gradient of the tangent. To find the equation of a tangent at x = a:

  1. Find f'(x)
  2. Substitute x = a to get the gradient m
  3. Find f(a) to get the y-coordinate
  4. Use point-gradient form: y - f(a) = m(x - a)

Optimisation

To find maximum and minimum values:

  1. Differentiate the function
  2. Set f'(x) = 0 and solve for x (these are stationary points)
  3. Use the second derivative test or a sign diagram to classify them
  4. Check endpoints if the domain is restricted

Second derivative test: If f''(x) < 0 at a stationary point, it's a maximum. If f''(x) > 0, it's a minimum. If f''(x) = 0, the test is inconclusive.

Rates of change

If a quantity is changing over time, its derivative with respect to time gives the rate of change. For example, if the radius of a circle is increasing at 2 cm/s, the rate of change of the area is dA/dt = dA/dr × dr/dt = 2πr × 2 = 4πr cm²/s.

This technique (multiplying rates using the chain rule) is called related rates and shows up in harder exam questions.

Common mistakes

Forgetting the chain rule

The number one error in HSC Maths. If you're differentiating anything with a function inside it (brackets with a power, trig of an expression, exponential of an expression), you need the chain rule.

Fix: Before writing your answer, ask: "Is there something inside this function?" If yes, multiply by the derivative of the inside.

Sign errors in the quotient rule

The order matters: v·du/dx minus u·dv/dx. Get the order wrong and your answer has the wrong sign.

Fix: Write "low D-high minus high D-low" at the top of the question.

Mixing up rules

Using the product rule when you should use the chain rule, or vice versa. The product rule is for two separate functions multiplied. The chain rule is for one function inside another.

Fix: Look at the structure. If it's f(g(x)), use chain rule. If it's f(x)·g(x), use product rule. If it's f(x)/g(x), use quotient rule.

Not simplifying

In exam questions, the final simplified form is often what earns the last mark. Leaving an unsimplified answer can lose a mark.

Fix: Always try to simplify. If you can factorise or cancel, do it.

How to practice

  1. Start with power rule questions until they're automatic
  2. Add chain rule questions (the most important)
  3. Practice product and quotient rules
  4. Do mixed questions that combine multiple rules
  5. Move to application questions (tangents, optimisation, rates)
  6. Finish with HSC exam questions

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Based on the NESA HSC Mathematics Advanced and Extension 1 syllabuses. PracticePapers.io is not affiliated with NESA.