Differentiation — Worked Solutions (HSC Maths Advanced)

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.

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.