Module 2 differentiated sums term by term and pulled constants out in front. A product of two functions of $x$ is the first thing that refuses to be done piece by piece. Write the two factors as $f$ and $g$, with $f'$ and $g'$ their derivatives (read aloud f prime and g prime). The product rule is $(fg)' = f'g + fg'$, read aloud: the derivative of f times g is f prime g plus f g prime. Each factor is differentiated in turn while the other is left alone, and the two results are added. What it is NOT is $f'g'$: the slope of $x \cdot x$ would then be $1 \cdot 1 = 1$, and $x^2$ has slope $2x$. That one slip is the trap this lesson is named for.

For $f(x) = (2x + 1)(x^2 - 3)$: the first bracket has derivative 2, the second has derivative $2x$, so $f'(x) = 2(x^2 - 3) + (2x + 1)(2x) = 6x^2 + 2x - 6$, and $f'(2) = 24 + 4 - 6 = 22$. Expanding first, $2x^3 + x^2 - 6x - 3$, gives the same $6x^2 + 2x - 6$, as it must; the rule is for the products that will not expand, which is most of them from here on. At a point, evaluate four numbers, $f, f', g, g'$, then form the two products and add: the exam's usual shape is exactly that, four numbers at one $x$ and one line of arithmetic.