The quotient rule

Differentiating a quotient y = u/v as (u'v - uv') / v² — getting the order in the numerator and the squared denominator right, and when to prefer rewriting as a product instead.

Pure · Differentiation

The quotient rule

Differentiating a quotient y = u/v as (u'v - uv') / v² — getting the order in the numerator and the squared denominator right, and when to prefer rewriting as a product instead.

Why it works

A quotient

y = \dfrac{u}{v}

is just a product in disguise:

y = u\,v^{-1}

. Differentiate it with the product and chain rules and the quotient rule falls out, so you never have to take it on faith.

With

y = u v^{-1}

, the product rule gives

\frac{dy}{dx} = u'\,v^{-1} + u\cdot(-1)v^{-2}v' = \frac{u'}{v} - \frac{uv'}{v^2}.

Put both terms over the common denominator

v^2

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.

That is the rule. In words: (derivative of the top times the bottom) minus (the top times the derivative of the bottom), all over the bottom squared.

Two things trip people up, and both come straight from the derivation:

The order in the numerator matters. It is $u'v - uv'$ , not $uv' - u'v$ —

the minus sign comes from differentiating

v^{-1}

, and swapping the terms flips the sign of the whole answer. (Contrast the product rule, where the two terms are added so order is harmless.) A reliable memory hook: start with the top's derivative, "low d-high minus high d-low".

The denominator is squared. It is $v^2$ , the bottom function squared — a

frequent slip is to leave it as just

v

.

When to skip it. If the denominator is a single power of

x

, rewriting as a sum of powers is faster and less error-prone. For

y = \dfrac{x^3 + 2x}{x}

, just split it:

y = x^2 + 2

, so

\frac{dy}{dx} = 2x

— no quotient rule needed. Save the rule for genuine quotients like

\dfrac{2x+1}{x^2 - 3}

where the bottom cannot be divided out.