2.4 Product & quotient rules

Achievable AP Calculus AB

2. Derivative basics

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Product & quotient rules

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What you’ll learn

Product and quotient rules: Differentiate products or quotients of 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, which 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 of two functions is the product rule.

Product rule:

To differentiate a product of two functions $f (x)$ and $g (x)$ ,

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

In other 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.

Examples

Differentiate

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

using the product rule.

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

Then applying the product rule,

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

Differentiate

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

in two different ways:

a) With the product rule.

b) With the power rule (expanding into individual terms with powers first).

Solutions

a) With the product rule

(spoiler)

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:

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

b) With the power rule

(spoiler)

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

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

Now apply the power rule term-by-term:

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

Quotient rule

Use the quotient rule to differentiate a quotient of two functions $f (x)$ and $g (x)$ :

Quotient rule:

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

The steps are:

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, and the “square of what’s below” means square the bottom function.

Examples

Find $h^{'} (x)$ for

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

(spoiler)

First, identify the top and bottom functions:

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

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:

a) With the quotient rule.

b) Without the quotient rule (by simplifying the rational expression).

Solutions

a) With the quotient rule

(spoiler)

Top: $f (x) = 3 x^{4} + 1 ⟹ f^{'} (x) = 12 x^{3}$
Bottom: $g (x) = x^{2} ⟹ g^{'} (x) = 2 x$ .

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:

$\frac{6 x ^{5}}{x ^{4}} - \frac{2 x}{x ^{4}} = 6 x - \frac{2}{x ^{3}}$

b) Without the quotient rule (by simplifying the rational expression).

(spoiler)

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

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

recognize that $\frac{π}{4}$ is just a constant coefficient. Use the constant multiple rule while applying 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”
Useful when expanding is impractical

Product rule examples

Identify $f (x)$ and $g (x)$ and their derivatives before applying the rule
Can confirm results by expanding and using the power rule term-by-term

Quotient rule

Differentiates quotient: $[\frac{f ( x )}{g ( x )}]^{'} = \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{[ g ( x ) ] ^{2}}$
Steps:
- Bottom × derivative of top
- Minus top × 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

Identify numerator and denominator functions and their derivatives
Sometimes easier to simplify the expression first and use the power rule instead
- Only split fractions when algebraically valid

AP tip and shortcuts

Split fractions to simplify when possible, but only when algebraically correct
Constants in numerators/denominators can be treated as coefficients (constant multiple rule)
Do not incorrectly split denominators (e.g., $\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$ )

Next | 2.5 Special derivatives

Product & quotient rules

What you’ll learn

Product and quotient rules: Differentiate products or quotients of 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, which 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 of two functions is the product rule.

Product rule:

To differentiate a product of two functions $f (x)$ and $g (x)$ ,

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

In other 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.

Examples

Differentiate

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

using the product rule.

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

Then applying the product rule,

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

Differentiate

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

in two different ways:

a) With the product rule.

b) With the power rule (expanding into individual terms with powers first).

Solutions

a) With the product rule

(spoiler)

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:

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

b) With the power rule

(spoiler)

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

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

Now apply the power rule term-by-term:

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

Quotient rule

Use the quotient rule to differentiate a quotient of two functions $f (x)$ and $g (x)$ :

Quotient rule:

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

The steps are:

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, and the “square of what’s below” means square the bottom function.

Examples

Find $h^{'} (x)$ for

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

(spoiler)

First, identify the top and bottom functions:

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

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:

a) With the quotient rule.

b) Without the quotient rule (by simplifying the rational expression).

Solutions

a) With the quotient rule

(spoiler)

Top: $f (x) = 3 x^{4} + 1 ⟹ f^{'} (x) = 12 x^{3}$
Bottom: $g (x) = x^{2} ⟹ g^{'} (x) = 2 x$ .

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:

$\frac{6 x ^{5}}{x ^{4}} - \frac{2 x}{x ^{4}} = 6 x - \frac{2}{x ^{3}}$

b) Without the quotient rule (by simplifying the rational expression).

(spoiler)

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

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

recognize that $\frac{π}{4}$ is just a constant coefficient. Use the constant multiple rule while applying the power rule to $x^{2}$ :

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

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Product & quotient rules

6 min read

Font

Discuss

Feedback

What you’ll learn

Product and quotient rules: Differentiate products or quotients of 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, which 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 of two functions is the product rule.

Product rule:

To differentiate a product of two functions $f (x)$ and $g (x)$ ,

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

In other 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.

Examples

Differentiate

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

using the product rule.

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

Then applying the product rule,

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

Differentiate

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

in two different ways:

a) With the product rule.

b) With the power rule (expanding into individual terms with powers first).

Solutions

a) With the product rule

(spoiler)

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:

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

b) With the power rule

(spoiler)

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

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

Now apply the power rule term-by-term:

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

Quotient rule

Use the quotient rule to differentiate a quotient of two functions $f (x)$ and $g (x)$ :

Quotient rule:

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

The steps are:

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, and the “square of what’s below” means square the bottom function.

Examples

Find $h^{'} (x)$ for

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

(spoiler)

First, identify the top and bottom functions:

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

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:

a) With the quotient rule.

b) Without the quotient rule (by simplifying the rational expression).

Solutions

a) With the quotient rule

(spoiler)

Top: $f (x) = 3 x^{4} + 1 ⟹ f^{'} (x) = 12 x^{3}$
Bottom: $g (x) = x^{2} ⟹ g^{'} (x) = 2 x$ .

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:

$\frac{6 x ^{5}}{x ^{4}} - \frac{2 x}{x ^{4}} = 6 x - \frac{2}{x ^{3}}$

b) Without the quotient rule (by simplifying the rational expression).

(spoiler)

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

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

recognize that $\frac{π}{4}$ is just a constant coefficient. Use the constant multiple rule while applying 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”
Useful when expanding is impractical

Product rule examples

Identify $f (x)$ and $g (x)$ and their derivatives before applying the rule
Can confirm results by expanding and using the power rule term-by-term

Quotient rule

Differentiates quotient: $[\frac{f ( x )}{g ( x )}]^{'} = \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{[ g ( x ) ] ^{2}}$
Steps:
- Bottom × derivative of top
- Minus top × 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

Identify numerator and denominator functions and their derivatives
Sometimes easier to simplify the expression first and use the power rule instead
- Only split fractions when algebraically valid

AP tip and shortcuts

Split fractions to simplify when possible, but only when algebraically correct
Constants in numerators/denominators can be treated as coefficients (constant multiple rule)
Do not incorrectly split denominators (e.g., $\frac{x ^{3}}{x + 1} \neq = \frac{x ^{3}}{x} + \frac{x ^{3}}{1}$ )

Next | 2.5 Special derivatives

Product & quotient rules

What you’ll learn

Product and quotient rules: Differentiate products or quotients of 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, which 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 of two functions is the product rule.

Product rule:

To differentiate a product of two functions $f (x)$ and $g (x)$ ,

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

In other 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.

Examples

Differentiate

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

using the product rule.

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

Then applying the product rule,

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

Differentiate

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

in two different ways:

a) With the product rule.

b) With the power rule (expanding into individual terms with powers first).

Solutions

a) With the product rule

(spoiler)

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:

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

b) With the power rule

(spoiler)

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

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

Now apply the power rule term-by-term:

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

Quotient rule

Use the quotient rule to differentiate a quotient of two functions $f (x)$ and $g (x)$ :

Quotient rule:

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

The steps are:

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, and the “square of what’s below” means square the bottom function.

Examples

Find $h^{'} (x)$ for

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

(spoiler)

First, identify the top and bottom functions:

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

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:

a) With the quotient rule.

b) Without the quotient rule (by simplifying the rational expression).

Solutions

a) With the quotient rule

(spoiler)

Top: $f (x) = 3 x^{4} + 1 ⟹ f^{'} (x) = 12 x^{3}$
Bottom: $g (x) = x^{2} ⟹ g^{'} (x) = 2 x$ .

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:

$\frac{6 x ^{5}}{x ^{4}} - \frac{2 x}{x ^{4}} = 6 x - \frac{2}{x ^{3}}$

b) Without the quotient rule (by simplifying the rational expression).

(spoiler)

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

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

Sidenote

Symbols as coefficients

If you see a function like

$y = \frac{π x ^{2}}{4}$

recognize that $\frac{π}{4}$ is just a constant coefficient. Use the constant multiple rule while applying the power rule to $x^{2}$ :

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