Math Core

Rational functions, $\dfrac{\sin x}{x}$ , $\dfrac{e^x}{x + 1}$ : many important functions are quotients. The quotient rule differentiates them directly. Like the product rule, it has two terms, but the order of those terms now matters, because subtraction is involved.

The rule

The quotient rule

If $f$ and $g$ are differentiable and $g(x) \ne 0$ , then

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\big[g(x)\big]^2}

A popular way to remember it, with "high" for the numerator and "low" for the denominator:

low d-high, minus high d-low, over the square of what's below.

"Low d-high" means $g \cdot f'$ , and it comes first. Because of the subtraction, swapping the two terms flips the sign of your answer.

Where it comes from

Write $Q(x) = \dfrac{f(x)}{g(x)}$ , so $f = Q \cdot g$ . Differentiate both sides with the product rule:

f' = Q' g + Q g'

Solve for $Q'$ and substitute $Q = \dfrac{f}{g}$ :

Q' = \frac{f' - Q g'}{g} = \frac{f' - \frac{f}{g} g'}{g} = \frac{f' g - f g'}{g^2}

(This argument assumes $Q$ is differentiable; a proof from the definition confirms it.)

Using the rule

As with products, name the pieces first: the numerator $f$ , the denominator $g$ , and their derivatives. Then assemble and simplify the numerator.

Worked example: A rational function

Differentiate $y = \dfrac{x^2}{x + 1}$ .

$f = x^2$ , $f' = 2x$ , $g = x + 1$ , $g' = 1$ .

\frac{dy}{dx} = \frac{2x(x + 1) - x^2(1)}{(x + 1)^2} = \frac{2x^2 + 2x - x^2}{(x + 1)^2} = \frac{x^2 + 2x}{(x + 1)^2}

Leave the denominator in factored form. Expanding it rarely helps.

Worked example: Exponential over a power

Differentiate $y = \dfrac{e^x}{x}$ and find where the tangent line is horizontal.

$f = e^x$ , $f' = e^x$ , $g = x$ , $g' = 1$ .

\frac{dy}{dx} = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x(x - 1)}{x^2}

A fraction is zero when its numerator is zero (and its denominator isn't). Since $e^x > 0$ , the derivative is zero only at $x = 1$ . The horizontal tangent is at $(1, e)$ .

y = eˣ/x has a horizontal tangent at x = 1, where y = e ≈ 2.718.Open in grapher →

Quotients from a table

Worked example: Using a table of values

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
4	2	3	$-1$	5

If $h(x) = \dfrac{f(x)}{g(x)}$ , find $h'(4)$ .

h'(4) = \frac{f'(4)\,g(4) - f(4)\,g'(4)}{\big[g(4)\big]^2} = \frac{(3)(-1) - (2)(5)}{(-1)^2} = \frac{-3 - 10}{1} = -13

Table questions like this are common on the AP exam because they test whether you know the rule's structure, not just the algebra. Write the formula with the function names first, then substitute. Keeping $f$ , $f'$ , $g$ and $g'$ in their correct slots is the whole problem, and a single swap changes the sign of the answer. Notice too that $\big[g(4)\big]^2 = (-1)^2 = 1$ is positive: the denominator of a quotient-rule derivative is a square, so it can never be negative.

When not to use the quotient rule

The quotient rule always works, but it isn't always the easiest route.

Constant numerator: $\dfrac{5}{x^3} = 5x^{-3}$ , so its derivative is $-15x^{-4}$ by the power rule.
Single-term denominator: $\dfrac{x^3 + 2x}{x} = x^2 + 2$ , so its derivative is $2x$ .
Constant denominator: $\dfrac{x^2 + 1}{7} = \dfrac{1}{7}(x^2 + 1)$ , a constant multiple.

Common mistake

The two most common errors: reversing the numerator (writing $fg' - f'g$ ), which gives the negative of the right answer, and forgetting to square the denominator. Also, don't cancel terms across the minus sign: in $\dfrac{2x(x + 1) - x^2}{(x + 1)^2}$ , the factor $(x + 1)$ in the first term does not cancel with the denominator, because the numerator has two terms.

Tip

Simplify the numerator only. After the numerator is simplified, check whether it shares a factor with the denominator; that's the only safe cancellation.

Practice

Practice 1

Find $f'(x)$ for $f(x) = \dfrac{3x + 1}{x - 2}$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

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

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Find $f'(x)$ for $f(x) = \dfrac{\sin x}{x}$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

The functions $f$ and $g$ are differentiable, with $f(2) = 6$ , $f'(2) = 1$ , $g(2) = 3$ and $g'(2) = -2$ . If $h(x) = \dfrac{f(x)}{g(x)}$ , find $h'(2)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Let $f(x) = \dfrac{x^2 - 1}{x^2 + 1}$ . Find $f'(1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find the value of $x$ where $f(x) = \dfrac{\ln x}{x}$ has a horizontal tangent line.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

What is $\dfrac{d}{dx}\left[\dfrac{5}{x^3}\right]$ ?

Practice 8

Find an equation of the tangent line to $y = \dfrac{e^x}{x + 1}$ at $x = 0$ .

Enter an expression, e.g. 3x^2 - 2x + 1