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Achievable AP Calculus AB

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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 first and then applying the power rule term by term.

Expanding works, but there’s usually a faster approach when you’re differentiating a product of functions: use the product rule.

Here’s the formula for the derivative of the product of two functions $f (x)$ and $g (x)$ :

Product rule:

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

Read it like this:

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

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.

Common pitfall: $(f g)^{'} \neq = f^{'} g^{'}$ . You cannot simply differentiate each factor separately and multiply them together. Always use the product rule formula: $(f g)^{'} = f \cdot g^{'} + f^{'} \cdot g$ .

If you’d like to see where the 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.

(spoiler)

Solution:

$y$ 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).

(spoiler)

Solution:

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:

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

ii. With power rule

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 to each term:

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

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

Find the derivative of

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

(spoiler)

Solution:

Answer:

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

Even though $f (x)$ is a product of three factors, you can treat it as a product of two functions by grouping:

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

Then $f (x) = L (x) R (x)$ . Remember: you’ll still need the product rule again to find $L^{'} (x)$ .

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

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 original function and use only the power rule. Try both approaches to confirm the same final derivative.

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?

(spoiler)

Solution:

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 $A$ .

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:

$\frac{d}{d t} [A (t)] = (2 t - 1) (5) f (t) g^{'} (t) + (2) (5 t + 3) f^{'} (t) g (t) = 10 t - 5 + 10 t + 6 = 20 t + 1$

So the rate of change of area at time $t$ is $A^{'} (t) = 20 t + 1$ . The units are $meters^{2} / second$ because you’re measuring “change in area per change in time,” $\frac{d A}{d t}$ .

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 keep the structure straight is to think in steps:

(Bottom)(derivative of top)
minus (top)(derivative of bottom)
all over (bottom) $^{2}$

The rhyme

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

where “low” represents the bottom function and “high” is the top function, “D” means derivative again, and the “square of what’s below” means square the bottom function, is a catchy way to remember the rule.

Common pitfall: The order in the numerator matters. The quotient rule is $\frac{g f ^{'} - f g ^{'}}{g ^{2}}$ , not $\frac{f g ^{'} - g f ^{'}}{g ^{2}}$ . Reversing the order flips the sign of your answer. When in doubt, think “bottom times derivative of top, minus top times derivative of bottom.”

The proof of the quotient rule from the limit definition of the derivative can also be found on Wikipedia.

Examples

Find the derivative of

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

(spoiler)

Solution:

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 $g^{'} (x) = 1$

Apply 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.

(spoiler)

Solution:

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:

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

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

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

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

Split the original fraction first:

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

Simplify:

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

Now use 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 if you can avoid the quotient rule. Just make sure you’re splitting correctly. For example,

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

To see why, try substituting numbers such as

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

which does not equal

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

you don’t need the quotient rule. The fraction $\frac{π}{4}$ is just a constant coefficient multiplying $x^{2}$ .

Keep the constant in front and apply the constant multiple rule and the power rule:

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

Product rule

Formula: $\frac{d}{d x} [f (x) g (x)] = f (x) g^{'} (x) + f^{'} (x) g (x)$
Mnemonic: “Left D right + right D left”
Use for derivatives of products of functions

Product rule examples

List each function and its derivative before substituting
Can use product rule or expand first, both give same result
For more than two factors, group and apply product rule recursively

Word problems with product rule

Area or other quantities as products: differentiate using product rule
Rate of change = derivative with respect to time (e.g., $\frac{d A}{d t}$ )

Quotient rule

Formula: $\frac{d}{d x} [\frac{f ( x )}{g ( x )}] = \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{[ g ( x ) ] ^{2}}$
Mnemonic: “Lo-D-Hi minus Hi-D-Lo, all over the square of what’s below”
Use for derivatives of quotients of functions

Quotient rule examples

Identify top ( $f (x)$ ) and bottom ( $g (x)$ ) functions and their derivatives
Substitute into quotient rule formula and simplify
Alternative: rewrite as sum of terms (if possible) to use power rule instead

AP tip and simplification

Split fractions into simpler terms before differentiating, if possible
Do not split denominators incorrectly (e.g., $\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$ )
Constants in numerators/denominators can be factored out and treated as coefficients

Testing details tag

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 first and then applying the power rule term by term.

Expanding works, but there’s usually a faster approach when you’re differentiating a product of functions: use the product rule.

Here’s the formula for the derivative of the product of two functions $f (x)$ and $g (x)$ :

Product rule:

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

Read it like this:

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

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’d like to see where the 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.

(spoiler)

Solution:

$y$ 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).

(spoiler)

Solution:

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:

ii. With power rule

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 to each term:

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

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

Find the derivative of

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

(spoiler)

Solution:

Answer:

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

Even though $f (x)$ is a product of three factors, you can treat it as a product of two functions by grouping:

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

Then $f (x) = L (x) R (x)$ . Remember: you’ll still need the product rule again to find $L^{'} (x)$ .

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

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 original function and use only the power rule. Try both approaches to confirm the same final derivative.

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?

(spoiler)

Solution:

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 $A$ .

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:

$\frac{d}{d t} [A (t)] = (2 t - 1) (5) f (t) g^{'} (t) + (2) (5 t + 3) f^{'} (t) g (t) = 10 t - 5 + 10 t + 6 = 20 t + 1$

So the rate of change of area at time $t$ is $A^{'} (t) = 20 t + 1$ . The units are $meters^{2} / second$ because you’re measuring “change in area per change in time,” $\frac{d A}{d t}$ .

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 keep the structure straight is to think in steps:

(Bottom)(derivative of top)
minus (top)(derivative of bottom)
all over (bottom) $^{2}$

The rhyme

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

The proof of the quotient rule from the limit definition of the derivative can also be found on Wikipedia.

Examples

Find the derivative of

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

(spoiler)

Solution:

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 $g^{'} (x) = 1$

Apply 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.

(spoiler)

Solution:

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:

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

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

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

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

Split the original fraction first:

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

Simplify:

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

Now use 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 if you can avoid the quotient rule. Just make sure you’re splitting correctly. For example,

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

To see why, try substituting numbers such as

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

which does not equal

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

you don’t need the quotient rule. The fraction $\frac{π}{4}$ is just a constant coefficient multiplying $x^{2}$ .

Keep the constant in front and apply the constant multiple rule and the power rule:

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

Achievable AP Calculus AB

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

Testing details tag

8 min read

Font

Discuss

Feedback

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 first and then applying the power rule term by term.

Expanding works, but there’s usually a faster approach when you’re differentiating a product of functions: use the product rule.

Here’s the formula for the derivative of the product of two functions $f (x)$ and $g (x)$ :

Product rule:

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

Read it like this:

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

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’d like to see where the 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.

(spoiler)

Solution:

$y$ 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).

(spoiler)

Solution:

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:

ii. With power rule

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 to each term:

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

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

Find the derivative of

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

(spoiler)

Solution:

Answer:

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

Even though $f (x)$ is a product of three factors, you can treat it as a product of two functions by grouping:

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

Then $f (x) = L (x) R (x)$ . Remember: you’ll still need the product rule again to find $L^{'} (x)$ .

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

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 original function and use only the power rule. Try both approaches to confirm the same final derivative.

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?

(spoiler)

Solution:

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 $A$ .

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:

$\frac{d}{d t} [A (t)] = (2 t - 1) (5) f (t) g^{'} (t) + (2) (5 t + 3) f^{'} (t) g (t) = 10 t - 5 + 10 t + 6 = 20 t + 1$

So the rate of change of area at time $t$ is $A^{'} (t) = 20 t + 1$ . The units are $meters^{2} / second$ because you’re measuring “change in area per change in time,” $\frac{d A}{d t}$ .

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 keep the structure straight is to think in steps:

(Bottom)(derivative of top)
minus (top)(derivative of bottom)
all over (bottom) $^{2}$

The rhyme

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

The proof of the quotient rule from the limit definition of the derivative can also be found on Wikipedia.

Examples

Find the derivative of

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

(spoiler)

Solution:

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 $g^{'} (x) = 1$

Apply 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.

(spoiler)

Solution:

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:

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

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

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

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

Split the original fraction first:

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

Simplify:

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

Now use 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 if you can avoid the quotient rule. Just make sure you’re splitting correctly. For example,

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

To see why, try substituting numbers such as

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

which does not equal

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

you don’t need the quotient rule. The fraction $\frac{π}{4}$ is just a constant coefficient multiplying $x^{2}$ .

Keep the constant in front and apply the constant multiple rule and the power rule:

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

Product rule

Formula: $\frac{d}{d x} [f (x) g (x)] = f (x) g^{'} (x) + f^{'} (x) g (x)$
Mnemonic: “Left D right + right D left”
Use for derivatives of products of functions

Product rule examples

List each function and its derivative before substituting
Can use product rule or expand first, both give same result
For more than two factors, group and apply product rule recursively

Word problems with product rule

Area or other quantities as products: differentiate using product rule
Rate of change = derivative with respect to time (e.g., $\frac{d A}{d t}$ )

Quotient rule

Formula: $\frac{d}{d x} [\frac{f ( x )}{g ( x )}] = \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{[ g ( x ) ] ^{2}}$
Mnemonic: “Lo-D-Hi minus Hi-D-Lo, all over the square of what’s below”
Use for derivatives of quotients of functions

Quotient rule examples

Identify top ( $f (x)$ ) and bottom ( $g (x)$ ) functions and their derivatives
Substitute into quotient rule formula and simplify
Alternative: rewrite as sum of terms (if possible) to use power rule instead

AP tip and simplification

Split fractions into simpler terms before differentiating, if possible
Do not split denominators incorrectly (e.g., $\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$ )
Constants in numerators/denominators can be factored out and treated as coefficients

Testing details tag

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 first and then applying the power rule term by term.

Expanding works, but there’s usually a faster approach when you’re differentiating a product of functions: use the product rule.

Here’s the formula for the derivative of the product of two functions $f (x)$ and $g (x)$ :

Product rule:

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

Read it like this:

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

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’d like to see where the 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.

(spoiler)

Solution:

$y$ 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).

(spoiler)

Solution:

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:

ii. With power rule

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 to each term:

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

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

Find the derivative of

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

(spoiler)

Solution:

Answer:

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

Even though $f (x)$ is a product of three factors, you can treat it as a product of two functions by grouping:

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

Then $f (x) = L (x) R (x)$ . Remember: you’ll still need the product rule again to find $L^{'} (x)$ .

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

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 original function and use only the power rule. Try both approaches to confirm the same final derivative.

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?

(spoiler)

Solution:

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 $A$ .

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:

$\frac{d}{d t} [A (t)] = (2 t - 1) (5) f (t) g^{'} (t) + (2) (5 t + 3) f^{'} (t) g (t) = 10 t - 5 + 10 t + 6 = 20 t + 1$

So the rate of change of area at time $t$ is $A^{'} (t) = 20 t + 1$ . The units are $meters^{2} / second$ because you’re measuring “change in area per change in time,” $\frac{d A}{d t}$ .

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 keep the structure straight is to think in steps:

(Bottom)(derivative of top)
minus (top)(derivative of bottom)
all over (bottom) $^{2}$

The rhyme

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

The proof of the quotient rule from the limit definition of the derivative can also be found on Wikipedia.

Examples

Find the derivative of

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

(spoiler)

Solution:

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 $g^{'} (x) = 1$

Apply 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.

(spoiler)

Solution:

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:

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

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

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

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

Split the original fraction first:

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

Simplify:

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

Now use 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 if you can avoid the quotient rule. Just make sure you’re splitting correctly. For example,

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

To see why, try substituting numbers such as

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

which does not equal

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

you don’t need the quotient rule. The fraction $\frac{π}{4}$ is just a constant coefficient multiplying $x^{2}$ .

Keep the constant in front and apply the constant multiple rule and the power rule:

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