The Quotient Rule

An enormous number of the quantities we actually care about are ratios: speed is distance over time, fuel economy is kilometres over litres, average cost is total cost over items made, a batting average is runs over dismissals. And in every interesting case, both the top and the bottom are changing at once. A factory's costs rise while its output rises too — so is the cost per item going up or down? That's a question about the derivative of a quotient: one function divided by another, like \dfrac{x^2}{x + 1}.

You might hope you could just differentiate top and bottom separately. You can't. Try it on \dfrac{x^2}{x}, which is simply x in disguise: the true derivative is 1, but "top' over bottom'" gives \tfrac{2x}{1} = 2x — wrong everywhere except one point. Quotients need their own rule, just as products needed the product rule.

And the quotient rule comes with a sting the product rule doesn't have: a minus sign. The product rule u'v + uv' is perfectly symmetric — swap u and v and nothing changes, which is only fair, since uv = vu. But \dfrac{u}{v} \neq \dfrac{v}{u}, and the rule knows it: its numerator is antisymmetric, so getting the order wrong doesn't just dent your answer — it flips the sign of every single term.

The rule

If u(x) and v(x) are differentiable and v(x) \neq 0, then \frac{d}{dx}\!\left[\frac{u}{v}\right] = \frac{u'\,v - u\,v'}{v^2}.
The numerator starts with the derivative of the top: u'v first, then minus uv'.
The denominator is v^2 — the original bottom function, squared.

Two things separate this from the product rule: the minus instead of a plus (so order matters), and the division by v^2. Generations of students have kept the order straight with a chant — "low d-high minus high d-low, over low-low", where high is the top u, low is the bottom v, and "d" means "the derivative of":

\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}

Notice that v\,u' - u\,v' and u'v - uv' are the same thing — multiplication commutes, so it doesn't matter whether you write the v before or after the u'. What you may never do is swap which function gets differentiated first: because of the minus sign, u'v - uv' is the exact negative of uv' - u'v. Always differentiate the top first.

The simplest quotient of all

Before anything fancy, feed the rule the humblest fraction there is: \dfrac{1}{x}. Here u = 1 and v = x, so u' = 0 and v' = 1:

\frac{d}{dx}\!\left[\frac{1}{x}\right] = \frac{(0)(x) - (1)(1)}{x^2} = -\frac{1}{x^2}.

The minus sign in the answer came straight from the minus sign in the rule — and it's telling the truth. Here is \dfrac{1}{x} (bold) with its derivative -\dfrac{1}{x^2} (dashed): on both branches the curve always slopes downhill, so its derivative is negative everywhere it exists — and very steeply negative near x = 0, where the curve plunges.

The same computation with any constant c on top gives a pattern worth memorising, since it appears constantly:

\frac{d}{dx}\!\left[\frac{c}{x}\right] = \frac{(0)(x) - (c)(1)}{x^2} = -\frac{c}{x^2}.

A worked example

Differentiate f(x) = \dfrac{x^2}{x + 1}. Both top and bottom genuinely change with x, so this is honest quotient-rule territory. Set u = x^2 and v = x + 1, so u' = 2x and v' = 1. Drop them into the formula — top-derivative first:

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, and best left factored. (Expanding it to x^2 + 2x + 1 gains you nothing and hides the structure.) A quick sanity check: at x = 1 the formula gives f'(1) = \tfrac{1 + 2}{4} = \tfrac{3}{4}, a gentle positive slope — and indeed f climbs from f(1) = \tfrac{1}{2} to f(2) = \tfrac{4}{3}, so a positive slope is exactly right.

See the slope

Here is f(x) = \dfrac{x^2}{x+1} (bold) plotted with the derivative we just computed, 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 quotient (and so the rule) simply doesn't apply, and both curves blow up beside the vertical asymptote.

Away from the break, the picture and the algebra agree beautifully. The numerator x^2 + 2x = x(x+2) is zero at x = 0 and x = -2 — and those are precisely the two places the bold curve levels off: a valley bottom on the right branch, a hilltop on the left. Between them (across the asymptote) the dashed derivative sits below zero and the function falls; outside them it sits above zero and the function climbs. The v^2 in the denominator also whispers something subtle: being a square, it is never negative — so the sign of f' is decided entirely by the numerator u'v - uv'.

Check it two ways

Should you trust a rule with this many moving parts? Test it against something you can already do. Take g(x) = \dfrac{x^2 + 1}{x}.

Way 1 — the quotient rule. With u = x^2 + 1, v = x, so u' = 2x and v' = 1:

g'(x) = \frac{(2x)(x) - (x^2 + 1)(1)}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2} = 1 - \frac{1}{x^2}.

Way 2 — no quotient rule at all. This particular fraction splits when you divide each term of the top by x:

g(x) = \frac{x^2 + 1}{x} = \frac{x^2}{x} + \frac{1}{x} = x + x^{-1}.

Now it's just two powers, and the power rule (which works for negative powers too) does everything:

g'(x) = 1 + (-1)\,x^{-2} = 1 - \frac{1}{x^2}.

The same answer, twice — that's how you build trust in a new rule. And it teaches a genuinely useful habit: before reaching for the quotient rule, look for a rewrite. Whenever the bottom is a single term, splitting the fraction turns a quotient-rule problem into an easy power-rule one, with far fewer chances to drop a minus sign.

When there's no way round it

The escape hatch doesn't always open. Differentiate s(x) = \dfrac{x^2}{x^2 + 1}. The bottom is a genuine sum, so no amount of term-splitting simplifies it — this is what the quotient rule is for. With u = x^2, v = x^2 + 1, so u' = 2x and v' = 2x:

s'(x) = \frac{(2x)(x^2 + 1) - (x^2)(2x)}{(x^2 + 1)^2} = \frac{2x^3 + 2x - 2x^3}{(x^2 + 1)^2} = \frac{2x}{(x^2 + 1)^2}.

Look how much cancelled: the two 2x^3 terms wiped each other out, leaving something tiny. That kind of collapse is common with the quotient rule — which is why you should always simplify the numerator before calling the answer done. The result also reads well: s' is negative for x < 0, zero at x = 0 and positive for x > 0, so s falls, bottoms out at 0, then rises — climbing towards (but never reaching) 1, since the top x^2 is always one unit shy of the bottom x^2 + 1.

Three sign-eating traps, in descending order of carnage:

You cannot differentiate top and bottom separately. \left(\tfrac{u}{v}\right)' \neq \tfrac{u'}{v'}. The one-line refutation from the top of the page: \tfrac{x^2}{x} = x has derivative 1, but \tfrac{u'}{v'} = \tfrac{2x}{1} = 2x. If differentiating were that easy, there'd be no rule to learn.
The numerator order matters: u'v - uv', top-derivative first. Swap the two terms and the minus sign flips every sign in your answer — you get exactly the negative of the truth, which is the most dangerous kind of wrong because the algebra still "works". Check yourself on \tfrac{1}{x}: the right order gives -\tfrac{1}{x^2} (downhill, matching the graph); the swapped order gives +\tfrac{1}{x^2}, claiming a falling curve rises.
The denominator is v^2 — the original bottom, squared. Not (v')^2 (don't square the derivative), not plain v (don't forget the square), and not the square of your answer's numerator. For \tfrac{x}{x^2+1} the denominator is (x^2+1)^2 — writing (2x)^2 there is a classic exam giveaway.

Where the rule comes from

The quotient rule isn't a new law of nature — it's the product rule wearing a disguise, and unmasking it takes two lines. Call the quotient q = \dfrac{u}{v}. Multiply both sides by v and the fraction disappears:

u = q \cdot v.

Now differentiate both sides. The left side is just u'; the right side is a product, so the product rule handles it:

u' = q'v + qv'.

Solve for the thing we want, q', then substitute q = \tfrac{u}{v} back in and clear the nested fraction by multiplying top and bottom by v:

q' = \frac{u' - qv'}{v} = \frac{u' - \frac{u}{v}\,v'}{v} = \frac{u'v - uv'}{v^2}.

There it is — minus sign, squared denominator and all, forced into existence by the product rule and a little algebra. This is also the best way to recover the rule in an exam if your memory of the chant deserts you: you can re-derive it faster than you can second-guess yourself.

Here's a small scandal: plenty of professionals almost never use the rule you've just learned. Their trick is to stop seeing quotients at all. Any quotient is secretly a product with a negative power:

\frac{u}{v} = u \cdot v^{-1},

and then the product rule — together with the chain rule, which you'll meet as a peer of this rule — differentiates it with no new formula to memorise and, crucially, no order to get backwards. Done that way, the minus sign appears on its own, falling out of the derivative of v^{-1}.

Meanwhile, the classroom fights a centuries-old mnemonic war over the same formula. Team "lo d-hi": "lo d-hi minus hi d-lo, over lo-lo". Team rhyme: "low d-high minus high d-low — draw the line and square below!" Team theatre sings it to the tune of old jingles, and some teachers act out "the bottom holding the derivative of the top first". They all encode the identical fact: bottom-times-derivative-of-top comes first, and the bottom gets squared. Pick whichever sticks — or join the professionals and quietly rewrite the fraction as a product. Both roads lead to \tfrac{u'v - uv'}{v^2}.

Watch Sal explain it

Sal walks through the rule and a worked example — including the "product rule applied to u \cdot v^{-1}" point of view from the vignette above: