Top Banner
Chapter 3 Limits and the Derivative Section 6 Differentials
12

Chapter 3 Limits and the Derivative

Jan 03, 2016

Download

Documents

valentine-elias

Chapter 3 Limits and the Derivative. Section 6 Differentials. Learning Objectives for Section 3.6 Differentials. The student will be able to apply the concept of increments. The student will be able to compute differentials. - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Chapter 3 Limits and the Derivative

Chapter 3

Limits and the Derivative

Section 6

Differentials

Page 2: Chapter 3 Limits and the Derivative

2

Learning Objectives for Section 3.6 Differentials

The student will be able to apply the concept of increments.

The student will be able to compute differentials.

The student will be able to calculate approximations using differentials.

Page 3: Chapter 3 Limits and the Derivative

3

Increments

In a previous section we defined the derivative of f at x as the limit of the difference quotient:

Increment notation will enable us to interpret the numerator and the denominator of the difference quotient separately.

f (x) lim

h 0

f (x h) f (x)

h

Page 4: Chapter 3 Limits and the Derivative

4

Example

Let y = f (x) = x3. If x changes from 2 to 2.1, then y will change from y = f (2) = 8 to y = f (2.1) = 9.261.

We can write this using increment notation. The change in x is called the increment in x and is denoted by x. is the Greek letter “delta”, which often stands for a difference or change. Similarly, the change in y is called the increment in y and is denoted by y.

In our example,

x = 2.1 – 2 = 0.1

y = f (2.1) – f (2) = 9.261 – 8 = 1.261.

Page 5: Chapter 3 Limits and the Derivative

5

Graphical Illustrationof Increments

For y = f (x)

x = x2 - x1 y = y2 - y1

x2 = x1 + x = f (x2) – f (x1) = f (x1 + x) – f (x1)

(x1, f (x1))

(x2, f (x2))

x1 x2

x

y■ y represents the

change in y corresponding to a x change in x.

■ x can be either positive or negative.

Page 6: Chapter 3 Limits and the Derivative

6

Assume that the limit exists.

For small x,

Multiplying both sides of this equation by x gives us

y f (x) x.

Here the increments x and y represent the actual changes in x and y.

Differentials

f (x) lim

x 0

y

x

f (x)

y

x

Page 7: Chapter 3 Limits and the Derivative

7

One of the notations for the derivative is

If we pretend that dx and dy are actual quantities, we get

We treat this equation as a definition, and call dx and dy differentials.

Differentials(continued)

f (x)

dy

dx

dy f (x) dx

Page 8: Chapter 3 Limits and the Derivative

8

x and dx are the same, and represent the change in x.

The increment y stands for the actual change in y resulting from the change in x.

The differential dy stands for the approximate change in y, estimated by using derivatives.

In applications, we use dy (which is easy to calculate) to estimate y (which is what we want).

Interpretation of Differentials

y dy f (x) dx

Page 9: Chapter 3 Limits and the Derivative

9

Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

Page 10: Chapter 3 Limits and the Derivative

10

Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

Solution:

dy = f (x) dx = (2x + 3) dx

When x = 2 and dx = 0.1, dy = [2(2) + 3] 0.1 = 0.7.

Page 11: Chapter 3 Limits and the Derivative

11

Example 2 Cost-Revenue

A company manufactures and sells x transistor radios per week. If the weekly cost and revenue equations are

find the approximate changes in revenue and profit if production is increased from 2,000 to 2,010 units/week.

000,80

000,110)(

2000,5)(2

x

xxxR

xxC

Page 12: Chapter 3 Limits and the Derivative

12

The profit is

We will approximate R and P with dR and dP, respectively, using x = 2,000 and dx = 2,010 – 2,000 = 10.

Example 2Solution

000,5000,1

8)()()(2

x

xxCxRxP

dR R (x) dx (10 x

500) dx

(10 2,000

500) 10 $60 per week

dP P (x) dx (8 x

500) dx

(8 2,000

500) 10 $40 per week