The rule where order matters AB & BC

For differentiable $u$ and $v$ with $v \neq 0$,

$$\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^{2}}.$$

Three structural facts, each of which is a place errors enter:

1. The numerator's derivative comes first. $u'v$ leads, $uv'$ follows, and they are subtracted in that order.
2. The whole denominator is squared. Not $v$, and not $v'$ squared.
3. The numerator has the same shape as the product rule, with a minus in place of the plus. That resemblance is useful for recall and dangerous for order.

Reversing the two terms gives $\frac{uv' - u'v}{v^2}$, which is exactly the negative of the correct answer. A sign error of this kind never announces itself: the expression still looks plausible.

Take $f(x) = \dfrac{2x + 1}{x - 3}$, with $u = 2x + 1$ and $v = x - 3$, so $u' = 2$ and $v' = 1$:

$$f'(x) = \frac{2(x - 3) - (2x + 1)(1)}{(x - 3)^{2}} = \frac{2x - 6 - 2x - 1}{(x-3)^{2}} = \frac{-7}{(x - 3)^{2}}.$$

The distribution of the minus sign across $(2x + 1)$ is the step that goes wrong most often. Writing the numerator with brackets and expanding deliberately costs one line and prevents it.

The answer is negative everywhere it is defined, since the denominator is a square. A quick sanity check: $f(4) = 9$ and $f(5) = 5.5$, so the function is indeed falling, which agrees.

A second worked one, $\dfrac{\sin x}{x}$, gives $\dfrac{x\cos x - \sin x}{x^{2}}$. Note that the leading term is $x \cos x$: the numerator's derivative, $\cos x$, is multiplied by the denominator.

The quotient rule always works and is often the long route. Two cases where rewriting is faster, and where the rule tends to produce mistakes for no benefit:

1. A constant denominator. $\dfrac{x^{3} + 2x}{5} = \frac{1}{5}\left(x^{3} + 2x\right)$, whose derivative is $\frac{3x^{2} + 2}{5}$. The quotient rule gives the same thing after cancelling a factor of 5.
2. A single power underneath. $\dfrac{1}{x^{3}} = x^{-3}$, whose derivative is $-3x^{-4}$ by the power rule. Running the quotient rule gives $\frac{0 \cdot x^{3} - 1 \cdot 3x^{2}}{x^{6}} = \frac{-3}{x^{4}}$, the same answer with more opportunities to slip.

What cannot be rewritten away is a denominator with more than one term. $\frac{1}{x + 3}$ does not split, and the quotient rule is the tool for it until the chain rule arrives in Unit 3.

When the numerator is a constant, do not forget that $u' = 0$ kills the first term and the answer is $\frac{-uv'}{v^2}$, carrying a minus sign that has nowhere to hide.

The product and quotient rules are close enough in shape to be confused, and the confusion always runs in the same direction: the plus becomes a minus, or the order gets swapped.

$$\left(uv\right)' = u'v + uv' \qquad \text{versus} \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}.$$

Two anchors. First, a quotient rule answer must have a squared denominator; if yours does not, you used the wrong rule. Second, the term that leads is always the one that differentiates the numerator, which matches the order the expression is read in.

Neither rule tolerates the shortcut version. $\frac{u'}{v'}$ fails immediately on $u = x^2$, $v = x$: the quotient is $x$ with derivative 1, while $\frac{u'}{v'} = \frac{2x}{1} = 2x$.