Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Product & quotient rules

8 min read

Font

Discuss

Feedback

What you’ll learn:

How to differentiate products or quotients of two or more functions

Product rule

In the previous section, we found the derivative of

$y = (x + 1) (x^{3} - 2)$

by fully expanding and then using the power rule on each term. That gave

$y^{'} = 4 x^{3} + 3 x^{2} - 2.$

Expanding works, but it can get messy quickly. A more direct method for differentiating a product is the product rule.

The formula for differentiating the product of two functions $f (x)$ and $g (x)$ is:

Product rule:

$\frac{d}{d x} [f (x) g (x)] = f (x) g^{'} (x) + f^{'} (x) g (x)$

Here’s the idea in words:

Multiply the first function by the derivative of the second.
Add the derivative of the first function times the second.

The mnemonic

$"Left D right + right D left"$

where “left” is the left (1st) function, “D” means “derivative,” and “right” is the right (2nd) function, may help you remember the product rule.

If you want to see where the product rule comes from, the proof using the limit definition of the derivative is on Wikipedia.

Examples

Let’s redo the same problem using the product rule.

Find the derivative of

$y = (x + 1) (x^{3} - 2)$

using the product rule.

Solution

(spoiler)

$y$ is the product of two functions:

$f (x) g (x) = x + 1 = x^{3} - 2$

It helps to list each function and its derivative before substituting into the formula.

Function	Expression
$f (x)$	$x + 1$
$f^{'} (x)$	$1$
$g (x)$	$x^{3} - 2$
$g^{'} (x)$	$3 x^{2}$

Now substitute into the product rule:

$\frac{d}{d x} [f (x) g (x)] = (x + 1) (3 x^{2}) f (x) g^{'} (x) + (1) (x^{3} - 2) f^{'} (x) g (x) = 3 x^{3} + 3 x^{2} + x^{3} - 2 = 4 x^{3} + 3 x^{2} - 2$

This matches the result from expanding first.

Differentiate

$h (t) = t (t + 1)$

in two different ways:

i. With the product rule. ii. With the power rule (expanding into individual terms with powers first).

Solution

(spoiler)

Answer: $h^{'} (t) = \frac{3}{2} t^{1/2} + \frac{1}{2} t^{- 1/2}$

i. With product rule

Write $h (t)$ as a product $f (t) g (t)$ :

Function	Expression
$f (t)$	$t^{1/2}$
$f^{'} (t)$	$\frac{1}{2} t^{- 1/2}$
$g (t)$	$t + 1$
$g^{'} (t)$	$1$

Apply the product rule:

$h^{'} (t) = (t^{1/2}) (1) f (t) g^{'} (t) + (\frac{1}{2} t^{- 1/2}) (t + 1 f^{'} (t) g (t)) = t^{1/2} + \frac{1}{2} t^{1/2} + \frac{1}{2} t^{- 1/2} = \frac{3}{2} t^{1/2} + \frac{1}{2 t ^{1/2}}$

ii. With power rule

First rewrite $t$ as $t^{1/2}$ and distribute:

$h (t) = t^{1/2} (t + 1) = t^{3/2} + t^{1/2} .$

Now apply the power rule term-by-term:

$h^{'} (t) = \frac{3}{2} t^{1/2} + 1/2 t^{- 1/2} .$

Next, let’s find the derivative of a function that is the product of 3 functions.

Find the derivative of

$f (x) = (x^{2} + 1) (2 x + 1) (3 - x)$

Solution

(spoiler)

Answer: $f^{'} (x) = - 8 x^{3} + 15 x^{2} + 2 x + 5$

Even though $f (x)$ is a product of three factors, you can still use the product rule by grouping two factors into one function. For example, define

$L (x) R (x) = [(x^{2} + 1) (2 x + 1)] = (3 - x)$

Then $f (x) = L (x) R (x)$ .

Keep in mind: to find $L^{'} (x)$ , you also need the product rule (since $L (x)$ is itself a product).

Function	Expression
$L (x)$	$(x^{2} + 1) (2 x + 1)$
$L^{'} (x)$	$6 x^{2} + 2 x + 2$
$R (x)$	$3 - x$
$R^{'} (x)$	$- 1$

Now apply the product rule to $f (x) = L (x) R (x)$ :

$f^{'} (x) = (x^{2} + 1) (2 x + 1) (- 1) + (6 x^{2} + 2 x + 2) (3 - x)$

After expanding and simplifying,

$f^{'} (x) = - 8 x^{3} + 15 x^{2} + 2 x + 5$

Alternatively, you could expand $L (x) = (x^{2} + 1) (2 x + 1)$ into a polynomial, use the power rule to find $L^{'} (x)$ , and then combine with $R (x) = (3 - x)$ using the product rule.

You could also fully expand the entire function and use only the power rule.

Let’s do a word problem that involves derivatives.

A rectangle has length $(2 t - 1)$ and height $(5 t + 3)$ , where $t$ is the time in seconds and the dimensions are in meters. What is the rate of change of the area with respect to time?

Solution

(spoiler)

Answer: $\frac{d A}{d t} = 20 t + 1$

“Rate of change” tells you to take a derivative. Here, the quantity changing is the area $A$ , and the variable is time $t$ , so we want $\frac{d A}{d t}$ .

Start by writing the area as a function of time. Since area = (length)(height),

$A (t) = (2 t - 1) (5 t + 3) .$

Differentiate using the product rule:

$\frac{d}{d t} [A (t)] = (2 t - 1) (5) + 2 (5 t + 3) = 10 t - 5 + 10 t + 6 = 20 t + 1$

This derivative gives the rate of change of area at time $t$ . The units are $meters^{2} / second$ because it represents (change in area)/(change in time).

Quotient rule

When a function is written as a quotient of two functions, the quotient rule can be used.

Here’s the formula to differentiate a quotient $\frac{f ( x )}{g ( x )}$ :

Quotient rule:

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

A good way to read the formula is as a sequence of steps:

Multiply the bottom function by the derivative of the top.
Subtract the top function times the derivative of the bottom.
Divide by the square of the bottom function.

A catchy way to remember the rule is the rhyme

$"Lo-D-Hi minus Hi-D-Lo, all over the square of what’s below."$

“Lo” represents the lower/bottom function, “Hi” is the higher/top function, “D” means derivative again, and the “square of what’s below” means square the bottom function.

The proof of the quotient rule from the limit definition of the derivative is also on Wikipedia.

Examples

Find the derivative of

$h (x) = \frac{x + 1}{x - 1}$

Solution

(spoiler)

Answer: $h^{'} (x) = - \frac{2}{( x - 1 ) ^{2}}$

Identify the top and bottom functions:

Top: $f (x) = (x + 1)$ , so $f^{'} (x) = 1$ .
Bottom: $g (x) = (x - 1)$ , so $g^{'} (x) = 1$ .

Substitute into the quotient rule:

$h^{'} (x) = \frac{( x - 1 ) ( 1 ) g ( x ) f ^{'} ( x ) - ( x + 1 ) ( 1 ) f ( x ) g ^{'} ( x )}{( g ( x ) ) ^{2} ( x - 1 ) ^{2}} = \frac{x - 1 - x - 1}{( x - 1 ) ^{2}} = \frac{- 2}{( x - 1 ) ^{2}}$

Find the derivative of

$y = \frac{3 x ^{4} + 1}{x ^{2}}$

in two different ways:

i. With the quotient rule. ii. No quotient rule.

Solution

(spoiler)

Answer: $y^{'} = 6 x - \frac{2}{x ^{3}}$

i. Quotient rule

Let

$f (x) = 3 x^{4} + 1$ , so $f^{'} (x) = 12 x^{3}$ .
$g (x) = x^{2}$ , so $g^{'} (x) = 2 x$ .

Apply the quotient rule:

$y^{'} = \frac{( x ^{2} ) ( 12 x ^{3} ) - ( 3 x ^{4} + 1 ) ( 2 x )}{( x ^{2} ) ^{2}}$

$= \frac{12 x ^{5} - 6 x ^{5} - 2 x}{x ^{4}}$

$= \frac{6 x ^{5} - 2 x}{x ^{4}}$

Now split the fraction into two simpler terms:

$\frac{6 x ^{5}}{x ^{4}} - \frac{2 x}{x ^{4}}$

$= 6 x - \frac{2}{x ^{3}}$

ii. No quotient rule

Start by splitting the original fraction:

$y = \frac{3 x ^{4}}{x ^{2}} + \frac{1}{x ^{2}} .$

Simplify each term:

$y = 3 x^{2} + x^{- 2} .$

Differentiate using the power rule:

$y^{'} = 6 x - 2 x^{- 3}$

$= 6 x - \frac{2}{x ^{3}}$

AP tip:

Look for opportunities to split fractions to save time when the quotient rule can be avoided. However, be careful not to split a fraction incorrectly. For example,

$\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$

To see why, substitute numbers such as

$\frac{8}{2 + 1}$

which does not equal

$\frac{8}{2} + \frac{8}{1}$

Let’s find the derivative of

$f (x) = \frac{x ^{3}}{x + 1}$

using the quotient rule.

Solution

(spoiler)

$f^{'} (x) = \frac{( x + 1 ) ( 3 x ^{2} ) - ( x ^{3} ) ( 1 )}{( x + 1 ) ^{2}} = \frac{3 x ^{3} + 3 x ^{2} - x ^{3}}{( x + 1 ) ^{2}} = \frac{2 x ^{3} + 3 x ^{2}}{( x + 1 ) ^{2}}$

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

you don’t need the quotient rule. It’s faster to recognize that $\frac{π}{4}$ is just a constant coefficient. Keep it in front (constant multiple rule) and apply the power rule to $x^{2}$ .

$y^{'} = \frac{π}{4} (2 x) = \frac{π x}{2}$

Product rule

Differentiates product of two functions: $(f (x) g (x))^{'} = f (x) g^{'} (x) + f^{'} (x) g (x)$
Mnemonic: “Left D right + right D left”
Expand only if simpler; product rule is more direct for complex products

Product rule examples

For $y = (x + 1) (x^{3} - 2)$ :
- $y^{'} = 4 x^{3} + 3 x^{2} - 2$
For $h (t) = t (t + 1)$ :
- Product rule: $h^{'} (t) = \frac{3}{2} t^{1/2} + \frac{1}{2} t^{- 1/2}$
- Power rule (expand first): same result
For $f (x) = (x^{2} + 1) (2 x + 1) (3 - x)$ :
- Group as two functions, apply product rule (and within, another product rule)
- $f^{'} (x) = - 8 x^{3} + 15 x^{2} + 2 x + 5$
Word problem (rectangle area): If $A (t) = (2 t - 1) (5 t + 3)$ ,
- $\frac{d A}{d t} = 20 t + 1$
- Units: $meters^{2} / second$

Quotient rule

Differentiates quotient: $(\frac{f ( x )}{g ( x )})^{'} = \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{[ g ( x ) ] ^{2}}$
Steps:
- Bottom $\times$ derivative of top
- Minus top $\times$ derivative of bottom
- Divide by square of bottom
Mnemonic: “Lo-D-Hi minus Hi-D-Lo, all over the square of what’s below”

Quotient rule examples

For $h (x) = \frac{x + 1}{x - 1}$ :
- $h^{'} (x) = - \frac{2}{( x - 1 ) ^{2}}$
For $y = \frac{3 x ^{4} + 1}{x ^{2}}$ :
- Quotient rule: $y^{'} = 6 x - \frac{2}{x ^{3}}$
- Or, rewrite as $3 x^{2} + x^{- 2}$ and use power rule
For $f (x) = \frac{x ^{3}}{x + 1}$ :
- $f^{'} (x) = \frac{2 x ^{3} + 3 x ^{2}}{( x + 1 ) ^{2}}$
Do not split fractions incorrectly (e.g., $\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$ )
Coefficient-only denominators (e.g., $\frac{π x ^{2}}{4}$ ): treat as constant, use power rule directly

Product & quotient rules

What you’ll learn:

How to differentiate products or quotients of two or more functions

Product rule

In the previous section, we found the derivative of

$y = (x + 1) (x^{3} - 2)$

by fully expanding and then using the power rule on each term. That gave

$y^{'} = 4 x^{3} + 3 x^{2} - 2.$

Expanding works, but it can get messy quickly. A more direct method for differentiating a product is the product rule.

The formula for differentiating the product of two functions $f (x)$ and $g (x)$ is:

Product rule:

$\frac{d}{d x} [f (x) g (x)] = f (x) g^{'} (x) + f^{'} (x) g (x)$

Here’s the idea in words:

Multiply the first function by the derivative of the second.
Add the derivative of the first function times the second.

The mnemonic

$"Left D right + right D left"$

where “left” is the left (1st) function, “D” means “derivative,” and “right” is the right (2nd) function, may help you remember the product rule.

If you want to see where the product rule comes from, the proof using the limit definition of the derivative is on Wikipedia.

Examples

Let’s redo the same problem using the product rule.

Find the derivative of

$y = (x + 1) (x^{3} - 2)$

using the product rule.

Solution

(spoiler)

$y$ is the product of two functions:

$f (x) g (x) = x + 1 = x^{3} - 2$

It helps to list each function and its derivative before substituting into the formula.

Function	Expression
$f (x)$	$x + 1$
$f^{'} (x)$	$1$
$g (x)$	$x^{3} - 2$
$g^{'} (x)$	$3 x^{2}$

Now substitute into the product rule:

$\frac{d}{d x} [f (x) g (x)] = (x + 1) (3 x^{2}) f (x) g^{'} (x) + (1) (x^{3} - 2) f^{'} (x) g (x) = 3 x^{3} + 3 x^{2} + x^{3} - 2 = 4 x^{3} + 3 x^{2} - 2$

This matches the result from expanding first.

Differentiate

$h (t) = t (t + 1)$

in two different ways:

i. With the product rule. ii. With the power rule (expanding into individual terms with powers first).

Solution

(spoiler)

Answer: $h^{'} (t) = \frac{3}{2} t^{1/2} + \frac{1}{2} t^{- 1/2}$

i. With product rule

Write $h (t)$ as a product $f (t) g (t)$ :

Function	Expression
$f (t)$	$t^{1/2}$
$f^{'} (t)$	$\frac{1}{2} t^{- 1/2}$
$g (t)$	$t + 1$
$g^{'} (t)$	$1$

Apply the product rule:

$h^{'} (t) = (t^{1/2}) (1) f (t) g^{'} (t) + (\frac{1}{2} t^{- 1/2}) (t + 1 f^{'} (t) g (t)) = t^{1/2} + \frac{1}{2} t^{1/2} + \frac{1}{2} t^{- 1/2} = \frac{3}{2} t^{1/2} + \frac{1}{2 t ^{1/2}}$

ii. With power rule

First rewrite $t$ as $t^{1/2}$ and distribute:

$h (t) = t^{1/2} (t + 1) = t^{3/2} + t^{1/2} .$

Now apply the power rule term-by-term:

$h^{'} (t) = \frac{3}{2} t^{1/2} + 1/2 t^{- 1/2} .$

Next, let’s find the derivative of a function that is the product of 3 functions.

Find the derivative of

$f (x) = (x^{2} + 1) (2 x + 1) (3 - x)$

Solution

(spoiler)

Answer: $f^{'} (x) = - 8 x^{3} + 15 x^{2} + 2 x + 5$

Even though $f (x)$ is a product of three factors, you can still use the product rule by grouping two factors into one function. For example, define

$L (x) R (x) = [(x^{2} + 1) (2 x + 1)] = (3 - x)$

Then $f (x) = L (x) R (x)$ .

Keep in mind: to find $L^{'} (x)$ , you also need the product rule (since $L (x)$ is itself a product).

Function	Expression
$L (x)$	$(x^{2} + 1) (2 x + 1)$
$L^{'} (x)$	$6 x^{2} + 2 x + 2$
$R (x)$	$3 - x$
$R^{'} (x)$	$- 1$

Now apply the product rule to $f (x) = L (x) R (x)$ :

$f^{'} (x) = (x^{2} + 1) (2 x + 1) (- 1) + (6 x^{2} + 2 x + 2) (3 - x)$

After expanding and simplifying,

$f^{'} (x) = - 8 x^{3} + 15 x^{2} + 2 x + 5$

Alternatively, you could expand $L (x) = (x^{2} + 1) (2 x + 1)$ into a polynomial, use the power rule to find $L^{'} (x)$ , and then combine with $R (x) = (3 - x)$ using the product rule.

You could also fully expand the entire function and use only the power rule.

Let’s do a word problem that involves derivatives.

A rectangle has length $(2 t - 1)$ and height $(5 t + 3)$ , where $t$ is the time in seconds and the dimensions are in meters. What is the rate of change of the area with respect to time?

Solution

(spoiler)

Answer: $\frac{d A}{d t} = 20 t + 1$

“Rate of change” tells you to take a derivative. Here, the quantity changing is the area $A$ , and the variable is time $t$ , so we want $\frac{d A}{d t}$ .

Start by writing the area as a function of time. Since area = (length)(height),

$A (t) = (2 t - 1) (5 t + 3) .$

Differentiate using the product rule:

$\frac{d}{d t} [A (t)] = (2 t - 1) (5) + 2 (5 t + 3) = 10 t - 5 + 10 t + 6 = 20 t + 1$

This derivative gives the rate of change of area at time $t$ . The units are $meters^{2} / second$ because it represents (change in area)/(change in time).

Quotient rule

When a function is written as a quotient of two functions, the quotient rule can be used.

Here’s the formula to differentiate a quotient $\frac{f ( x )}{g ( x )}$ :

Quotient rule:

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

A good way to read the formula is as a sequence of steps:

Multiply the bottom function by the derivative of the top.
Subtract the top function times the derivative of the bottom.
Divide by the square of the bottom function.

A catchy way to remember the rule is the rhyme

$"Lo-D-Hi minus Hi-D-Lo, all over the square of what’s below."$

“Lo” represents the lower/bottom function, “Hi” is the higher/top function, “D” means derivative again, and the “square of what’s below” means square the bottom function.

The proof of the quotient rule from the limit definition of the derivative is also on Wikipedia.

Examples

Find the derivative of

$h (x) = \frac{x + 1}{x - 1}$

Solution

(spoiler)

Answer: $h^{'} (x) = - \frac{2}{( x - 1 ) ^{2}}$

Identify the top and bottom functions:

Top: $f (x) = (x + 1)$ , so $f^{'} (x) = 1$ .
Bottom: $g (x) = (x - 1)$ , so $g^{'} (x) = 1$ .

Substitute into the quotient rule:

$h^{'} (x) = \frac{( x - 1 ) ( 1 ) g ( x ) f ^{'} ( x ) - ( x + 1 ) ( 1 ) f ( x ) g ^{'} ( x )}{( g ( x ) ) ^{2} ( x - 1 ) ^{2}} = \frac{x - 1 - x - 1}{( x - 1 ) ^{2}} = \frac{- 2}{( x - 1 ) ^{2}}$

Find the derivative of

$y = \frac{3 x ^{4} + 1}{x ^{2}}$

in two different ways:

i. With the quotient rule. ii. No quotient rule.

Solution

(spoiler)

Answer: $y^{'} = 6 x - \frac{2}{x ^{3}}$

i. Quotient rule

Let

$f (x) = 3 x^{4} + 1$ , so $f^{'} (x) = 12 x^{3}$ .
$g (x) = x^{2}$ , so $g^{'} (x) = 2 x$ .

Apply the quotient rule:

$y^{'} = \frac{( x ^{2} ) ( 12 x ^{3} ) - ( 3 x ^{4} + 1 ) ( 2 x )}{( x ^{2} ) ^{2}}$

$= \frac{12 x ^{5} - 6 x ^{5} - 2 x}{x ^{4}}$

$= \frac{6 x ^{5} - 2 x}{x ^{4}}$

Now split the fraction into two simpler terms:

$\frac{6 x ^{5}}{x ^{4}} - \frac{2 x}{x ^{4}}$

$= 6 x - \frac{2}{x ^{3}}$

ii. No quotient rule

Start by splitting the original fraction:

$y = \frac{3 x ^{4}}{x ^{2}} + \frac{1}{x ^{2}} .$

Simplify each term:

$y = 3 x^{2} + x^{- 2} .$

Differentiate using the power rule:

$y^{'} = 6 x - 2 x^{- 3}$

$= 6 x - \frac{2}{x ^{3}}$

AP tip:

Look for opportunities to split fractions to save time when the quotient rule can be avoided. However, be careful not to split a fraction incorrectly. For example,

$\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$

To see why, substitute numbers such as

$\frac{8}{2 + 1}$

which does not equal

$\frac{8}{2} + \frac{8}{1}$

Let’s find the derivative of

$f (x) = \frac{x ^{3}}{x + 1}$

using the quotient rule.

Solution

(spoiler)

$f^{'} (x) = \frac{( x + 1 ) ( 3 x ^{2} ) - ( x ^{3} ) ( 1 )}{( x + 1 ) ^{2}} = \frac{3 x ^{3} + 3 x ^{2} - x ^{3}}{( x + 1 ) ^{2}} = \frac{2 x ^{3} + 3 x ^{2}}{( x + 1 ) ^{2}}$

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

you don’t need the quotient rule. It’s faster to recognize that $\frac{π}{4}$ is just a constant coefficient. Keep it in front (constant multiple rule) and apply the power rule to $x^{2}$ .

$y^{'} = \frac{π}{4} (2 x) = \frac{π x}{2}$

Key points

Product rule

Differentiates product of two functions: $(f (x) g (x))^{'} = f (x) g^{'} (x) + f^{'} (x) g (x)$
Mnemonic: “Left D right + right D left”
Expand only if simpler; product rule is more direct for complex products

Product rule examples

For $y = (x + 1) (x^{3} - 2)$ :
- $y^{'} = 4 x^{3} + 3 x^{2} - 2$
For $h (t) = t (t + 1)$ :
- Product rule: $h^{'} (t) = \frac{3}{2} t^{1/2} + \frac{1}{2} t^{- 1/2}$
- Power rule (expand first): same result
For $f (x) = (x^{2} + 1) (2 x + 1) (3 - x)$ :
- Group as two functions, apply product rule (and within, another product rule)
- $f^{'} (x) = - 8 x^{3} + 15 x^{2} + 2 x + 5$
Word problem (rectangle area): If $A (t) = (2 t - 1) (5 t + 3)$ ,
- $\frac{d A}{d t} = 20 t + 1$
- Units: $meters^{2} / second$

Quotient rule

Differentiates quotient: $(\frac{f ( x )}{g ( x )})^{'} = \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{[ g ( x ) ] ^{2}}$
Steps:
- Bottom $\times$ derivative of top
- Minus top $\times$ derivative of bottom
- Divide by square of bottom
Mnemonic: “Lo-D-Hi minus Hi-D-Lo, all over the square of what’s below”

Quotient rule examples

For $h (x) = \frac{x + 1}{x - 1}$ :
- $h^{'} (x) = - \frac{2}{( x - 1 ) ^{2}}$
For $y = \frac{3 x ^{4} + 1}{x ^{2}}$ :
- Quotient rule: $y^{'} = 6 x - \frac{2}{x ^{3}}$
- Or, rewrite as $3 x^{2} + x^{- 2}$ and use power rule
For $f (x) = \frac{x ^{3}}{x + 1}$ :
- $f^{'} (x) = \frac{2 x ^{3} + 3 x ^{2}}{( x + 1 ) ^{2}}$
Do not split fractions incorrectly (e.g., $\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$ )
Coefficient-only denominators (e.g., $\frac{π x ^{2}}{4}$ ): treat as constant, use power rule directly