Two terms, never one AB & BC

For differentiable $u$ and $v$,

$$\frac{d}{dx}\left[u\,v\right] = u'v + uv'.$$

Two terms. In each one, exactly one factor is differentiated and the other is copied unchanged.

Before using it, see why nothing simpler works. Take $u = x$ and $v = x$. Then $uv = x^2$, whose derivative is $2x$. But $u'v' = 1 \cdot 1 = 1$. Those disagree at every $x$ except $\frac{1}{2}$, so there is no rule of that shape.

The two-term formula does check out on the same example: $u'v + uv' = 1 \cdot x + x \cdot 1 = 2x$, exactly as the power rule says.

The rule is short and the errors are clerical, so a fixed layout is worth the seconds it takes. Before differentiating anything, write down four things:

1. $u$ and $v$, the two factors.
2. $u'$ and $v'$, each found on its own.
3. Assemble $u'v + uv'$, reading straight off the list.
4. Simplify only afterwards, and only if it helps.

Worked, on $f(x) = x^2 \sin x$. Here $u = x^2$ and $v = \sin x$, so $u' = 2x$ and $v' = \cos x$:

$$f'(x) = 2x\sin x + x^{2}\cos x.$$

Both terms survive to the answer. Neither can be combined with the other, and the most common single error in this topic is writing the first term and stopping.

The order of the two terms does not matter, because addition commutes. That is a real difference from the quotient rule in Topic 2.9, where the order is load bearing.

The rule does not care what the factors are, only that each derivative is available.

1. $\frac{d}{dx}\left[(3x + 1)e^{x}\right] = 3e^{x} + (3x+1)e^{x} = (3x + 4)e^{x}$.
2. $\frac{d}{dx}\left[x\ln x\right] = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$, for $x > 0$.
3. $\frac{d}{dx}\left[4x^{2}\cos x\right] = 8x\cos x - 4x^{2}\sin x$.

In the third, the minus sign arrives from the cosine's derivative and is easy to lose because it appears in the middle of an assembled expression rather than at the front.

A constant factor is the degenerate case. In $\frac{d}{dx}[5x^3]$, taking $u = 5$ gives $u' = 0$, so the first term vanishes and the second is $5 \cdot 3x^2$. That is the constant multiple rule from Topic 2.6, recovered rather than contradicted.

The product rule is always correct, and it is sometimes the slow road. Two situations where expanding first is better:

1. Both factors are powers of $x$. $x^2 \cdot x^3 = x^5$, so the derivative is $5x^4$ in one step. The product rule gives $2x \cdot x^3 + x^2 \cdot 3x^2 = 2x^4 + 3x^4 = 5x^4$, which agrees, having taken four times as long.
2. A short polynomial times a short polynomial. $(x + 2)(x - 5) = x^2 - 3x - 10$, whose derivative is $2x - 3$.

Expanding and then differentiating is a genuine check on a product rule answer, and on a free-response question it is worth doing when the algebra is small.

Where expanding is not available, as with $x^2 \sin x$ or $x \ln x$, the rule is the only route and both of its terms are mandatory.