A quotient $\frac{f}{g}$ is a product in disguise, so the The Product Rule leads straight to a rule for quotients.
Key idea
$$\left(\frac{f}{g}\right)' = \frac{f' g - f g'}{g^2} \qquad (g \ne 0).$$ Derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared. The order matters because of the minus sign.
Why it works
Let $q = \frac{f}{g}$, so $f = q g$. The product rule gives $f' = q'g + qg'$. Solve for $q'$: $q' = \frac{f' - qg'}{g} = \frac{f' - \frac{f}{g}g'}{g}$. Multiply top and bottom by $g$: $q' = \frac{f'g - fg'}{g^2}$. (This assumes $q$ is differentiable; a limit argument confirms it.)
Worked example: Linear over linear
Differentiate $\dfrac{2x + 1}{x - 3}$.
- Top $f = 2x + 1$, $f' = 2$. Bottom $g = x - 3$, $g' = 1$.
- $$\frac{2(x - 3) - (2x + 1)(1)}{(x - 3)^2} = \frac{2x - 6 - 2x - 1}{(x - 3)^2}$$.
- $= \frac{-7}{(x - 3)^2}$.
Answer: $-\frac{7}{(x - 3)^2}$: always negative, so the function falls on each side of its asymptote.
Worked example: A rational function
Differentiate $\dfrac{x^2}{x^2 + 1}$.
- $f = x^2$, $f' = 2x$; $g = x^2 + 1$, $g' = 2x$.
- $$\frac{2x(x^2 + 1) - x^2(2x)}{(x^2 + 1)^2} = \frac{2x^3 + 2x - 2x^3}{(x^2 + 1)^2}$$.
Answer: $\frac{2x}{(x^2 + 1)^2}$
Worked example: From a table
$f(1) = 4$, $f'(1) = 2$, $g(1) = 2$, $g'(1) = -3$. Find $\left(\frac{f}{g}\right)'(1)$.
- $$\frac{f'(1)g(1) - f(1)g'(1)}{g(1)^2} = \frac{(2)(2) - (4)(-3)}{2^2}$$.
- $= \frac{4 + 12}{4} = 4$.
Answer: $4$
Choosing a rule
- A constant on the bottom: use the constant multiple rule. $\frac{x^3}{7} = \frac{1}{7}x^3$ has derivative $\frac{3}{7}x^2$.
- A constant on top and a power below: rewrite with a negative exponent. $\frac{3}{x^2} = 3x^{-2}$ has derivative $-6x^{-3}$.
- Anything else: the quotient rule. Simplify the numerator at the end, but leave the denominator squared and factored.
Common mistake
Swapping the order in the numerator, $fg' - f'g$, which flips the sign of the answer. Check with a quotient you can do another way: $\frac{x}{1}$ has derivative 1, and $\frac{1 \cdot 1 - x \cdot 0}{1^2} = 1$ is right, while the swapped order gives $-1$. A rhyme some people like: "low d-high minus high d-low, over low squared."
Sometimes simplifying beats the quotient rule: $\frac{x^3 + x}{x} = x^2 + 1$ for $x \ne 0$, with derivative $2x$.