Chapter 3 Limits and the Derivative
-
Upload
valentine-elias -
Category
Documents
-
view
34 -
download
2
description
Transcript of Chapter 3 Limits and the Derivative
![Page 1: Chapter 3 Limits and the Derivative](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/1.jpg)
Chapter 3
Limits and the Derivative
Section 6
Differentials
![Page 2: Chapter 3 Limits and the Derivative](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/2.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/3.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/4.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/5.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/6.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/7.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/8.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/9.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/10.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/11.jpg)
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](https://reader036.fdocuments.in/reader036/viewer/2022062314/5681366d550346895d9dfc82/html5/thumbnails/12.jpg)
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