Achievable AP Calculus AB

2. Derivative basics

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

Definition of the derivative

12 min read

Font

Discuss

Feedback

What you’ll learn:

Average vs. instantaneous rate of change
Limit definition of derivative (2 forms)
Recognizing limit expressions as derivative definitions

Average rate of change

To describe how fast something is changing, we use a rate of change. For example, think about your phone’s battery percentage throughout the day. If you check it at two different times, time $a$ and time $b$ , then the average rate of change is the change in battery percentage divided by the time that passed:

$Average rate of change:$

$= \frac{f ( b ) - f ( a )}{b - a}$

This tells you how quickly the battery drained on average between time $a$ and time $b$ . With that information, you can estimate how long the battery might last before needing a recharge (a simplified version of what your phone does).

From algebra, this is the slope: change in $y$ divided by change in $x$ for the line connecting two points on $f (x)$ . You may have seen slope written as

$m = \frac{Δ y}{Δ x}$

where the delta symbol $Δ$ means “change.”

Instantaneous rate of change

In real life, the battery doesn’t deplete at a steady rate. Watching videos uses more power, while leaving the phone idle slows the drain. So how do we describe how quickly the battery is draining at one specific moment?

The average rate of change gives a general picture, but the instantaneous rate of change focuses on what’s happening right at a particular time. The idea is to “zoom in” on a very small time interval.

This is where limits come in. Instead of using two fixed points $a$ and $b$ , we:

Fix the point $a$ .
Let another point $x$ move closer and closer to $a$ .
Track what happens to the average rate of change as $x \to a$ .

As $x$ approaches $a$ , the time interval shrinks toward zero, and the limiting value gives the instantaneous rate of change at $a$ . This is also called the derivative of $f (x)$ at $x = a$ :

$Derivative: 1^{st} definition$

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

The notation $f^{'} (a)$ is read as “ $f$ prime of $a$ .” The prime symbol ( $^{'}$ ) is commonly used to indicate a derivative. Later, you’ll also see second derivatives, written as $f^{''} (a)$ (“ $f$ double prime of $a$ ”).

AP tip:

When asked for the average rate of change, use the basic slope formula with two points - no limits needed.

If it’s the instantaneous rate of change, you must use the derivative or the limit definition. Watch out for multiple-choice traps that mix these up!

Other limit definition of the derivative

The first definition gives the derivative at a specific point $a$ . We can also define a derivative function $f^{'} (x)$ , which gives the instantaneous rate of change at any input value $x$ .

In this form, we:

Fix a point $x$ .
Move a small amount $h$ away from it.
Look at the average rate of change over that small interval.

That average rate of change is called the difference quotient:

$Difference quotient:$

$\frac{f ( x + h ) - f ( x )}{h}$

In some problems, $h$ is replaced with $Δ x$ , but the meaning is the same: it’s the slope formula applied over a small interval of width $h$ .

To get the instantaneous rate of change at $x$ , we make the interval as small as possible by letting $h \to 0$ . That gives the second limit definition of the derivative:

$Derivative: 2^{nd} definition$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

This limit process creates a new function $f^{'} (x)$ . Plugging in a value of $x$ gives the instantaneous rate of change at that point.

AP tip:

If a problem asks for the instantaneous rate of change or the derivative at a specific point, the 1st form (with $x \to a$ ) is often easier to work with, particularly if $f$ is a polynomial. Since $a$ is the fixed point of interest, the result will be a single number that represents the instantaneous rate of change at that point.

On the other hand, if a question asks you to “find the derivative using the limit definition of the derivative,” it’s sometimes better to work with the 2nd form to find an expression that represents the derivative in terms of $x$ .

Either one can be used, but depending on the function, one may be easier to work with than the other.

Examples

Find the derivative of $f (x) = x^{2}$ using the limit definition.

Let’s use both forms to show the difference:

2nd form:

$f^{'} (x) f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{( x + h ) ^{2} - x ^{2}}{h} = h \to 0 lim \frac{x ^{2} + 2 x h + h ^{2} - x ^{2}}{h} = h \to 0 lim \frac{2 x h + h ^{2}}{h} = h \to 0 lim \frac{h ( 2 x + h )}{h} = h \to 0 lim (2 x + h) = 2 x$

1st form:

$f^{'} (a) f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} = x \to a lim \frac{x ^{2} - a ^{2}}{x - a} = x \to a lim \frac{( x + a ) ( x - a )}{( x - a )} = x \to a lim (x + a) = a + a = 2 a$

The results match; the only difference is the variable name. The 2nd form produces a general derivative function in terms of $x$ . The 1st form focuses on a specific input $a$ . If $x = a$ , then $2 x$ and $2 a$ represent the same instantaneous rate of change.

If $f (x) = x$ , find $f^{'} (9)$ .

Solution

(spoiler)

Answer: $f^{'} (9) = \frac{1}{6}$

Let’s use the 1st form, since the point of interest is $a = 9$ .

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

Where $a = 9$ .

$f^{'} (9) = x \to 9 lim \frac{x - 9}{x - 9}$

$= x \to 9 lim \frac{x - 3}{x - 9}$

Direct substitution gives the indeterminate form $\frac{0}{0}$ , so we need algebraic manipulation. Rationalize the numerator:

$f^{'} (9) = x \to 9 lim \frac{x - 3}{x - 9} \cdot \frac{x + 3}{x + 3}$

$= x \to 9 lim \frac{x - 9}{( x - 9 ) ( x + 3 )}$

$= x \to 9 lim \frac{1}{x + 3}$

$= \frac{1}{6}$

As a bonus exercise, try finding the derivative of $f (x) = x$ using the second form of the limit definition. The answer should be $f^{'} (x) = \frac{1}{2 x}$ , and plugging in $x = 9$ gives the same instantaneous rate of change as the 1st limit definition.

Working backwards

Most derivative problems ask you to find $f^{'} (x)$ given a function $f (x)$ . However, some AP problems give a limit expression for a derivative and ask you to identify the original function $f (x)$ and the point $a$ where it’s evaluated.

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

$or$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the limit represents $f^{'} (a)$ :

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Answer: $f (x) = (3 + x)^{2}, a = 3$

First determine whether the limit notation specifies $x \to a$ or $h \to 0$ . Since it’s the latter, the 2nd definition is being used.

Next, $(3 + h)^{2}$ matches the role of $f (x + h)$ , with $3$ playing the role of $x$ and the output being squared. That suggests $f (x + h) = (x + h)^{2}$ , so $f (x) = x^{2}$ .

Testing $x = 3$ gives $f (3) = (3)^{2} = 9$ , and $9$ is in the position where $f (x)$ appears in the limit definition.

Find $f (x)$ and a number $a$ such that the limit represents $f^{'} (a)$ :

$x \to 5 lim \frac{- 1 + e ^{(3 x - 15)}}{x - 5}$

Solution

(spoiler)

Answer: $f (x) = e^{(3 x - 15)}, a = 5$

Since the limit notation specifies $x \to a$ , the 1st definition is being used, and $a = 5$ .

Next, even though the terms are out of the usual order, $e^{(3 x - 15)}$ is the part containing the variable, so it corresponds to $f (x)$ .

Rewriting the numerator as $e^{(3 x - 15)} - 1$ makes the match to $f (x) - f (a)$ clearer.

Lastly, check that $f (a) = 1$ when $a = 5$ :

$f (5) = e^{(3 (5) - 15)} = e^{0} = 1$

Using tables

Some questions present a table and ask you to find the average rate of change over an interval.

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Estimate the average rate of change over the time interval $4 \leq t \leq 7$ based on the table.

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

In AP Calculus, time is often the independent variable, written as $t$ . Here, water depth is a function of time, so we can write it as $D (t)$ .

Because the water level is increasing over time, the rate of change should be positive.

The average rate of change is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

Depth is measured in inches (numerator) and time is measured in minutes (denominator), so the units are inches per minute. This means that, on average, the water depth increased by $\frac{8}{3}$ inches per minute between minute $4$ and $7$ .

You may also be asked to “estimate the instantaneous rate of change” from a table. With only a few data points, the usual approach is to use an average rate of change over a small interval near the time of interest.

Using the same table, estimate the instantaneous rate of change of the water depth at $3$ minutes and $9$ minutes.

Solution

(spoiler)

Answers:

$2.5$ in/min at $3$ minutes
$1.5$ in/min at $9$ minutes

For $t = 3$ , use the closest values around 3 minutes: $t = 2$ and $t = 4$ .

Rate of change

$\frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2}$

$= \frac{5}{2} in/min$

So the best estimate for how quickly the water level was changing at exactly 3 minutes is 2.5 inches per minute. The estimate is positive, which matches the fact that the water level is increasing.

For $t = 9$ , the closest earlier time in the table is $t = 7$ .

Rate of change

$\frac{D ( 9 ) - D ( 7 )}{9 - 7} = \frac{19 - 16}{2}$

$= \frac{3}{2} in/min$

Sidenote

A note on notation

In addition to $f^{'} (x)$ , you may also see the derivative written as $\frac{d y}{d x}$ , or you may see the operator $\frac{d}{d x}$ .

$f^{'} (x)$

Lagrange’s notation, typically used for derivatives of functions.

$\frac{d y}{d x}$

Leibniz’s notation - represents the rate of change of $y$ with respect to $x$ , when $y$ is a function of $x$ . It shows how $y$ changes as $x$ changes. Later on you’ll see letters other than $x$ and $y$ used.
E.g. $d g / d t$ represents the rate of change of $g$ with respect to $t$ .

$\frac{d}{d x}$

Indicates the operation of taking a derivative. It works like a verb: we apply the derivative operator to differentiate an expression. For example, if asked to differentiate $f (x) = 2 x$ , you could write $\frac{d}{d x} [2 x]$ .

Average rate of change

Formula: $\frac{f ( b ) - f ( a )}{b - a}$
Represents slope between two points; “change in $y$ over change in $x$ ”
Used for estimating change over an interval (no limits needed)

Instantaneous rate of change (Derivative)

Describes change at a single point; requires limits
1st limit definition: $f^{'} (a) = lim_{x \to a} \frac{f ( x ) - f ( a )}{x - a}$
$f^{'} (a)$ (“ $f$ prime of $a$ ”) gives instantaneous rate at $a$

Other limit definition of the derivative

2nd limit definition: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$
Difference quotient: $\frac{f ( x + h ) - f ( x )}{h}$
Produces a general derivative function $f^{'} (x)$

Recognizing limit expressions as derivative definitions

Identify which form is used: $x \to a$ (1st), $h \to 0$ (2nd)
Match limit expression structure to $f (x)$ and $a$
Rearranging numerator may clarify correspondence

Examples

Derivative of $f (x) = x^{2}$ :
- 2nd form yields $f^{'} (x) = 2 x$
- 1st form yields $f^{'} (a) = 2 a$
Derivative of $f (x) = x$ at $x = 9$ :
- $f^{'} (9) = \frac{1}{6}$

Using tables

Average rate of change: use $\frac{f ( b ) - f ( a )}{b - a}$ with table values
Instantaneous rate of change: estimate using average rate over smallest interval near point
- E.g., use closest surrounding values for estimation

Notation for derivatives

$f^{'} (x)$ : Lagrange notation (function prime)
$\frac{d y}{d x}$ : Leibniz notation (rate of change of $y$ with respect to $x$ )
$\frac{d}{d x}$ : Derivative operator (indicates differentiation)

Definition of the derivative

What you’ll learn:

Average vs. instantaneous rate of change
Limit definition of derivative (2 forms)
Recognizing limit expressions as derivative definitions

Average rate of change

$Average rate of change:$

$= \frac{f ( b ) - f ( a )}{b - a}$

From algebra, this is the slope: change in $y$ divided by change in $x$ for the line connecting two points on $f (x)$ . You may have seen slope written as

$m = \frac{Δ y}{Δ x}$

where the delta symbol $Δ$ means “change.”

Instantaneous rate of change

This is where limits come in. Instead of using two fixed points $a$ and $b$ , we:

Fix the point $a$ .
Let another point $x$ move closer and closer to $a$ .
Track what happens to the average rate of change as $x \to a$ .

As $x$ approaches $a$ , the time interval shrinks toward zero, and the limiting value gives the instantaneous rate of change at $a$ . This is also called the derivative of $f (x)$ at $x = a$ :

$Derivative: 1^{st} definition$

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

AP tip:

When asked for the average rate of change, use the basic slope formula with two points - no limits needed.

If it’s the instantaneous rate of change, you must use the derivative or the limit definition. Watch out for multiple-choice traps that mix these up!

Other limit definition of the derivative

The first definition gives the derivative at a specific point $a$ . We can also define a derivative function $f^{'} (x)$ , which gives the instantaneous rate of change at any input value $x$ .

In this form, we:

Fix a point $x$ .
Move a small amount $h$ away from it.
Look at the average rate of change over that small interval.

That average rate of change is called the difference quotient:

$Difference quotient:$

$\frac{f ( x + h ) - f ( x )}{h}$

In some problems, $h$ is replaced with $Δ x$ , but the meaning is the same: it’s the slope formula applied over a small interval of width $h$ .

To get the instantaneous rate of change at $x$ , we make the interval as small as possible by letting $h \to 0$ . That gives the second limit definition of the derivative:

$Derivative: 2^{nd} definition$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

This limit process creates a new function $f^{'} (x)$ . Plugging in a value of $x$ gives the instantaneous rate of change at that point.

AP tip:

Either one can be used, but depending on the function, one may be easier to work with than the other.

Examples

Find the derivative of $f (x) = x^{2}$ using the limit definition.

Let’s use both forms to show the difference:

2nd form:

1st form:

$f^{'} (a) f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} = x \to a lim \frac{x ^{2} - a ^{2}}{x - a} = x \to a lim \frac{( x + a ) ( x - a )}{( x - a )} = x \to a lim (x + a) = a + a = 2 a$

If $f (x) = x$ , find $f^{'} (9)$ .

Solution

(spoiler)

Answer: $f^{'} (9) = \frac{1}{6}$

Let’s use the 1st form, since the point of interest is $a = 9$ .

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

Where $a = 9$ .

$f^{'} (9) = x \to 9 lim \frac{x - 9}{x - 9}$

$= x \to 9 lim \frac{x - 3}{x - 9}$

Direct substitution gives the indeterminate form $\frac{0}{0}$ , so we need algebraic manipulation. Rationalize the numerator:

$f^{'} (9) = x \to 9 lim \frac{x - 3}{x - 9} \cdot \frac{x + 3}{x + 3}$

$= x \to 9 lim \frac{x - 9}{( x - 9 ) ( x + 3 )}$

$= x \to 9 lim \frac{1}{x + 3}$

$= \frac{1}{6}$

Working backwards

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

$or$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the limit represents $f^{'} (a)$ :

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Answer: $f (x) = (3 + x)^{2}, a = 3$

First determine whether the limit notation specifies $x \to a$ or $h \to 0$ . Since it’s the latter, the 2nd definition is being used.

Next, $(3 + h)^{2}$ matches the role of $f (x + h)$ , with $3$ playing the role of $x$ and the output being squared. That suggests $f (x + h) = (x + h)^{2}$ , so $f (x) = x^{2}$ .

Testing $x = 3$ gives $f (3) = (3)^{2} = 9$ , and $9$ is in the position where $f (x)$ appears in the limit definition.

Find $f (x)$ and a number $a$ such that the limit represents $f^{'} (a)$ :

$x \to 5 lim \frac{- 1 + e ^{(3 x - 15)}}{x - 5}$

Solution

(spoiler)

Answer: $f (x) = e^{(3 x - 15)}, a = 5$

Since the limit notation specifies $x \to a$ , the 1st definition is being used, and $a = 5$ .

Next, even though the terms are out of the usual order, $e^{(3 x - 15)}$ is the part containing the variable, so it corresponds to $f (x)$ .

Rewriting the numerator as $e^{(3 x - 15)} - 1$ makes the match to $f (x) - f (a)$ clearer.

Lastly, check that $f (a) = 1$ when $a = 5$ :

$f (5) = e^{(3 (5) - 15)} = e^{0} = 1$

Using tables

Some questions present a table and ask you to find the average rate of change over an interval.

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Estimate the average rate of change over the time interval $4 \leq t \leq 7$ based on the table.

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

In AP Calculus, time is often the independent variable, written as $t$ . Here, water depth is a function of time, so we can write it as $D (t)$ .

Because the water level is increasing over time, the rate of change should be positive.

The average rate of change is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

Using the same table, estimate the instantaneous rate of change of the water depth at $3$ minutes and $9$ minutes.

Solution

(spoiler)

Answers:

$2.5$ in/min at $3$ minutes
$1.5$ in/min at $9$ minutes

For $t = 3$ , use the closest values around 3 minutes: $t = 2$ and $t = 4$ .

Rate of change

$\frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2}$

$= \frac{5}{2} in/min$

So the best estimate for how quickly the water level was changing at exactly 3 minutes is 2.5 inches per minute. The estimate is positive, which matches the fact that the water level is increasing.

For $t = 9$ , the closest earlier time in the table is $t = 7$ .

Rate of change

$\frac{D ( 9 ) - D ( 7 )}{9 - 7} = \frac{19 - 16}{2}$

$= \frac{3}{2} in/min$

Sidenote

A note on notation

In addition to $f^{'} (x)$ , you may also see the derivative written as $\frac{d y}{d x}$ , or you may see the operator $\frac{d}{d x}$ .

$f^{'} (x)$

Lagrange’s notation, typically used for derivatives of functions.

$\frac{d y}{d x}$

$\frac{d}{d x}$

Key points

Average rate of change

Formula: $\frac{f ( b ) - f ( a )}{b - a}$
Represents slope between two points; “change in $y$ over change in $x$ ”
Used for estimating change over an interval (no limits needed)

Instantaneous rate of change (Derivative)

Describes change at a single point; requires limits
1st limit definition: $f^{'} (a) = lim_{x \to a} \frac{f ( x ) - f ( a )}{x - a}$
$f^{'} (a)$ (“ $f$ prime of $a$ ”) gives instantaneous rate at $a$

Other limit definition of the derivative

2nd limit definition: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$
Difference quotient: $\frac{f ( x + h ) - f ( x )}{h}$
Produces a general derivative function $f^{'} (x)$

Recognizing limit expressions as derivative definitions

Identify which form is used: $x \to a$ (1st), $h \to 0$ (2nd)
Match limit expression structure to $f (x)$ and $a$
Rearranging numerator may clarify correspondence

Examples

Derivative of $f (x) = x^{2}$ :
- 2nd form yields $f^{'} (x) = 2 x$
- 1st form yields $f^{'} (a) = 2 a$
Derivative of $f (x) = x$ at $x = 9$ :
- $f^{'} (9) = \frac{1}{6}$

Using tables

Average rate of change: use $\frac{f ( b ) - f ( a )}{b - a}$ with table values
Instantaneous rate of change: estimate using average rate over smallest interval near point
- E.g., use closest surrounding values for estimation

Notation for derivatives

$f^{'} (x)$ : Lagrange notation (function prime)
$\frac{d y}{d x}$ : Leibniz notation (rate of change of $y$ with respect to $x$ )
$\frac{d}{d x}$ : Derivative operator (indicates differentiation)

What you’ll learn:

Average vs. instantaneous rate of change
Limit definition of derivative (2 forms)
Recognizing limit expressions as derivative definitions

Average rate of change

$Average rate of change:$

$= \frac{f ( b ) - f ( a )}{b - a}$

From algebra, this is the slope: change in $y$ divided by change in $x$ for the line connecting two points on $f (x)$ . You may have seen slope written as

$m = \frac{Δ y}{Δ x}$

where the delta symbol $Δ$ means “change.”

Instantaneous rate of change

This is where limits come in. Instead of using two fixed points $a$ and $b$ , we:

Fix the point $a$ .
Let another point $x$ move closer and closer to $a$ .
Track what happens to the average rate of change as $x \to a$ .

As $x$ approaches $a$ , the time interval shrinks toward zero, and the limiting value gives the instantaneous rate of change at $a$ . This is also called the derivative of $f (x)$ at $x = a$ :

$Derivative: 1^{st} definition$

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

AP tip:

When asked for the average rate of change, use the basic slope formula with two points - no limits needed.

If it’s the instantaneous rate of change, you must use the derivative or the limit definition. Watch out for multiple-choice traps that mix these up!

Other limit definition of the derivative

The first definition gives the derivative at a specific point $a$ . We can also define a derivative function $f^{'} (x)$ , which gives the instantaneous rate of change at any input value $x$ .

In this form, we:

Fix a point $x$ .
Move a small amount $h$ away from it.
Look at the average rate of change over that small interval.

That average rate of change is called the difference quotient:

$Difference quotient:$

$\frac{f ( x + h ) - f ( x )}{h}$

In some problems, $h$ is replaced with $Δ x$ , but the meaning is the same: it’s the slope formula applied over a small interval of width $h$ .

To get the instantaneous rate of change at $x$ , we make the interval as small as possible by letting $h \to 0$ . That gives the second limit definition of the derivative:

$Derivative: 2^{nd} definition$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

This limit process creates a new function $f^{'} (x)$ . Plugging in a value of $x$ gives the instantaneous rate of change at that point.

AP tip:

Either one can be used, but depending on the function, one may be easier to work with than the other.

Examples

Find the derivative of $f (x) = x^{2}$ using the limit definition.

Let’s use both forms to show the difference:

2nd form:

1st form:

$f^{'} (a) f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} = x \to a lim \frac{x ^{2} - a ^{2}}{x - a} = x \to a lim \frac{( x + a ) ( x - a )}{( x - a )} = x \to a lim (x + a) = a + a = 2 a$

If $f (x) = x$ , find $f^{'} (9)$ .

Solution

(spoiler)

Answer: $f^{'} (9) = \frac{1}{6}$

Let’s use the 1st form, since the point of interest is $a = 9$ .

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

Where $a = 9$ .

$f^{'} (9) = x \to 9 lim \frac{x - 9}{x - 9}$

$= x \to 9 lim \frac{x - 3}{x - 9}$

Direct substitution gives the indeterminate form $\frac{0}{0}$ , so we need algebraic manipulation. Rationalize the numerator:

$f^{'} (9) = x \to 9 lim \frac{x - 3}{x - 9} \cdot \frac{x + 3}{x + 3}$

$= x \to 9 lim \frac{x - 9}{( x - 9 ) ( x + 3 )}$

$= x \to 9 lim \frac{1}{x + 3}$

$= \frac{1}{6}$

Working backwards

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

$or$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the limit represents $f^{'} (a)$ :

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Answer: $f (x) = (3 + x)^{2}, a = 3$

First determine whether the limit notation specifies $x \to a$ or $h \to 0$ . Since it’s the latter, the 2nd definition is being used.

Next, $(3 + h)^{2}$ matches the role of $f (x + h)$ , with $3$ playing the role of $x$ and the output being squared. That suggests $f (x + h) = (x + h)^{2}$ , so $f (x) = x^{2}$ .

Testing $x = 3$ gives $f (3) = (3)^{2} = 9$ , and $9$ is in the position where $f (x)$ appears in the limit definition.

Find $f (x)$ and a number $a$ such that the limit represents $f^{'} (a)$ :

$x \to 5 lim \frac{- 1 + e ^{(3 x - 15)}}{x - 5}$

Solution

(spoiler)

Answer: $f (x) = e^{(3 x - 15)}, a = 5$

Since the limit notation specifies $x \to a$ , the 1st definition is being used, and $a = 5$ .

Next, even though the terms are out of the usual order, $e^{(3 x - 15)}$ is the part containing the variable, so it corresponds to $f (x)$ .

Rewriting the numerator as $e^{(3 x - 15)} - 1$ makes the match to $f (x) - f (a)$ clearer.

Lastly, check that $f (a) = 1$ when $a = 5$ :

$f (5) = e^{(3 (5) - 15)} = e^{0} = 1$

Using tables

Some questions present a table and ask you to find the average rate of change over an interval.

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Estimate the average rate of change over the time interval $4 \leq t \leq 7$ based on the table.

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

In AP Calculus, time is often the independent variable, written as $t$ . Here, water depth is a function of time, so we can write it as $D (t)$ .

Because the water level is increasing over time, the rate of change should be positive.

The average rate of change is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

Using the same table, estimate the instantaneous rate of change of the water depth at $3$ minutes and $9$ minutes.

Solution

(spoiler)

Answers:

$2.5$ in/min at $3$ minutes
$1.5$ in/min at $9$ minutes

For $t = 3$ , use the closest values around 3 minutes: $t = 2$ and $t = 4$ .

Rate of change

$\frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2}$

$= \frac{5}{2} in/min$

So the best estimate for how quickly the water level was changing at exactly 3 minutes is 2.5 inches per minute. The estimate is positive, which matches the fact that the water level is increasing.

For $t = 9$ , the closest earlier time in the table is $t = 7$ .

Rate of change

$\frac{D ( 9 ) - D ( 7 )}{9 - 7} = \frac{19 - 16}{2}$

$= \frac{3}{2} in/min$

Sidenote

A note on notation

In addition to $f^{'} (x)$ , you may also see the derivative written as $\frac{d y}{d x}$ , or you may see the operator $\frac{d}{d x}$ .

$f^{'} (x)$

Lagrange’s notation, typically used for derivatives of functions.

$\frac{d y}{d x}$

$\frac{d}{d x}$