The Quotient Rule
To differentiate a quotient — one function divided by another, like
\dfrac{x^2}{x + 1} — there is a rule that looks like the
product rule but with a twist.
For u(x) over v(x):
\frac{d}{dx}\!\left[\frac{u}{v}\right] = \frac{u'\,v - u\,v'}{v^2}
Two things make this rule different from the product rule: there's a minus
sign instead of a plus (so order matters), and you divide everything by
v^2, the square of the bottom function.
A memory hook
The numerator follows a fixed pattern — "low d-high minus high
d-low", where high is the top
u and low is the bottom v:
\frac{d}{dx}\!\left[\frac{u}{v}\right] = \frac{(\text{low})(\text{d-high}) - (\text{high})(\text{d-low})}{(\text{low})^2} = \frac{v\,u' - u\,v'}{v^2}
The order is what trips people up. Because of the minus sign,
u'v - uv' is not the same as
uv' - u'v — swapping them flips the sign of the whole answer.
Always differentiate the top first.
The simplest quotient of all is \dfrac{1}{x}. Here it is (bold)
with its derivative -\dfrac{1}{x^2} (dashed): the curve always
slopes downhill, so its derivative is negative everywhere.
A worked example
Differentiate f(x) = \dfrac{x^2}{x + 1}. Set
u = x^2 and v = x + 1, so
u' = 2x and v' = 1. Drop them into the
formula:
f'(x) = \frac{u'v - uv'}{v^2} = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2}
Expand the top and tidy up:
= \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}
Leave the denominator as (x+1)^2 — squared, exactly as the rule
demands.
See the slope
Here is f(x) = \dfrac{x^2}{x+1} (bold) plotted with its derivative
f'(x) = \dfrac{x^2 + 2x}{(x+1)^2} (dashed). The function has a
break at x = -1 where the bottom is zero — there the rule (and the
function) simply don't apply. Away from that break, notice the derivative is
0 exactly where the bold curve levels off.
Watch Sal explain it
Many people find the quotient rule easiest to remember as the product rule applied to
u \cdot v^{-1}. Sal walks through both the rule and a worked
example: