Differentiation — Worked Solutions (HSC Maths Advanced)
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Worked examples for HSC Maths Advanced differentiation. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Advanced — Differentiation. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Stationary points
Question
The curve is $y = x^3 - 6x^2 + 9x$. Find the coordinates of the stationary points and determine their nature.
Solution
Stationary points are where the gradient is zero, so differentiate and solve $y' = 0$.
$y' = 3x^2 - 12x + 9 = 3(x-1)(x-3)$, so $x = 1$ or $x = 3$.
Substitute back: $y(1) = 4$ and $y(3) = 0$, giving $(1, 4)$ and $(3, 0)$.
Decide the nature with the second derivative: $y'' = 6x - 12$. $y''(1) = -6 < 0$ → maximum; $y''(3) = 6 > 0$ → minimum.
So $(1, 4)$ is a maximum and $(3, 0)$ a minimum. Always justify the nature with the $y''$ test — a sketch alone won't get the mark.
Let's start with what a stationary point is — a place where the curve momentarily flattens, so its gradient is zero. That means we differentiate and set $y' = 0$.
$y' = 3x^2 - 12x + 9$. Factoring out the 3: $3(x^2 - 4x + 3) = 3(x-1)(x-3)$, so $x = 1$ and $x = 3$.
We need full coordinates, so substitute back into the original: $y(1) = 1 - 6 + 9 = 4$ and $y(3) = 27 - 54 + 27 = 0$. Our points are $(1, 4)$ and $(3, 0)$.
For their nature, the second derivative tells us the shape: $y'' = 6x - 12$. At $x = 1$, $y'' = -6$ (negative → concave down → a maximum); at $x = 3$, $y'' = 6$ (positive → concave up → a minimum).
So $(1, 4)$ is a maximum and $(3, 0)$ a minimum. The $y''$ test works because concavity tells you whether you're at the top of a hill or the bottom of a valley.
Stationary points: $y' = 0$.
- $y' = 3x^2 - 12x + 9 = 3(x-1)(x-3)$
- $x = 1,\ 3$
- $y(1) = 4$, $y(3) = 0$ → $(1, 4)$, $(3, 0)$
Nature: $y'' = 6x - 12$.
- $y''(1) = -6 < 0$ → maximum
- $y''(3) = 6 > 0$ → minimum
$(1, 4)$ max, $(3, 0)$ min.
Where the marks go
- 1 mark: Correct first derivative $y' = 3x^2 - 12x + 9$
- 1 mark: Solves $y' = 0$ to get $x = 1$ and $x = 3$
- 1 mark: Correct coordinates $(1, 4)$ and $(3, 0)$
- 1 mark: Determines nature using the second derivative
Key idea
Stationary points are where $y' = 0$; the sign of $y''$ tells you maximum (negative) or minimum (positive).
Example 2 — Product and chain rule
Question
Differentiate $y = x^2 e^{3x}$.
Solution
This is a product, $x^2$ times $e^{3x}$, so use the product rule — and the chain rule on $e^{3x}$.
$u = x^2,\ u' = 2x$. $v = e^{3x},\ v' = 3e^{3x}$.
$y' = u'v + uv' = 2x e^{3x} + 3x^2 e^{3x}$.
Factor: $y' = x e^{3x}(2 + 3x)$.
Don't skip the factoring — a clean form earns the last mark and sets you up if the next part asks for stationary points.
Notice this is a product of two functions, $x^2$ and $e^{3x}$, so the product rule is the way in: $(uv)' = u'v + uv'$.
Let $u = x^2$, so $u' = 2x$, and $v = e^{3x}$. For $v'$ we use the chain rule — the derivative of $e^{\text{something}}$ is itself times the derivative of that something, so $v' = 3e^{3x}$.
Putting it together: $y' = 2x\,e^{3x} + x^2 \cdot 3e^{3x} = 2x e^{3x} + 3x^2 e^{3x}$.
Both terms share $x e^{3x}$, so we factor: $y' = x e^{3x}(2 + 3x)$. Factoring isn't just tidy — it makes any next step, like solving $y' = 0$, much easier.
Product of $x^2$ and $e^{3x}$. Product rule; chain rule on the exponential.
- $u = x^2,\ u' = 2x$
- $v = e^{3x},\ v' = 3e^{3x}$
- $y' = u'v + uv' = 2x e^{3x} + 3x^2 e^{3x}$
- Factor: $y' = x e^{3x}(2 + 3x)$
Where the marks go
- 1 mark: Applies the product rule with a correct setup
- 1 mark: Correct chain-rule derivative of $e^{3x}$ (i.e. $3e^{3x}$)
- 1 mark: Correct simplified/factored derivative
Key idea
A product of functions → product rule; an exponential like $e^{3x}$ → the chain rule gives the extra factor of 3.
Frequently asked questions
Step-by-step solutions to exam-style questions on Differentiation in HSC Maths Advanced, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Advanced syllabus for Differentiation, using the methods and notation expected in the exam.