The Quotient Rule (AP Calculus AB 2.9)

A quotient such as $\dfrac{3x - 1}{x^2 + 2}$ can't be split into separate terms, and dividing the derivatives gives the wrong answer for the same reason that multiplying them fails for products (topic 2.8). The quotient rule gives the right derivative. It follows from the product rule in a few lines of algebra, and its one awkward feature, a subtraction whose order matters, comes straight out of that derivation.

Where the rule comes from

Let $q(x) = \dfrac{f(x)}{g(x)}$ , where $g(x) \neq 0$ . Multiplying both sides by $g(x)$ turns the quotient into a product:

f(x) = q(x)\,g(x).

Assuming $q$ is differentiable (a proof straight from the limit definition shows it is, wherever $f$ and $g$ are differentiable and $g \neq 0$ ), differentiate both sides with the product rule:

f'(x) = q'(x)\,g(x) + q(x)\,g'(x).

Solve for $q'(x)$ , then replace $q(x)$ with $\frac{f(x)}{g(x)}$ :

q'(x) = \frac{f'(x) - q(x)\,g'(x)}{g(x)} = \frac{f'(x) - \dfrac{f(x)}{g(x)}g'(x)}{g(x)}.

Multiplying the top and bottom by $g(x)$ clears the inner fraction and gives the rule.

Unlike the product rule, the order in the numerator matters, because subtraction isn't commutative: swapping the two terms changes the sign of the answer. The derivation shows which term comes first. The term with $f'$ , the derivative of the top, is the one that was already there before you solved for $q'$ , so it is the positive one. The sign also makes sense: if the top grows (positive $f'$ ) while the bottom stays fixed, the quotient grows, so the $f'$ term must be positive; if the bottom grows (positive $g'$ ) while the top stays fixed, the quotient shrinks, so the $g'$ term must be negative.

A quick test: for $\frac{x^2}{x} = x$ with $x \neq 0$ , the rule gives $\frac{2x\cdot x - x^2\cdot 1}{x^2} = \frac{x^2}{x^2} = 1$ , as it should.

Two consequences: reciprocals and negative powers

When the top is the constant 1, its derivative is 0 and the rule shortens to the reciprocal rule:

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

With $g(x) = x^n$ for a positive integer $n$ , this proves the power rule (topic 2.5) for negative integer exponents:

\frac{d}{dx}\,x^{-n} = \frac{d}{dx}\left[\frac{1}{x^n}\right] = -\frac{n x^{n-1}}{x^{2n}} = -n\,x^{-n-1}.

That is exactly $nx^{n-1}$ with $n$ replaced by $-n$ , so the power rule's pattern holds for negative integer exponents too.

When not to use it

The quotient rule always works when its conditions hold, but it is often the long way around. Before using it, check whether the quotient can be rewritten:

Function	Faster route	Derivative
$\dfrac{5}{x^3}$	constant over a power: $5x^{-3}$	$-15x^{-4}$
$\dfrac{x^2 + \sin x}{4}$	constant denominator: $\frac{1}{4}\left(x^2 + \sin x\right)$	$\dfrac{2x + \cos x}{4}$
$\dfrac{x^3 - 2\sqrt{x}}{x}$	single power below: $x^2 - 2x^{-1/2}$	$2x + x^{-3/2}$
$\dfrac{x^2 - 9}{x - 3}$	cancel: $x + 3$ for $x \neq 3$	$1$ for $x \neq 3$

In the last row, the original function is undefined at $x = 3$ , so its derivative is undefined there too, even though $x + 3$ is defined everywhere. Cancelling changes the formula, not the domain.

Use the quotient rule when no rewrite gets rid of the division. That happens when the denominator is a sum that doesn't divide out or a non-power function ( $\frac{x}{x^2 + 1}$ , $\frac{\sin x}{1 + \cos x}$ , $\frac{x^2}{\sin x}$ ). It also happens when the denominator is a single power but the numerator is not a sum of powers ( $\frac{e^x}{x}$ , $\frac{\ln x}{x}$ ): rewriting $\frac{e^x}{x}$ as $e^x \cdot x^{-1}$ leaves a product of two functions of $x$ , which the power rule can't handle.

Worked examples

Worked exampleExample 1: a rational function

Find $\dfrac{dy}{dx}$ for $y = \dfrac{3x - 1}{x^2 + 2}$ .

Show solution

Name the pieces first: top $f = 3x - 1$ with $f' = 3$ ; bottom $g = x^2 + 2$ with $g' = 2x$ . Then

\frac{dy}{dx} = \frac{3\left(x^2 + 2\right) - (3x - 1)(2x)}{\left(x^2 + 2\right)^2}.

Expand the numerator, distributing the minus sign over the whole second product:

3x^2 + 6 - \left(6x^2 - 2x\right) = -3x^2 + 2x + 6,

\frac{dy}{dx} = \frac{-3x^2 + 2x + 6}{\left(x^2 + 2\right)^2}.

Leave the denominator squared and unexpanded. It is always positive here, so the sign of the derivative is the sign of the numerator $-3x^2 + 2x + 6$ .

Worked exampleExample 2: a logarithm over a power

Let $f(x) = \dfrac{\ln x}{x}$ for $x > 0$ .

(a) Find $f'(x)$ .

(b) Find the point where the graph of $f$ has a horizontal tangent line.

Show solution

(a) Top $\ln x$ with derivative $\frac{1}{x}$ ; bottom $x$ with derivative 1:

f'(x) = \frac{\frac{1}{x}\cdot x - \ln x\cdot 1}{x^2} = \frac{1 - \ln x}{x^2}.

(b) $f'(x) = 0$ when the numerator is 0, since $x^2 > 0$ : $1 - \ln x = 0$ , so $x = e$ . The point is $\left(e, f(e)\right) = \left(e, \frac{1}{e}\right)$ .

Rewriting as $x^{-1}\ln x$ and using the product rule would also work; the quotient rule is just as short here.

Worked exampleExample 3: quotients from values (no calculator)

The functions $f$ and $g$ are differentiable, with $f(2) = 4$ , $f'(2) = -1$ , $g(2) = -2$ and $g'(2) = 3$ .

(a) Let $h(x) = \dfrac{f(x)}{g(x)}$ . Find $h'(2)$ .

(b) Let $k(x) = \dfrac{f(x) - 1}{x\,g(x)}$ . Find $k'(2)$ .

Show solution

(a) Write the rule, then substitute:

h'(2) = \frac{f'(2)\,g(2) - f(2)\,g'(2)}{\big[g(2)\big]^2} = \frac{(-1)(-2) - (4)(3)}{(-2)^2} = \frac{2 - 12}{4} = -\frac{5}{2}.

(b) The bottom is itself a product, so first find its value and derivative at 2. Let $d(x) = x\,g(x)$ . Then $d(2) = 2(-2) = -4$ , and by the product rule $d'(x) = g(x) + x\,g'(x)$ , so $d'(2) = -2 + 2(3) = 4$ . The top $n(x) = f(x) - 1$ has $n(2) = 3$ and $n'(2) = -1$ . Now apply the quotient rule:

k'(2) = \frac{n'(2)\,d(2) - n(2)\,d'(2)}{\big[d(2)\big]^2} = \frac{(-1)(-4) - (3)(4)}{16} = \frac{4 - 12}{16} = -\frac{1}{2}.

Computing the four numbers $n(2)$ , $n'(2)$ , $d(2)$ , $d'(2)$ separately keeps the product rule and quotient rule from tangling.

Worked exampleExample 4: an exam-style free-response question (no calculator)

A particle moves along the $x$ -axis. Its position at time $t \ge 0$ is $x(t) = \dfrac{4t}{t^2 + 4}$ . Velocity is the derivative of position (topic 4.2 develops motion in detail).

(a) Find the velocity $v(t)$ of the particle.

(b) For $t > 0$ , at what time is the particle at rest?

Show solution

(a) By the quotient rule, with top $4t$ and bottom $t^2 + 4$ :

v(t) = x'(t) = \frac{4\left(t^2 + 4\right) - 4t(2t)}{\left(t^2 + 4\right)^2} = \frac{16 - 4t^2}{\left(t^2 + 4\right)^2}.

(b) The particle is at rest when $v(t) = 0$ . The denominator is never 0, so set the numerator to 0: $16 - 4t^2 = 0$ gives $t = 2$ (since $t > 0$ ).

(c) $v(3) = \dfrac{16 - 36}{13^2} = -\dfrac{20}{169}$ . Since $v(3) < 0$ , the particle is moving to the left at $t = 3$ .

The reason in (c) has to cite the sign of the velocity. The position $x(3) = \frac{12}{13}$ is positive, but that says where the particle is, not which way it is going.

Where students lose points

On the exam

The quotient rule appears with formulas and with tables of values. On free-response parts, an unsimplified derivative is acceptable unless the question asks for something that depends on simplifying, such as where the derivative is 0. Topic 2.10 uses the quotient rule to find the derivatives of $\tan x$ , $\cot x$ , $\sec x$ and $\csc x$ .

Check yourself

Question 1
If $y = \dfrac{x}{x + 1}$ , then $\dfrac{dy}{dx} =$
1. A
  $\dfrac{1}{(x+1)^2}$
2. B
  $-\dfrac{1}{(x+1)^2}$
3. C
  $1$
4. D
  $\dfrac{2x + 1}{(x+1)^2}$
Show answer
A
$\dfrac{1\cdot(x+1) - x\cdot 1}{(x+1)^2} = \dfrac{1}{(x+1)^2}$ . Choice B reverses the subtraction, choice C divides the derivatives $\frac{1}{1}$ , and choice D adds the two products instead of subtracting.
Question 2
Find the slope of the graph of $y = \dfrac{\sin x}{e^x}$ at $x = 0$ .
Show answer
$1$
$\dfrac{dy}{dx} = \dfrac{\cos x\cdot e^x - \sin x\cdot e^x}{e^{2x}} = \dfrac{\cos x - \sin x}{e^x}$ , which is $\dfrac{1 - 0}{1} = 1$ at $x = 0$ .
Question 3
A differentiable function $g$ has $g(a) = 4$ and $g'(a) = 2$ . What is the derivative of $\dfrac{1}{g(x)}$ at $x = a$ ?
Show answer
$-\dfrac{1}{8}$
The quotient rule with top 1 gives $-\dfrac{g'(x)}{[g(x)]^2}$ , so the value is $-\dfrac{2}{16} = -\dfrac{1}{8}$ . It is negative because a growing denominator makes the reciprocal shrink.
Question 4
Which function's derivative can't be found by rewriting it and using only the power, constant multiple, sum and basic-function rules (topics 2.5 to 2.7), so that it needs the quotient rule?
1. A
  $\dfrac{x^3 + 2x}{x}$
2. B
  $\dfrac{7}{x^4}$
3. C
  $\dfrac{\sin x}{3}$
4. D
  $\dfrac{x}{x^2 + 1}$
Show answer
D
A simplifies to $x^2 + 2$ (for $x \neq 0$ ) and B is $7x^{-4}$ , both handled by the power rule. C is the constant multiple $\frac{1}{3}\sin x$ , so its derivative $\frac{1}{3}\cos x$ needs only the constant multiple rule and the derivative of $\sin x$ , even though $\sin x$ is not a power of $x$ . In D the denominator $x^2 + 1$ is a sum, so it doesn't divide into the numerator term by term, and the quotient rule is required.

Go deeper

OpenStax, Calculus Volume 1, 3.3 Differentiation Rules