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Chapter 3
Limits and the Derivative
Section 4
The Derivative
(Part 1)
2Barnett/Ziegler/Byleen Business Calculus 12e
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.
3
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.
Barnett/Ziegler/Byleen Business Calculus 12e
4
Slopes
Slope of a secant Slope of a tangent
Barnett/Ziegler/Byleen Business Calculus 12e
5
Example 1Revenue 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.
Barnett/Ziegler/Byleen Business Calculus 12e
𝑅 (𝑥 )=20 𝑥− 0.02𝑥2
6
Example 1 (continued)
When production increases from 100 to 400 widgets the change in revenue is:
= $3000 The average change in revenue is:
So the average change in revenue is $10 per widget when production increases from 100 to 400 widgets.
Barnett/Ziegler/Byleen Business Calculus 12e
𝑅 (400 )−𝑅(100)400 −100
=$ 3000
300𝑤𝑖𝑑𝑔𝑒𝑡𝑠=$ 10𝑝𝑒𝑟𝑤𝑖𝑑𝑔𝑒𝑡
7
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.
Barnett/Ziegler/Byleen Business Calculus 12e
8Barnett/Ziegler/Byleen Business Calculus 12e
The Rate of Change
For y = f (x), the average rate of change from x = a to x = a + h is
0,)()(
hh
afhaf
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.
9Barnett/Ziegler/Byleen Business Calculus 12e
Graphical Interpretation
Average rate of change = slope of the secant line
𝑦2 −𝑦 1
𝑥2 −𝑥1
=𝑓 (𝑎+h ) − 𝑓 (𝑎)
(𝑎+h ) −𝑎
¿𝑓 (𝑎+h ) − 𝑓 (𝑎)
h
10
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?
Barnett/Ziegler/Byleen Business Calculus 12e
Answer: h approaches zero
11Barnett/Ziegler/Byleen Business Calculus 12e
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.
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.
h
afhafh
)()(lim
0
12Barnett/Ziegler/Byleen Business Calculus 12e
Visual Interpretation
h
afhaf
h
)()(
0
lim
Slope of tangent at x = a is theinstantaneous rate of change.
Tangent line at x=a
13Barnett/Ziegler/Byleen Business Calculus 12e
Given y = f (x), the instantaneous rate of change at x = a is
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 is increasing at x=a.
If the slope is negative, then is decreasing at x=a.
h
afhafh
)()(lim
0
Instantaneous Rate of Change
14
Example 3A
Find the avg. rate of change if x changes from 1 to 4.
Answer:
This is equal to the slope of the secant line through (1, 3) and (4, 0).
Barnett/Ziegler/Byleen Business Calculus 12e
𝑓 ( 4 )− 𝑓 (1)4 −1
=0 − 3
3=− 1
15
Example 3B
Find the instantaneous rate of change of f(x) at x = 1
Barnett/Ziegler/Byleen Business Calculus 12e
limh →0
𝑓 (1+h )− 𝑓 (1)h ¿
limh → 0
[4 (1+h ) − (1+h )2 ] −(4 − 12)
h
¿limh → 0
[4+4h−(1+2h+h2)]− 3
h
¿limh → 0
[4+4 h −1 −2 h− h2 ]− 3
h
¿limh → 0
2 h− h2
h
Continued on next slide…
16
Example 3B - continued
Find the instantaneous rate of change of f(x) at x = 1
This is equal to the slope of the tangent line at x=1.
Barnett/Ziegler/Byleen Business Calculus 12e
¿ limh→0
(2− h)
¿2
¿limh → 0
2 h− h2
h
¿limh → 0
h(2− h)
h
17
Example 3C
Find the equation of the tangent line at x=1.
When x=1, y=3 and slope = 2
Barnett/Ziegler/Byleen Business Calculus 12e
𝑦=𝑚𝑥+𝑏3=2(1)+𝑏𝑏=1𝑦=2 𝑥+1
18
Application - Velocity
A watermelon that is dropped from the Eiffel Tower will fall a distance of y feet in x seconds.
Find the average velocity from 2 to 5 seconds.• Answer:
Barnett/Ziegler/Byleen Business Calculus 12e
𝑦=16 𝑥2
400 − 645 − 2
𝑓 (5 )− 𝑓 (2)5 −2
=¿
¿336 𝑓𝑡3𝑠𝑒𝑐
=112 𝑓𝑡 /𝑠𝑒𝑐
19
Velocity(continued)
Find the instantaneous velocity at x = 2 seconds.
Barnett/Ziegler/Byleen Business Calculus 12e
𝐼𝑛𝑠𝑡𝑎𝑛𝑡 .𝑣𝑒𝑙 .=limh →0
𝑓 (2+h ) − 𝑓 (2)h
¿ limh→0
(64+16 h)
¿64 𝑓𝑡 /𝑠𝑒𝑐
¿limh → 0
16(2+h)2− 16(2)2
h
¿limh → 0
16(4+4 h+h2)−16 (4)
h
¿limh → 0
64+64 h+16 h2 −64
h
¿limh → 0
h(64+16 h)
h
𝑦=16 𝑥2
20
Summary
Slope of a secant Average rate of change Average velocity
Slope of a tangent Instantaneous rate of
change Instantaneous velocity
Barnett/Ziegler/Byleen Business Calculus 12e
𝑓 (𝑎+h )− 𝑓 (𝑎)h
limh → 0
𝑓 (𝑎+h )− 𝑓 (𝑎)
h
21
Homework
#3-4A
Pg 175
(1-4, 27-30)
Barnett/Ziegler/Byleen Business Calculus 12e
Chapter 3
Limits and the Derivative
Section 4
The Derivative
(Part 2)
23Barnett/Ziegler/Byleen Business Calculus 12e
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.
24
Introduction
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.
Barnett/Ziegler/Byleen Business Calculus 12e
limh →0
𝑓 (𝑎+h )− 𝑓 (𝑎)h
25Barnett/Ziegler/Byleen Business Calculus 12e
The Derivative
For y = f (x), we define the derivative of f at x, denoted f (x), to be
if the limit exists.
I refer to as a “slope machine”. It will allow me to find the slope at any x value.
f (x) lim
h 0
f (x h) f (x)
h
26Barnett/Ziegler/Byleen Business Calculus 12e
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.
27Barnett/Ziegler/Byleen Business Calculus 12e
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
Step 4. Find f (x) =
h
xfhxf )()(
h
xfhxfh
)()(lim
0
*Feel free to go directly to Step 4 when you’ve got the process down!
28Barnett/Ziegler/Byleen Business Calculus 12e
Find the derivative of f (x) = x 2 – 3x.
Step 1: Find f(x+h)
Step 2: Find f(x+h) – f(x)
Step 3: Find
Example 1
¿ (𝑥+h)2−3 (𝑥+h)¿ 𝑥2+2 h𝑥 +h2 −3 𝑥−3 h
¿ 𝑥2+2 h𝑥 +h2 −3 𝑥−3 h−(𝑥¿¿2 −3 𝑥)¿¿ 𝑥2+2 h𝑥 +h2 −3 𝑥−3 h− 𝑥2+3 𝑥¿2 h𝑥 +h2− 3 h
¿ 2 h𝑥 +h2 −3 h h
¿2 𝑥+h−3
29
Example 1 (continued)
Step 4: Find f (x) =
Barnett/Ziegler/Byleen Business Calculus 12e
limh →0
𝑓 (𝑥+h ) − 𝑓 (𝑥)h
𝑓 (𝑥)=limh → 0
(2 𝑥+h−¿3)¿
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
30Barnett/Ziegler/Byleen Business Calculus 12e
Find f (x) where f (x) = 2x – 3x2 using the four-step process.
Step 1: Find f(x+h)
Step 2: Find f(x+h) – f(x)
Step 3: Find
Example 2
¿2 (𝑥+h ) −3 (𝑥+h)2
¿2 𝑥+2h− 3(𝑥2+2 h𝑥 +h2)¿2 𝑥+2 h− 3 𝑥2 −6 h𝑥 −3 h2
¿2 𝑥+2h− 3 𝑥2 −6 h𝑥 −3 h2−(2 𝑥− 3𝑥2)¿2 𝑥+2 h− 3 𝑥2 −6 h𝑥 −3 h2− 2𝑥+3 𝑥2
¿2 h −6 h𝑥 − 3 h2
¿ 2 h−6 h𝑥 −3 h2 h
¿2 −6 𝑥−3 h
31
Example 2 (continued)
Step 4: Find f (x) =
Barnett/Ziegler/Byleen Business Calculus 12e
limh →0
𝑓 (𝑥+h ) − 𝑓 (𝑥)h
𝑓 (𝑥)=limh → 0
(2 −6 𝑥−3 h¿)¿
𝑓 (𝑥 )=2− 6 𝑥
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
32Barnett/Ziegler/Byleen Business Calculus 12e
Find f (x) where using the four-step process.
Step 1: Find f(x+h)
Step 2: Find f(x+h) – f(x)
Step 3: Find
Example 3
¿√𝑥+h+2¿√𝑥+h+2−(√𝑥+2)¿√𝑥+h−√𝑥¿ √𝑥+h−√𝑥
h
33
Example 3 (continued)
Step 4: Find f (x) =
Barnett/Ziegler/Byleen Business Calculus 12e
limh →0
𝑓 (𝑥+h ) − 𝑓 (𝑥)h
¿1
(√𝑥+√𝑥 )
𝑓 (𝑥 )= 12√𝑥
∙ √𝑥√𝑥
=√𝑥2𝑥
¿1
2√𝑥
f ( x )=limh→ 0
√𝑥+h −√𝑥h
∙ √𝑥+h+√𝑥√𝑥+h+√𝑥
f ( x )=limh→ 0
𝑥+h−𝑥h (√𝑥+h+√𝑥 )
¿ limh→0
h
h (√𝑥+h+√𝑥 )f ( x )=lim
h→ 0
1(√𝑥+h+√𝑥 )
34Barnett/Ziegler/Byleen Business Calculus 12e
Nonexistence of the Derivative
The existence of a derivative at x = a depends on the existence of the limit
If the limit does not exist, we say that the function is nondifferentiable at x = a, or f (a) does not exist.
f (a) lim
h 0
f (a h) f (a)
h
35Barnett/Ziegler/Byleen Business Calculus 12e
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.
36
Examples of Nonexistent Derivatives
Barnett/Ziegler/Byleen Business Calculus 12e
In each graph, f is nondifferentiable at x=a.
37
Application – Profit
The profit (in dollars) from the sale of x infant car seats is given by: 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.
Barnett/Ziegler/Byleen Business Calculus 12e
38
Application – Profit(continued)
The profit (in dollars) from the sale of x infant car seats is given by: 0 x 2400
A) Find the average change in profit if production increases from 800 to 850 car seats.
Barnett/Ziegler/Byleen Business Calculus 12e
𝐴𝑣𝑔 h𝑐 𝑎𝑛𝑔𝑒𝑖𝑛𝑝𝑟𝑜𝑓𝑖𝑡=𝑃 (850 ) −𝑃 (800)
850 −800
¿15187.5 −15000
50¿3.75
The avg change in profit when production increases from 800 to 850 car seats is $3.75 per seat.
39
Application – Profit(continued)
The profit (in dollars) from the sale of x infant car seats is given by: 0 x 2400
B) Use the 4-step process to find P(x)
Barnett/Ziegler/Byleen Business Calculus 12e
𝑺𝒕𝒆𝒑𝟏 :𝑃 (𝑥+h )=45 (𝑥+h ) −0.025 (𝑥+h )2 −5000
¿ 45 𝑥+45 h − 0.025 (𝑥2+2 h𝑥 +h2 ) −5000
¿ 45 𝑥+45 h − 0.025𝑥2 −0.05 h𝑥 −0.025 h2− 5000𝑺𝒕𝒆𝒑𝟐 :𝑃 (𝑥+h )− 𝑃 (𝑥)
¿ 45 h −0.05 h𝑥 − 0.025 h2
40
Application – Profit(continued)
Barnett/Ziegler/Byleen Business Calculus 12e
𝑺𝒕𝒆𝒑𝟑 :𝑃 (𝑥+h ) −𝑃 (𝑥)
h¿ 45 h− 0.05 h𝑥 −0.025h2
h¿ 45 −0.05 𝑥− 0.025 h
𝑺𝒕𝒆𝒑𝟒 : limh → 0
𝑃 (𝑥+h )−𝑃 (𝑥 ) h
¿ limh→0
(45− 0.05 𝑥−0.025 h)
¿ 45 − .05 𝑥
41
Application – Profit(continued)
The profit (in dollars) from the sale of x infant car seats is given by: 0 x 2400
C) Find P(800) and P(800) and explain their meaning.
Barnett/Ziegler/Byleen Business Calculus 12e
𝑃 (800 )=45 (800 ) −0.025 (800 )2− 5000𝑃 (800 )=$ 15,000𝑃 (800 )=45 − 0.05(800)𝑃 (800 )=$ 5
At a production level of 800 car seats, the profit is $15,000and it is increasing at a rate of $5 per car seat.
42
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
#3-4B Pg 176(7, 21, 25,
31-41, 61, 63)
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