Unit 2 · Topic 2.9 · about 30 minutes · 7 parts · 11 scored checks, 2 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 2.9: The Quotient Rule

Q: Do I need the quotient rule when the bottom is a number?

No. f(x)c is 1cf(x), so the constant multiple rule gives 1cf'(x). The quotient rule gives the same answer with more writing.

Q: Can I write the quotient as a product and use the product rule instead?

Yes, f(x)g(x) = f(x)[g(x)]-1, and both ways give the same derivative. But differentiating [g(x)]-1 for a general g takes the chain rule, which comes in Unit 3, so for now the quotient rule is the tool.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 2: 10 to 15% of the multiple-choice section

What you'll be able to do

Calculate the derivative of a quotient of two differentiable functions with the quotient rule, from formulas or from a table of values.

Key vocabulary

Term	What it means	The exam trap
Quotient rule	The rule that the derivative of $\frac{f (x)}{g (x)}$ is $\frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{[g (x)]^{2}}$ , for differentiable $f$ and $g$ wherever $g (x) \neq 0$ .	Write the rule's structure with the function names before you put in any formulas or numbers.
Quotient of functions	The function $\frac{f (x)}{g (x)}$ , defined at each input where $f$ and $g$ are defined and $g (x) \neq 0$ .	Before differentiating, find where $g (x) = 0$ : the quotient and its derivative do not exist there.
Differentiable function	A function whose derivative exists at each input being considered.	A function can be continuous at a point without being differentiable there, so check differentiability before using any derivative rule.

Graph of q(x) = x / (x squared + 1) for x from -4 to 4, with tangent lines at (0, 0) of slope 1, at (1, 0.5) of slope 0, and at (2, 0.4) of slope -0.12.

Part 1 Think first Sweeter or weaker?

Commit to a guess first. Even a wrong one makes the reveal easier to remember.

Predict first

A drink machine is filling a jug. Right now the jug holds 4 liters of liquid with 200 grams of sugar dissolved in it, so the sugar concentration is 50 grams per liter. Sugar water is flowing in at 0.5 liter per minute, and it carries 10 grams of sugar per minute. What is the concentration in the jug doing right now?

In short. The quotient rule says that if $f$ and $g$ are differentiable and $g (x) \neq 0$ , then $\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{[g (x)]^{2}}$ .

The derivative of a quotient is not the quotient of the derivatives.
The numerator is a difference, so its two terms keep a fixed order: the bottom times the derivative of the top comes first.
The bottom of the derivative is the original bottom squared, and the rule applies only where the original bottom is not 0.

Part 2 Learn it The rule, then the two ways you will use it

2a. The quotient rule

If $f$ and $g$ are differentiable and $g (x) \neq 0$ , the derivative of their quotient is

\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{[g (x)]^{2}} .

In words: the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared.

Two things trip people up. The numerator is a difference, so its order matters: swap the two terms and the answer changes sign. And the derivative of a quotient is not the quotient of the derivatives, in the same way that the derivative of a product is not the product of the derivatives.

The graph at the top is $q (x) = \frac{x}{x^{2} + 1}$ . The top is $x$ , with derivative 1, and the bottom is $x^{2} + 1$ , with derivative $2 x$ , so

q^{'} (x) = \frac{(x^{2} + 1) (1) - x (2 x)}{(x^{2} + 1)^{2}} = \frac{1 - x^{2}}{(x^{2} + 1)^{2}} .

At $x = 1$ this is $\frac{0}{4} = 0$ , the flat tangent line in the picture. Dividing the derivatives instead gives $\frac{1}{2 x}$ , which is $\frac{1}{2}$ at $x = 1$ , and the picture shows that cannot be the slope there.

Quick check · Multiple choice

No calculator. Which expression is the derivative of $\frac{x^{2}}{2 x + 1}$ ?

2b. Values from a table: write the structure first

Often you get no formulas at all, only the values of two functions and their derivatives at one input. That is enough. Write the rule with the function names, then put in the numbers.

Suppose $w (x) = \frac{u (x)}{v (x)}$ , with $u (1) = 3$ , $u^{'} (1) = - 2$ , $v (1) = 5$ and $v^{'} (1) = 4$ . Then

w^{'} (1) = \frac{v (1) u^{'} (1) - u (1) v^{'} (1)}{[v (1)]^{2}} = \frac{(5) (- 2) - (3) (4)}{5^{2}} = \frac{- 10 - 12}{25} = - 0.88.

The first expression, the rule written in terms of $v (1)$ , $u^{'} (1)$ , $u (1)$ and $v^{'} (1)$ , is the line that shows anyone reading your work which rule you used. Write it before any arithmetic.

Keep every value in parentheses, and keep the second product together. The minus sign in the middle applies to the whole product $u (1) v^{'} (1)$ . Suppose instead $v^{'} (1) = - 4$ . Then the numerator is $(5) (- 2) - (3) (- 4) = - 10 + 12 = 2$ . Writing $- 10 - 12$ there, by losing the parentheses around $(- 4)$ , is the slip that turns a right setup into a wrong answer.

Quick check · Calculate

No calculator. Let $h (x) = \frac{f (x)}{g (x)}$ , where $f (3) = 2$ , $f^{'} (3) = 5$ , $g (3) = 4$ and $g^{'} (3) = - 1$ . What is $h^{'} (3)$ ? Give an exact decimal.

Your answer

2c. Any differentiable top and bottom, wherever the bottom is not 0

The rule works for any differentiable top and bottom, including the functions whose derivatives you learned in 2.7. For $r (x) = \frac{e^{x}}{x}$ , the top's derivative is $e^{x}$ and the bottom's is 1, so

r^{'} (x) = \frac{x e^{x} - e^{x} (1)}{x^{2}} = \frac{e^{x} (x - 1)}{x^{2}} .

The rule needs $g (x) \neq 0$ . Where the bottom is 0 the quotient has no value, so it has no derivative there either. Here $r$ and $r^{'}$ are both not defined at $x = 0$ .

When the bottom is a constant, the quotient rule still works, but you do not need it. $\frac{x^{3}}{4}$ is $\frac{1}{4} x^{3}$ , and the constant multiple rule gives $\frac{3}{4} x^{2}$ at once; the quotient rule gives $\frac{4 (3 x^{2}) - x^{3} (0)}{16}$ , the same thing. A constant top works the same way: $\frac{5}{x^{2}} = 5 x^{- 2}$ , and the power rule gives $- 10 x^{- 3}$ , which is what the quotient rule's $\frac{x^{2} (0) - 5 (2 x)}{x^{4}}$ simplifies to.

Quick check · Fill the statement

No calculator. Tap a blank, then tap an answer.

Let $q (x) = \frac{\ln x}{x}$ for $x > 0$ . The derivative of the top, $\ln x$ , is , and the derivative of the bottom, $x$ , is . So the quotient rule gives $q^{'} (x)$ as a fraction over $x^{2}$ whose numerator is the bottom times the top's derivative, minus the top times the bottom's derivative. That numerator simplifies to .

Common mistakes

The mistake	Why it is wrong	Do this instead
Dividing the derivatives: writing $\frac{f^{'} (x)}{g^{'} (x)}$ for the derivative of $\frac{f (x)}{g (x)}$ .	A quotient changes when either part changes, and how much depends on the sizes of both parts, not only on their rates. For $\frac{x}{x^{2} + 1}$ at $x = 1$ the slope is 0, while $\frac{1}{2 x}$ gives $\frac{1}{2}$ .	Use the full rule: $\frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{[g (x)]^{2}}$ .
Reversing the numerator to $f (x) g^{'} (x) - g (x) f^{'} (x)$ .	Subtraction does not let its terms trade places. The reversed numerator is the negative of the right one, so every value comes out with the wrong sign.	Start with the bottom: bottom times the derivative of the top, then minus the top times the derivative of the bottom.
Dropping the parentheses around the second product, or forgetting to square the bottom.	The minus sign applies to the whole product $f (x) g^{'} (x)$ , so a negative value there becomes a plus. And the rule divides by $[g (x)]^{2}$ , not $g (x)$ .	Write each product in parentheses and the bottom as $[g (x)]^{2}$ in the structure line, before you put in any numbers.

Part 3 Watch it done A quotient rule derivative, start to finish

Open the steps one at a time, then finish the last one on your own.

Let $k (x) = \frac{x^{2} + 1}{x - 3}$ . Find $k^{'} (5)$ .

Step 1 · Name the top and the bottom
The top is $x^{2} + 1$ , with derivative $2 x$ . The bottom is $x - 3$ , with derivative 1. Both are differentiable everywhere, and the bottom is 0 only at $x = 3$ , so the rule applies at $x = 5$ .
Step 2 · Write the rule's structure
$k^{'} (x) = \frac{(x - 3) (2 x) - (x^{2} + 1) (1)}{(x - 3)^{2}}$ .
Why this step?
Why does $(x - 3) (2 x)$ come first in the numerator, and not $(x^{2} + 1) (1)$ ?
The numerator is a difference, and a difference changes sign when its two terms swap. The rule puts the bottom times the derivative of the top first. With the terms the other way around, every value of $k^{'}$ would have the wrong sign.
Step 3 · Simplify the numerator
$(2 x^{2} - 6 x) - (x^{2} + 1) = x^{2} - 6 x - 1$ , so $k^{'} (x) = \frac{x^{2} - 6 x - 1}{(x - 3)^{2}}$ for $x \neq 3$ .
Your step · Calculate

No calculator. Finish it. For $k (x) = \frac{x^{2} + 1}{x - 3}$ , the quotient rule gives $k^{'} (x) = \frac{x^{2} - 6 x - 1}{(x - 3)^{2}}$ . What is $k^{'} (5)$ ? Give an exact decimal.

Your answer

Part 4 Try it A rate in context, and a shortcut that fails

Try it · Calculate

No calculator. The concentration of a medicine in a patient's blood $t$ hours after a dose is $C (t) = \frac{5 t}{t^{2} + 4}$ milligrams per liter. At what rate, in milligrams per liter per hour, is the concentration changing at $t = 1$ ? Give an exact decimal.

Your answer

Try it · Decide and justify

No calculator. Let $f (x) = \frac{x^{5}}{x^{2}}$ for $x \neq 0$ . A classmate says $f^{'} (x) = \frac{5 x^{4}}{2 x} = 2.5 x^{3}$ , found by dividing the derivative of the top by the derivative of the bottom. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

Free response is written by hand on the exam, so do this one on paper. Nothing you type here leaves the page. When you are done, open the model answer and give yourself each point you earned.

Free response is 50% of the exam score. This one is shaped like Question 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. A town's total water use is $W (t)$ thousand gallons per day, and its population is $P (t)$ thousand people, $t$ years after the start of 2020. Both $W$ and $P$ are differentiable, and $P (t) > 0$ . Selected values are in the table. The town's per capita water use, in gallons per person per day, is $G (t) = \frac{W (t)}{P (t)}$ .

\begin{matrix} t & W (t) & W^{'} (t) & P (t) & P^{'} (t) \\ 2 & 3600 & 60 & 40 & 1 \\ 5 & 3780 & 50 & 42 & 0.5 \end{matrix}

(a) Find $G^{'} (2)$ . Show the work that leads to your answer, and indicate units of measure.

(b) A council member says that because $W$ and $P$ are both increasing at $t = 2$ , per capita water use must also be increasing at $t = 2$ . Use the quotient rule to explain why this reasoning is wrong.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $G^{'} (2) = \frac{P (2) W^{'} (2) - W (2) P^{'} (2)}{[P (2)]^{2}} = \frac{(40) (60) - (3600) (1)}{40^{2}} = \frac{2400 - 3600}{1600} = - 0.75$ gallons per person per day per year.

(b) The denominator $[P (2)]^{2}$ is positive, so $G^{'} (2)$ has the sign of its numerator, $P (2) W^{'} (2) - W (2) P^{'} (2)$ . Both $W^{'} (2)$ and $P^{'} (2)$ are positive, but $W (2) P^{'} (2) = 3600$ is larger than $P (2) W^{'} (2) = 2400$ , so the numerator is negative. Citing $G^{'} (2) = - 0.75 < 0$ from part (a) also answers this part. The population is growing fast enough, compared with water use, that use per person falls even though both totals rise.

(c) $G^{'} (5) = \frac{P (5) W^{'} (5) - W (5) P^{'} (5)}{[P (5)]^{2}} = \frac{(42) (50) - (3780) (0.5)}{42^{2}} = \frac{2100 - 1890}{1764} = \frac{210}{1764}$ . The numerator 210 and the denominator are both positive, so $G^{'} (5) > 0$ , and per capita water use is increasing at $t = 5$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $G^{'} (2) = \frac{P (2) W^{'} (2) - W (2) P^{'} (2)}{[P (2)]^{2}}$ , with the terms in this order, before substituting.
Gets $G^{'} (2) = - 0.75$ , with units of gallons per person per day per year.
Explains that $G^{'} (2) < 0$ , either from part (a) or because its numerator $P (2) W^{'} (2) - W (2) P^{'} (2) = 2400 - 3600$ is negative and $[P (2)]^{2} > 0$ , so $G$ decreases at $t = 2$ although $W$ and $P$ both increase.
Computes the numerator at $t = 5$ as $2100 - 1890 = 210$ , or $G^{'} (5) = \frac{210}{1764}$ .
Concludes that per capita water use is increasing at $t = 5$ because $G^{'} (5) > 0$ .

Part 6 Check yourself Two functions in a table

Both case questions read from one table, so keep it in view while you work.

The functions $f$ and $g$ are differentiable for all real numbers. The table gives their values and the values of their derivatives at two inputs.

Values of f, g and their derivatives
$x$	$f (x)$	$f^{'} (x)$	$g (x)$	$g^{'} (x)$
1	4	$- 2$	2	3
3	6	1	$- 3$	2

Case question 1 · Calculate

No calculator. Let $h (x) = \frac{f (x)}{g (x)}$ . What is $h^{'} (1)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Let $k (x) = \frac{g (x)}{f (x)}$ . What is $k^{'} (3)$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the derivative of $g (x) = \frac{2 x + 1}{x^{2} + 3}$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. Tap a function, then its derivative.

Part 7 Keep it Bring back earlier ideas

One question from the lesson just before this one, and one from Unit 1. Recalling them now is what keeps them.

Bring it back · Calculate

No calculator. Let $m (x) = (x^{2} - 3) (2 x + 5)$ . What is $m^{'} (2)$ ?

Your answer

From 1.12 Confirming Continuity over an Interval · Multiple choice

No calculator. On which of these intervals is $g (x) = \frac{\ln x}{x - 4}$ continuous?

In three lines

The derivative of $\frac{f (x)}{g (x)}$ is $\frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{[g (x)]^{2}}$ , wherever $f$ and $g$ are differentiable and $g (x) \neq 0$ .
The numerator's order is fixed and the bottom is squared, and the result is never just $f^{'} (x)$ divided by $g^{'} (x)$ .
From a table, write the rule with function names first, then substitute with every value in parentheses.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a table gives values of $f$ , $g$ and their derivatives, write the rule's structure with the function names, such as $h^{'} (2) = \frac{g (2) f^{'} (2) - f (2) g^{'} (2)}{[g (2)]^{2}}$ , before you substitute, and put every negative value in parentheses. That line shows which rule you used before any arithmetic can go wrong.

Questions students ask

How do I remember the order of the numerator?

Say it the same way every time: bottom times derivative of top, minus top times derivative of bottom. If you are unsure, test it on $\frac{x^{2}}{x}$ , which is $x$ for $x \neq 0$ and has derivative 1. The right order gives $\frac{x (2 x) - x^{2} (1)}{x^{2}} = 1$ ; the reversed order gives $- 1$ .

Do I need the quotient rule when the bottom is a number?

No. $\frac{f (x)}{c}$ is $\frac{1}{c} f (x)$ , so the constant multiple rule gives $\frac{1}{c} f^{'} (x)$ . The quotient rule gives the same answer with more writing.

Can I write the quotient as a product and use the product rule instead?

Yes, $\frac{f (x)}{g (x)} = f (x) [g (x)]^{- 1}$ , and both ways give the same derivative. But differentiating $[g (x)]^{- 1}$ for a general $g$ takes the chain rule, which comes in Unit 3, so for now the quotient rule is the tool.

Should I simplify the derivative?

To find its value at one input, you can substitute straight into the structure line; every extra step of algebra is another chance for a slip. Simplify when you need the derivative as a function, for example to see where it is 0.

Where does the quotient rule not apply?

Where the bottom is 0, because the quotient has no value there, and where the top or the bottom is not differentiable.

Course alignment, for teachers

AP Calculus AB topic 2.9, Unit 2: Differentiation: Definition and Fundamental Properties.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.