The Quotient Rule — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

Fuel economy is distance divided by fuel; a medicine's concentration is the amount of drug divided by the volume of blood it spreads through. When the top and the bottom of a fraction both change, the fraction changes in a way neither part shows alone. The quotient rule gives that rate.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\frac{d}{dx}\left( f(x)g(x) \right)$ is

$f'(x)g(x) + f(x)g'(x)$
$f'(x)g'(x)$
$f'(x)g(x) - f(x)g'(x)$
$f(x)g'(x)$

Show the answer

The product rule: each factor takes a turn being differentiated, and the terms add.

Correct: $f'(x)g(x) + f(x)g'(x)$ :
$f'(x)g'(x)$ :
$f'(x)g(x) - f(x)g'(x)$ :
$f(x)g'(x)$ :

2. $\frac{d}{dx}\left( x^2 + 1 \right)$ is

$2x$
$2x + 1$
$x^2$
$2$

Show the answer

Power rule on $x^2$ ; the constant has derivative 0.

Correct: $2x$ :
$2x + 1$ :
$x^2$ :
$2$ :

Part 4 · See it

See it first

Top: the graph of f(x) = x over (x squared plus 1), with a low point at (minus 1, minus 0.5) and a high point at (1, 0.5). Level tangent lines are drawn at x = minus 1 and 1, and a tangent line of slope 1 at x = 0. Bottom: the graph of f prime of x = (1 minus x squared) over (x squared plus 1) squared, which crosses 0 at x = minus 1 and 1 and has height 1 at x = 0. — The quotient $f(x) = \frac{x}{x^2 + 1}$ (top) and its derivative from the quotient rule, $\frac{1 - x^2}{(x^2 + 1)^2}$ (bottom): zero where the tangent lines are level, at $x = \pm 1$ , and 1 at $x = 0$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Write $q = \frac{f}{g}$ as $f = q\,g$ the product rule gives $f' = q'g + qg'$
Solve for $q'$ and replace $q$ with $\frac{f}{g}$ $q' = \frac{f'g - fg'}{g^2}$
The numerator is a differencethe order matters: derivative of the top times the bottom comes first
The denominator is a single power of $x$ rewriting as a sum of powers is often faster than the quotient rule

Part 6 · Key ideas

Key ideas

The quotient rule: $\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right)$ $= \frac{f'(x)g(x) - f(x)g'(x)}{\left( g(x) \right)^2}$ , where $g(x) \ne 0$ .
Memory aid: "low d-high minus high d-low, over low squared."
The order in the numerator matters: swapping it flips the sign of the answer.
The derivative of a quotient is NOT the quotient of the derivatives.
For $\frac{3x^2 - 5}{x}$ , divide first: $3x - 5x^{-1}$ , then use the power rule.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{d}{dx}\left( \frac{f}{g} \right)$ $= \frac{fg' - f'g}{g^2}$ , or $\frac{f'}{g'}$ .

What actually happens: The numerator starts with the derivative of the top: $f'g - fg'$ . Reversing it gives the negative of the right answer. And $\frac{f'}{g'}$ ignores how each part's size affects the other: $\frac{x^2}{x} = x$ has derivative 1, but $\frac{2x}{1} = 2x$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of two functions and their derivatives

The functions

f

and

g

have derivatives for all

x

.

Selected values of $f$ , $f'$ , $g$ and $g'$
$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
2	4	1	2	$-3$
5	$-1$	6	3	2

1. Let $q(x) = \frac{f(x)}{g(x)}$ . Find $q'(2)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$q'(2)$ $= \frac{f'(2)g(2) - f(2)g'(2)}{\left( g(2) \right)^2}$ $= \frac{(1)(2) - (4)(-3)}{2^2}$ $= \frac{14}{4}$ $= \frac{7}{2}$ .

Answer: $\frac{7}{2}$ ≈ 3.500

2. Let $r(x) = \frac{g(x)}{f(x)}$ . Find $r'(5)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$r'(5)$ $= \frac{g'(5)f(5) - g(5)f'(5)}{\left( f(5) \right)^2}$ $= \frac{-2 - 18}{(-1)^2}$ $= -20$ .

Answer: -20

3. Let $s(x) = \frac{x}{g(x)}$ . Find $s'(5)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$s'(5)$ $= \frac{(1)g(5) - 5g'(5)}{\left( g(5) \right)^2}$ $= \frac{3 - 10}{9}$ $= -\frac{7}{9}$ .

Answer: $\frac{-7}{9}$ ≈ -0.778

4. Which is an equation of the line tangent to the graph of $q(x) = \frac{f(x)}{g(x)}$ at $x = 2$ ?

$y = 2 + \frac{7}{2}(x - 2)$
$y = 2 - \frac{7}{2}(x - 2)$
$y = \frac{7}{2} + 2(x - 2)$
$y = 2 - \frac{1}{3}(x - 2)$

Show the answer

$q(2) = \frac{4}{2} = 2$ and $q'(2) = \frac{7}{2}$ .

Correct: $y = 2 + \frac{7}{2}(x - 2)$ : Height 2, slope $\frac{7}{2}$ .
$y = 2 - \frac{7}{2}(x - 2)$ : The slope's sign comes from a reversed numerator.
$y = \frac{7}{2} + 2(x - 2)$ : The height and the slope are swapped.
$y = 2 - \frac{1}{3}(x - 2)$ : Slope $-\frac{1}{3}$ is $\frac{f'(2)}{g'(2)}$ , not the quotient rule.

5. A student writes $\frac{d}{dx}\left( \frac{x}{x + 1} \right)$ $= \frac{(1)(x + 1) - x(1)}{x + 1}$ $= \frac{1}{x + 1}$ . What is the error?

The denominator should be $(x + 1)^2$ .
The numerator should be $x(1) - (1)(x + 1)$ .
The answer should be $\frac{1}{1}$ , the derivatives of the top and bottom.
There is no error.

Show the answer

The quotient rule divides by the square of the bottom: the derivative is $\frac{1}{(x + 1)^2}$ .

Correct: The denominator should be $(x + 1)^2$ .: The bottom is squared.
The numerator should be $x(1) - (1)(x + 1)$ .: The student's numerator order is right: derivative of the top comes first.
The answer should be $\frac{1}{1}$ , the derivatives of the top and bottom.: The quotient of the derivatives is not the derivative of the quotient.
There is no error.: The missing square changes the answer at every $x$ except where $x + 1 = \pm 1$ .

6. Which formula is the quotient rule for $\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right)$ ?

$\frac{f'(x)g(x) - f(x)g'(x)}{\left( g(x) \right)^2}$
$\frac{f(x)g'(x) - f'(x)g(x)}{\left( g(x) \right)^2}$
$\frac{f'(x)g(x) - f(x)g'(x)}{\left( f(x) \right)^2}$
$\frac{f'(x)g(x) + f(x)g'(x)}{\left( g(x) \right)^2}$

Show the answer

Low d-high minus high d-low, over low squared.

Correct: $\frac{f'(x)g(x) - f(x)g'(x)}{\left( g(x) \right)^2}$ : Derivative of the top times the bottom first; the bottom squared.
$\frac{f(x)g'(x) - f'(x)g(x)}{\left( g(x) \right)^2}$ : The numerator is reversed, which flips the sign.
$\frac{f'(x)g(x) - f(x)g'(x)}{\left( f(x) \right)^2}$ : The denominator is the bottom squared, not the top.
$\frac{f'(x)g(x) + f(x)g'(x)}{\left( g(x) \right)^2}$ : A quotient has a minus sign in the numerator; the plus belongs to the product rule.

Part 9 · Summary

Summary

The derivative of

\frac{f}{g}

is

\frac{f'g - fg'}{g^2}

, derived from the product rule by writing

f = \frac{f}{g} \cdot g

. Keep the numerator in order and square the denominator. When the denominator is a single term, dividing first and using the power rule is often simpler.

Part 10 · Up next

What comes next

Part 11 · Connections