Chapter 3
Limits and the
Derivative
Section 4
The Derivative
(Part 1)
Learning Objectives for Section 3.4
The Derivative
■ Part One
■ The student will be able to:
■Calculate slope of the secant
line.
■Calculate average rate of
change.
■Calculate slope of the tangent
line.
■Calculate instantaneous rate
of change.
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Introduction
 In Calculus, we study how a change in one variable affects
another variable.
 In studying this, we will make use of the limit concepts we
learned in the previous lessons of this chapter.
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Slopes
 Slope of a secant
Barnett/Ziegler/Byleen Business Calculus 12e
 Slope of a tangent
4
Example 1
Revenue Analysis
 The graph below shows the revenue (in dollars) from the
sale of x widgets.
 When 100 widgets are sold, the revenue is $1800.
 If we increase production by an additional 300 widgets, the
revenue increases to $4800.
𝑅 𝑥 = 20𝑥 − 0.02𝑥 2
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Example 1 (continued)
 When production increases from 100 to 400 widgets the
change in revenue is:
𝑅 400 − 𝑅 100 = $4800 − $1800
= $3000
 The average change in revenue is:
𝑅 400 − 𝑅(100)
$3000
=
= $10 𝑝𝑒𝑟 𝑤𝑖𝑑𝑔𝑒𝑡
400 − 100
300 𝑤𝑖𝑑𝑔𝑒𝑡𝑠
 So the average change in revenue is $10 per widget when
production increases from 100 to 400 widgets.
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Rate of Change
 This is an example of the “rate of change” concept.
 The average rate of change is the ratio of the change in y
over the change in x.
 You know this as the “slope” between two points.
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The Rate of Change
For y = f (x), the average rate of change from x = a to
x = a + h is
f ( a  h)  f ( a )
,h  0
h
The above expression is also called a difference quotient.
It can be interpreted as the slope of a secant.
See the picture on the next slide for illustration.
Barnett/Ziegler/Byleen Business Calculus 12e
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Graphical Interpretation
Average rate of change =
slope of the secant line
𝑦2 − 𝑦1 𝑓 𝑎 + ℎ − 𝑓(𝑎)
=
𝑥2 − 𝑥1
𝑎+ℎ −𝑎
𝑓 𝑎 + ℎ − 𝑓(𝑎)
=
ℎ
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What if…
 Suppose the 2nd point (a+h, f(a+h)) gets closer and closer
to the first point (a, f(a)). What happens to the value of h?
Answer: h approaches zero
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The Instantaneous
Rate of Change
If we find the slope of the secant line as h approaches zero, that’s
the same as the limit shown below.
lim
h0
f ( a  h)  f ( a )
h
Now, instead of the average rate of change, this limit gives us the
instantaneous rate of change of f(x) at x = a.
And instead of the slope of a secant, it’s the slope of a tangent.
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Visual Interpretation
Tangent line at x=a
Slope of tangent at x = a is the
instantaneous rate of change.
lim f (a  h)  f (a )
h0
h
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Instantaneous Rate of Change
Given y = f (x), the instantaneous rate of change at x = a is
f ( a  h)  f ( a )
lim
h0
h
provided that the limit exists. It can be interpreted as the slope of
the tangent at the point (a, f (a)).
If the slope is positive, then 𝑓(𝑥) is increasing at x=a.
If the slope is negative, then 𝑓(𝑥) is decreasing at x=a.
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Example 3A
 𝑓 𝑥 = 4𝑥 − 𝑥 2
 Find the avg. rate of change if x changes from 1 to 4.
Answer: 𝑓 4 − 𝑓(1) 0 − 3
=
= −1
4−1
3
This is equal to the slope of the secant line through (1, 3) and
(4, 0).
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Example 3B
 𝑓 𝑥 = 4𝑥 − 𝑥 2
 Find the instantaneous rate of change of f(x) at x = 1
𝑓 1 + ℎ − 𝑓(1)
4 1 + ℎ − 1 + ℎ 2 − (4 − 12 )
lim
= lim
ℎ→0
ℎ→0
ℎ
ℎ
4 + 4ℎ − (1 + 2ℎ + ℎ2 ) − 3
= lim
ℎ→0
ℎ
4 + 4ℎ − 1 − 2ℎ − ℎ2 − 3
= lim
ℎ→0
ℎ
2ℎ − ℎ2
= lim
ℎ→0
ℎ
Continued on next slide…
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Example 3B - continued
 𝑓 𝑥 = 4𝑥 − 𝑥 2
 Find the instantaneous rate of change of f(x) at x = 1
2ℎ − ℎ2
= lim
ℎ→0
ℎ
ℎ(2 − ℎ)
= lim
ℎ→0
ℎ
= lim (2 − ℎ)
ℎ→0
=2
This is equal to the slope of the tangent line at x=1.
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Example 3C
 𝑓 𝑥 = 4𝑥 − 𝑥 2
 Find the equation of the tangent line at x=1.
When x=1, y=3 and slope = 2
𝑦 = 𝑚𝑥 + 𝑏
3 = 2(1) + 𝑏
𝑏=1
𝑦 = 2𝑥 + 1
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Application - Velocity
 A watermelon that is dropped from the Eiffel Tower will
fall a distance of y feet in x seconds.
𝑦 = 16𝑥 2
 Find the average velocity from 2 to 5 seconds.
• Answer:
400 − 64
𝑓 5 − 𝑓(2)
=
5−2
5−2
336 𝑓𝑡
=
= 112 𝑓𝑡/𝑠𝑒𝑐
3 𝑠𝑒𝑐
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Velocity
(continued)
𝑦 = 16𝑥 2
 Find the instantaneous velocity at x = 2 seconds.
𝑓 2 + ℎ − 𝑓(2)
𝐼𝑛𝑠𝑡𝑎𝑛𝑡. 𝑣𝑒𝑙. = lim
ℎ→0
ℎ
16(2 + ℎ)2 −16(2)2
= lim
ℎ→0
ℎ
2
16(4 + 4ℎ + ℎ ) − 16(4)
= lim
ℎ→0
ℎ
ℎ(64 + 16ℎ)
= lim
ℎ→0
ℎ
= lim (64 + 16ℎ)
ℎ→0
= 64 𝑓𝑡/𝑠𝑒𝑐
64 + 64ℎ + 16ℎ2 − 64
= lim
ℎ→0
ℎ
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Summary
 Slope of a secant
 Average rate of change
 Average velocity
 Slope of a tangent
 Instantaneous rate of
change
 Instantaneous velocity
𝑓 𝑎 + ℎ − 𝑓(𝑎)
ℎ
𝑓 𝑎 + ℎ − 𝑓(𝑎)
ℎ→0
ℎ
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Homework
#3-4A
Pg 175
(1-4, 27-30)
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Chapter 3
Limits and the
Derivative
Section 4
The Derivative
(Part 2)
Learning Objectives for Section 3.4
The Derivative
■ Part Two
■ The student will be able to:
■Calculate the derivative.
■Identify the nonexistence of a
derivative.
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Introduction
𝑓 𝑎 + ℎ − 𝑓(𝑎)
lim
ℎ→0
ℎ
 In Part 1, we learned that the limit of a difference quotient,
can be interpreted as:
• instantaneous rate of change at x=a
• slope of the tangent line at x=a
• instantaneous velocity at x=a
 In this part of the lesson, we will take a closer look at this
limit where we replace a with x.
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The Derivative
For y = f (x), we define the derivative of f at x, denoted
f (x), to be
f (x)  lim
h0
f (x  h)  f (x)
h
if the limit exists.
I refer to 𝑓′(𝑥) as a “slope machine”. It will allow me to find the
slope at any x value.
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Same Meaning as Before
If f is a function, then f  has the following interpretations:
■ For each x in the domain of f , f (x) is the slope of the line
tangent to the graph of f at the point (x, f (x)).
■ For each x in the domain of f , f (x) is the instantaneous
rate of change of y = f (x) with respect to x.
■ If f (x) is the position of a moving object at time x, then
v = f (x) is the instantaneous velocity of the object with
respect to time.
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Finding the Derivative
To find f (x), we use a four-step process:
Step 1. Find f (x + h)
Step 2. Find f (x + h) – f (x)
Step 3. Find f ( x  h)  f ( x)
h
Step 4. Find f (x) = hlim
0
f ( x  h)  f ( x )
h
*Feel free to go directly to Step 4 when you’ve got
the process down!
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Example 1
Find the derivative of f (x) = x 2 – 3x.
Step 1: Find f(x+h) = (𝑥 + ℎ)2 −3(𝑥 + ℎ)
= 𝑥 2 + 2𝑥ℎ + ℎ2 − 3𝑥 − 3ℎ
Step 2: Find f(x+h) – f(x) = 𝑥 2 + 2𝑥ℎ + ℎ2 − 3𝑥 − 3ℎ − (𝑥 2 −3𝑥)
= 𝑥 2 + 2𝑥ℎ + ℎ2 − 3𝑥 − 3ℎ − 𝑥 2 + 3𝑥
= 2𝑥ℎ + ℎ2 − 3ℎ
Step 3: Find
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
2𝑥ℎ + ℎ2 − 3ℎ
=
ℎ
= 2𝑥 + ℎ − 3
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Example 1 (continued)
𝑓 𝑥 + ℎ − 𝑓(𝑥)
Step 4: Find f  (x) = lim
ℎ→0
ℎ
𝑓(𝑥) = lim (2𝑥 + ℎ − 3)
ℎ→0
𝑓(𝑥) = 2𝑥 − 3
For x=a, where a is in the domain of f(x),
f (a) is the slope of the line tangent to f(x) at x=a.
Find the slope of the line tangent to the graph of f (x) at
x = 0, x = 2, and x = 3.
f (0) = -3
f (2) = 1
f (3) = 3
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Example 2
Find f (x) where f (x) = 2x – 3x2 using the four-step process.
Step 1: Find f(x+h) = 2 𝑥 + ℎ − 3(𝑥 + ℎ)2
= 2𝑥 + 2ℎ − 3(𝑥 2 + 2𝑥ℎ + ℎ2 )
= 2𝑥 + 2ℎ − 3𝑥 2 − 6𝑥ℎ − 3ℎ2
Step 2: Find f(x+h) – f(x)
= 2𝑥 + 2ℎ − 3𝑥 2 − 6𝑥ℎ − 3ℎ2 − (2𝑥 − 3𝑥 2 )
= 2𝑥 + 2ℎ − 3𝑥 2 − 6𝑥ℎ − 3ℎ2 − 2𝑥 + 3𝑥 2
= 2ℎ −6𝑥ℎ − 3ℎ2
2
2ℎ
−
6𝑥ℎ
−
3ℎ
𝑓 𝑥+ℎ −𝑓(𝑥)
=
Step 3: Find
ℎ
ℎ
= 2 − 6𝑥 − 3ℎ
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Example 2 (continued)
𝑓 𝑥 + ℎ − 𝑓(𝑥)
Step 4: Find f  (x) = lim
ℎ→0
ℎ
𝑓(𝑥) = lim (2 − 6𝑥 − 3ℎ)
ℎ→0
𝑓 𝑥 = 2 − 6𝑥
Find the slope of the line tangent to the graph of f (x) at
x = -2, x = 0, and x = 1.
f (-2) = 14
f (0) = 2
f (1) = -4
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Example 3
Find f (x) where 𝑓 𝑥 = 𝑥 + 2 using the four-step process.
Step 1: Find f(x+h) = 𝑥 + ℎ + 2
Step 2: Find f(x+h) – f(x) = 𝑥 + ℎ + 2 − ( 𝑥 + 2)
= 𝑥+ℎ− 𝑥
Step 3: Find
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
𝑥+ℎ− 𝑥
=
ℎ
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 3 (continued)
𝑓 𝑥 + ℎ − 𝑓(𝑥)
ℎ→0
ℎ
𝑥+ℎ− 𝑥 𝑥+ℎ+ 𝑥
f  (x) = lim
∙
ℎ→0
ℎ
𝑥+ℎ+ 𝑥
Step 4: Find f  (x) = lim
f  (x) = lim
ℎ→0
f  (x) = lim
ℎ→0
𝑥+ℎ−𝑥
ℎ
𝑥+ℎ+ 𝑥
1
𝑥+ℎ+ 𝑥
= lim
ℎ
ℎ→0
=
ℎ 𝑥+ℎ+ 𝑥
1
1
=
𝑥+ 𝑥
2 𝑥
1
𝑥
𝑥
𝑓 𝑥 =
∙
=
2 𝑥 𝑥 2𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
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Nonexistence of the Derivative
The existence of a derivative at x = a depends on the
existence of the limit
f (a)  lim
h0
f (a  h)  f (a)
h
If the limit does not exist, we say that the function is
nondifferentiable at x = a, or f (a) does not exist.
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Nonexistence of the Derivative
(continued)
Some of the reasons why the derivative of a function may
not exist at x = a are
■ The graph of f has a hole or break at x = a, or
■ The graph of f has a sharp corner at x = a, or
■ The graph of f has a vertical tangent at x = a.
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Examples of Nonexistent
Derivatives
In each graph, f is nondifferentiable at x=a.
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Application – Profit
 The profit (in dollars) from the sale of x infant car seats is
given by: 𝑃 𝑥 = 45𝑥 − 0.025𝑥 2 − 5000 0  x  2400
 A) Find the average change in profit if production
increases from 800 to 850 car seats.
 B) Use the 4-step process to find P(x)
 C) Find P(800) and P(800) and explain their meaning.
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Application – Profit
(continued)
 The profit (in dollars) from the sale of x infant car seats is
given by: 𝑃 𝑥 = 45𝑥 − 0.025𝑥 2 − 5000 0  x  2400
 A) Find the average change in profit if production
increases from 800 to 850 car seats.
𝑃 850 − 𝑃(800)
𝐴𝑣𝑔 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑜𝑓𝑖𝑡 =
850 − 800
15187.5 − 15000
=
50
= 3.75
The avg change in profit when production increases from
800 to 850 car seats is $3.75 per seat.
Barnett/Ziegler/Byleen Business Calculus 12e
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Application – Profit
(continued)
 The profit (in dollars) from the sale of x infant car seats is
given by: 𝑃 𝑥 = 45𝑥 − 0.025𝑥 2 − 5000 0  x  2400
 B) Use the 4-step process to find P(x)
𝑺𝒕𝒆𝒑 𝟏: 𝑃 𝑥 + ℎ = 45 𝑥 + ℎ − 0.025 𝑥 + ℎ
2
− 5000
= 45𝑥 + 45ℎ − 0.025 𝑥 2 + 2𝑥ℎ + ℎ2 − 5000
= 45𝑥 + 45ℎ − 0.025𝑥 2 − 0.05𝑥ℎ − 0.025ℎ2 − 5000
𝑺𝒕𝒆𝒑 𝟐: 𝑃 𝑥 + ℎ − 𝑃(𝑥)
= 45𝑥 + 45ℎ − 0.025𝑥 2 − 0.05𝑥ℎ − 0.025ℎ2 − 5000
−(45x − 0.025𝑥 2 − 5000)
= 45ℎ − 0.05𝑥ℎ − 0.025ℎ2
Barnett/Ziegler/Byleen Business Calculus 12e
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Application – Profit
(continued)
45ℎ − 0.05𝑥ℎ − 0.025ℎ2
𝑃 𝑥 + ℎ − 𝑃(𝑥)
=
𝑺𝒕𝒆𝒑 𝟑:
ℎ
ℎ
= 45 − 0.05𝑥 − 0.025ℎ
𝑃 𝑥 + ℎ − 𝑃(𝑥)
𝑺𝒕𝒆𝒑 𝟒: lim
ℎ→0
ℎ
= lim (45 − 0.05𝑥 − 0.025ℎ)
ℎ→0
= 45 − .05𝑥
Barnett/Ziegler/Byleen Business Calculus 12e
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Application – Profit
(continued)
 The profit (in dollars) from the sale of x infant car seats is
given by: 𝑃 𝑥 = 45𝑥 − 0.025𝑥 2 − 5000 0  x  2400
 C) Find P(800) and P(800) and explain their meaning.
𝑃 800 = 45 800 − 0.025 800
2
− 5000
𝑃 800 = $15,000
𝑃 800 = 45 − 0.05(800)
𝑃 800 = $5
At a production level of 800 car seats, the profit is $15,000
and it is increasing at a rate of $5 per car seat.
Barnett/Ziegler/Byleen Business Calculus 12e
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Homework
#3-4B Pg 176
(7, 21, 25,
31-41, 61, 63)
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