If $y = x^4 - 2x^3$, find the value of $\dfrac{d^2y}{dx^2}$ when $x = 1$.

$\dfrac{dy}{dx} = 4x^3 - 6x^2$ and $\dfrac{d^2y}{dx^2} = 12x^2 - 12x$. At $x = 1$: $12 - 12 = 0$. Substituting into the first derivative instead gives $-2$, and forgetting the second term gives 12.