The Quotient Rule

A quotient written as a product

Let u and v be functions of x, and write u' for $\frac{du}{dx}$ and v' for $\frac{dv}{dx}$. To differentiate $y = \frac{u}{v}$, write the quotient as a product: $\frac{u}{v} = u \times \left(\frac{1}{v}\right)$.

The product rule, (uv)' = u'v + uv', then applies to u and $\frac{1}{v}$. It needs the derivative of each factor, and the one that is new is the derivative of $\frac{1}{v}$.

The derivative of $\frac{1}{v}$

Write v for v(x) and V for v(x + h). The change in $\frac{1}{v}$ is $\frac{1}{V} - \frac{1}{v} = \frac{{v - V}}{{v \cdot V}}$. Divide by h: the difference quotient is $-\left(\frac{{V - v}}{h}\right) \times \frac{1}{{v \cdot V}}$.

As $h \to 0$, $\frac{{V - v}}{h}$ tends to v', and V tends to v. So the derivative of $\frac{1}{v}$ is $- \frac{1}{v^{2}} \times v'$. The minus sign is there because v − V is the change in v taken the other way round: when v grows, $\frac{1}{v}$ shrinks.

With v = x this gives $- \frac{1}{x^{2}}$, the power rule’s answer for $x^{-1}$. With $v = x^{2} + 1$ it gives $- \frac{2x}{\left(x^{2} + 1\right)^{2}}$, which is $- \frac{2}{4} = -0.5$ at x = 1; the chord from x = 1 to x = 1.01 on $y = \frac{1}{{x^{2} + 1}}$ has gradient −0.4975.

The gold curve $y = \frac{1}{x}$, whose gradient is $- \frac{1}{x^{2}}$. The gold line is its tangent at (1, 1), with gradient −1, and the dashed line is its tangent at (2, 0.5), with gradient $- \frac{1}{4}$.

The rule

Put this into the product rule: $\left(u \times \left(\frac{1}{v}\right)\right)' = u' \times \left(\frac{1}{v}\right) - u \times \left(\frac{1}{v^{2}}\right) \times v'$.

Over the common denominator $v^{2}$, the first term is $\frac{u'v}{v^{2}}$ and the second is $\frac{{uv'}}{v^{2}}$. So $\left(\frac{u}{v}\right)' = \frac{{u'v - uv'}}{v^{2}}$. In words: the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all over the denominator squared. It holds wherever $v \ne 0$.

Check it on $\frac{x^{3}}{x}$, which is $x^{2}$, with derivative 2x. Here $u = x^{3}$ and v = x, so the rule gives $\frac{{3x^{2} \cdot x - x^{3} \cdot 1}}{x^{2}} = \frac{2x^{3}}{x^{2}} = 2x$.

An example, checked

Take $y = \frac{x^{2}}{{x + 1}}$, for $x \ne -1$. The numerator is $u = x^{2}$, so u' = 2x; the denominator is v = x + 1, so v' = 1.

The rule gives $\frac{dy}{dx} = \frac{{2x\left(x + 1\right) - x^{2} \cdot 1}}{\left(x + 1\right)^{2}} = \frac{{x^{2} + 2x}}{\left(x + 1\right)^{2}}$. At x = 1 that is $\frac{{1 + 2}}{4} = 0.75$, and at x = 2 it is $\frac{8}{9}$.

Check at x = 1 with a chord. $y\left(1\right) = \frac{1}{2} = 0.5$ and $y\left(1.01\right) = \frac{1.0201}{2.01} = 0.5075124$, so the chord gradient is $\frac{0.0075124}{0.01} = 0.7512$, close to 0.75.

The gold curve $y = \frac{x^{2}}{{x + 1}}$ and, at (1, 0.5), the gold tangent with gradient 0.75 from $\frac{{u'v - uv'}}{v^{2}}$. The dashed line has gradient −0.75, from the numerator taken in the other order: it cuts across a curve that is rising.

The order matters

In the product rule the two terms are added, so their order makes no difference. In the quotient rule they are subtracted. Swapping them gives $\frac{{uv' - u'v}}{v^{2}}$, which is exactly the negative of the derivative.

For $y = \frac{x^{2}}{{x + 1}}$ at x = 1, the swapped order gives $\frac{{1 - 4}}{4} = -0.75$. That says the curve is falling at x = 1, but y goes from 0.5 to 0.5075 between x = 1 and x = 1.01.

One way to hold the order: a growing denominator makes the quotient smaller, so the term with v′ in it is the one that is subtracted. The term with u′ in it comes first.

A constant over v

When the numerator is a constant k, u' = 0 and the rule leaves $\frac{{0 \times v - kv'}}{v^{2}}$, which is $- \frac{k}{v^{2}} \times v'$. For $y = \frac{3}{{x^{2} + 1}}$ that is $- \frac{6x}{\left(x^{2} + 1\right)^{2}}$, which is $- \frac{6}{4} = -1.5$ at x = 1.

When the denominator is a single power of x, the power rule is quicker. $\frac{3}{x^{2}} = 3x^{-2}$, which differentiates to $-6x^{-3}$. The quotient rule gives the same: $\frac{{0 \cdot x^{2} - 3 \cdot 2x}}{x^{4}} = - \frac{6x}{x^{4}} = - \frac{6}{x^{3}}$.

The usual mistakes

Differentiating the numerator and the denominator separately. For $\frac{x^{2}}{{x + 1}}$ that gives $\frac{2x}{1}$, which is 2 at x = 1, not 0.75.

Writing the numerator as uv′ − u′v. That changes the sign of every answer.

Adding the two terms, as in the product rule. The term with v′ is subtracted.

Leaving the denominator as v. The derivative of $\frac{1}{v}$ is $- \frac{1}{v^{2}} \times v'$, so the denominator is $v^{2}$, the original denominator squared.

A drug in the blood

In the application below, a concentration is $\frac{20x}{{x^{2} + 4}}$. The quotient rule gives its rate of change, and the numerator of that derivative, set to zero, gives the time of the greatest concentration.