Product & quotient rules | Derivative basics

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

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

to be

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

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

Another way to differentiate a product of multiple functions is by using the product rule.

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

Product rule:

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

In other words, take the first function times the derivative of the second, then add the second function times the derivative of the first.

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 more easily.

If you’re curious, the proof of the product rule using the limit definition of the derivative can be found in the Wikipedia page.

Examples

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

1. Find the derivative of

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

using the product rule.

Solution

(spoiler)

$y$ is the product of the two functions:

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

It may be easier to keep track of the individual pieces first.

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

Into the formula they go:

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

The results match!

2. 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

$h (t)$ is the product of $f (t)$ and $g (t)$ .

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

Then applying 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

Turn $t = t^{1/2}$ and distribute that to $(t + 1)$ with the exponent addition rule so that

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

Then the power rule can be applied to each term and

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

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

3. Find the derivative of

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

Solution

(spoiler)

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

Although $f (x)$ looks to be the product of 3 functions, we can group two of them to form a block such that $f (x)$ is the product of two functions:

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

Keep in mind that the product rule is also applied when taking the derivative of $L (x)$ .

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

Using the product rule on $f (x)$ :

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

After expanding and simplifying,

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

Alternatively, you could expand $L (x) = (x^{2} + 1) (2 x + 1)$ so that it becomes a polynomial, then use the power rule to find $L^{'} (x)$ before combining it with $R (x) = (3 - x)$ using the product rule.

You could also fully expand the function to use only the power rule. Give these a try for extra practice, and to confirm the above answer!

Let’s do a word problem that involves derivatives.

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

The key word “rate of change” should clue you into taking the derivative. But of what function?

Since we want to know how the area changes for an infinitesimally small change in time, or $\frac{d A}{d t}$ , we first need to define an area function that takes in the input $t$ and gives the area $A$ . The area of a rectangle is its length $\times$ height, so

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

Using the product rule,

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

The rate of change of the area at any time $t$ can be found by inputting the value of $t$ into that derivative function $A^{'} (t)$ . The units for this rate of change will be in $meters^{2} / second$ since the rate is $\frac{d A}{d t}$ , or the change in area over a 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:

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

This may look intimidating, but the steps are: take the bottom (function) times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared.

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 can also be found in its Wikipedia page.

Examples

1. Find the derivative of

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

Solution

(spoiler)

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

The top function is $f (x) = (x + 1)$ with derivative $f^{'} (x) = 1$ .

The bottom function is $g (x) = (x - 1)$ with derivative $g^{'} (x) = 1$ .

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

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

The top function is $f (x) = 3 x^{4} + 1$ and

$f^{'} (x) = 12 x^{3}$

The bottom function is $g (x) = x^{2}$ and

$g^{'} (x) = 2 x$

Applying the quotient rule,

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

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

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

Which can be split into

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

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

ii. No quotient rule

Notice that the fraction can be split directly into

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

then simplified into

$3 x^{2} + x^{- 2}$

Then using the power rule,

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

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

AP tip:

Look for opportunities to split fractions up to save time, if 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 think about why this might be the case, substitute numbers like

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

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

$y = \frac{π x ^{2}}{4}$

avoid wasting time with the quotient rule to find the derivative. It’s doable and will result in the same answer, but it’s faster to recognize that $\frac{π}{4}$ is just a real-number coefficient in front of $x^{2}$ . So keep it in front using the constant multiple rule and apply the power rule on $x^{2}$ .

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

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

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

to be

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

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

Another way to differentiate a product of multiple functions is by using the product rule.

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

Product rule:

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

In other words, take the first function times the derivative of the second, then add the second function times the derivative of the first.

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 more easily.

If you’re curious, the proof of the product rule using the limit definition of the derivative can be found in the Wikipedia page.

Examples

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

1. Find the derivative of

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

using the product rule.

Solution

(spoiler)

$y$ is the product of the two functions:

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

It may be easier to keep track of the individual pieces first.

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

Into the formula they go:

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

The results match!

2. 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

$h (t)$ is the product of $f (t)$ and $g (t)$ .

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

Then applying 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

Turn $t = t^{1/2}$ and distribute that to $(t + 1)$ with the exponent addition rule so that

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

Then the power rule can be applied to each term and

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

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

3. Find the derivative of

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

Solution

(spoiler)

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

Although $f (x)$ looks to be the product of 3 functions, we can group two of them to form a block such that $f (x)$ is the product of two functions:

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

Keep in mind that the product rule is also applied when taking the derivative of $L (x)$ .

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

Using the product rule on $f (x)$ :

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

After expanding and simplifying,

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

You could also fully expand the function to use only the power rule. Give these a try for extra practice, and to confirm the above answer!

Let’s do a word problem that involves derivatives.

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

The key word “rate of change” should clue you into taking the derivative. But of what function?

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

Using the product rule,

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

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:

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

This may look intimidating, but the steps are: take the bottom (function) times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared.

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 can also be found in its Wikipedia page.

Examples

1. Find the derivative of

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

Solution

(spoiler)

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

The top function is $f (x) = (x + 1)$ with derivative $f^{'} (x) = 1$ .

The bottom function is $g (x) = (x - 1)$ with derivative $g^{'} (x) = 1$ .

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

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

The top function is $f (x) = 3 x^{4} + 1$ and

$f^{'} (x) = 12 x^{3}$

The bottom function is $g (x) = x^{2}$ and

$g^{'} (x) = 2 x$

Applying the quotient rule,

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

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

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

Which can be split into

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

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

ii. No quotient rule

Notice that the fraction can be split directly into

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

then simplified into

$3 x^{2} + x^{- 2}$

Then using the power rule,

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

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

AP tip:

Look for opportunities to split fractions up to save time, if 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 think about why this might be the case, substitute numbers like

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

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

$y = \frac{π x ^{2}}{4}$

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