chapter 3 limits and the derivative section 4 the derivative (part 1)

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.

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

Slopes

Slope of a secant Slope of a tangent

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.

𝑅 (𝑥 )=20 𝑥− 0.02𝑥2

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.

𝑅 (400 )−𝑅(100)400 −100

=$ 3000

300𝑤𝑖𝑑𝑔𝑒𝑡𝑠=$ 10𝑝𝑒𝑟𝑤𝑖𝑑𝑔𝑒𝑡

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.

The Rate of Change

For y = f (x), the average rate of change from x = a to x = a + h is

0,)()(

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.

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)

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

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.

afhafh

)()(lim

Visual Interpretation

Slope of tangent at x = a is theinstantaneous rate of change.

Tangent line at x=a

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.

afhafh

)()(lim

Instantaneous Rate of Change

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).

𝑓 ( 4 )− 𝑓 (1)4 −1

=0 − 3

3=− 1

Example 3B

Find the instantaneous rate of change of f(x) at x = 1

limh →0

𝑓 (1+h )− 𝑓 (1)h ¿

limh → 0

[4 (1+h ) − (1+h )2 ] −(4 − 12)

¿limh → 0

[4+4h−(1+2h+h2)]− 3

¿limh → 0

[4+4 h −1 −2 h− h2 ]− 3

¿limh → 0

2 h− h2

Continued on next slide…

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.

¿ limh→0

(2− h)

¿limh → 0

2 h− h2

¿limh → 0

h(2− h)

Example 3C

Find the equation of the tangent line at x=1.

When x=1, y=3 and slope = 2

𝑦=𝑚𝑥+𝑏3=2(1)+𝑏𝑏=1𝑦=2 𝑥+1

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:

𝑦=16 𝑥2

400 − 645 − 2

𝑓 (5 )− 𝑓 (2)5 −2

¿336 𝑓𝑡3𝑠𝑒𝑐

=112 𝑓𝑡 /𝑠𝑒𝑐

Velocity(continued)

Find the instantaneous velocity at x = 2 seconds.

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

¿limh → 0

16(4+4 h+h2)−16 (4)

¿limh → 0

64+64 h+16 h2 −64

¿limh → 0

h(64+16 h)

𝑦=16 𝑥2

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

limh → 0

𝑓 (𝑎+h )− 𝑓 (𝑎)

Homework

Pg 175

(1-4, 27-30)

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.

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.

limh →0

𝑓 (𝑎+h )− 𝑓 (𝑎)h

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

f (x h) f (x)

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.

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) =

xfhxf )()(

xfhxfh

)()(lim

*Feel free to go directly to Step 4 when you’ve got the process down!

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

Step 4: Find f (x) =

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

Find f (x) where f (x) = 2x – 3x2 using the four-step process.

Step 1: Find f(x+h)

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

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

Find f (x) where using the four-step process.

Step 1: Find f(x+h)

Step 3: Find

Example 3

¿√𝑥+h+2¿√𝑥+h+2−(√𝑥+2)¿√𝑥+h−√𝑥¿ √𝑥+h−√𝑥

limh →0

𝑓 (𝑥+h ) − 𝑓 (𝑥)h

(√𝑥+√𝑥 )

𝑓 (𝑥 )= 12√𝑥

∙ √𝑥√𝑥

=√𝑥2𝑥

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 (√𝑥+h+√𝑥 )f ( x )=lim

h→ 0

1(√𝑥+h+√𝑥 )

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

f (a h) f (a)

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.

Examples of Nonexistent Derivatives

In each graph, f is nondifferentiable at x=a.

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.

Application – Profit(continued)

A) Find the average change in profit if production increases from 800 to 850 car seats.

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.

B) Use the 4-step process to find P(x)

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

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

C) Find P(800) and P(800) and explain their meaning.

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.

Homework

#3-4B Pg 176(7, 21, 25,

31-41, 61, 63)

chapter 3 limits and the derivative section 4 the derivative (part 1)

slide slide

secant line slide

rate of change concept

e answer

secant slope

e learning objectives

slopes slope

x seconds

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