Upcoming SlideShare
×

# Lesson 9: The Product and Quotient Rule

4,724 views

Published on

These rules allow us to differentiate the product or quotient of functions whose derivatives are themselves known.

Published in: Education, Technology
1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
4,724
On SlideShare
0
From Embeds
0
Number of Embeds
48
Actions
Shares
0
50
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Lesson 9: The Product and Quotient Rule

1. 1. Section 3.2 The Product and Quotient Rules Math 1a February 22, 2008 Announcements Problem Sessions Sunday, Thursday, 7pm, SC 310 Oﬃce hours Tuesday, Wednesday 2–4pm SC 323 Midterm I Friday 2/29 in class (up to §3.2)
2. 2. Outline The Product Rule Derivation of the product rule Examples The Quotient Rule Derivation Examples More on the Power Rule Power Rule for nonnegative integers by induction Power Rule for negative integers
3. 3. Recollection and extension We have shown that if u and v are functions, that (u + v ) = u + v (u − v ) = u − v What about uv ? Is it u v ?
4. 4. Is the derivative of a product the product of the derivatives? NO!
5. 5. Is the derivative of a product the product of the derivatives? NO! Try this with u = x and v = x 2 .
6. 6. Is the derivative of a product the product of the derivatives? NO! Try this with u = x and v = x 2 . Then uv = x 3 , so (uv ) = 3x 2 . But u v = 1(2x) = 2x.
7. 7. Is the derivative of a product the product of the derivatives? NO! Try this with u = x and v = x 2 . Then uv = x 3 , so (uv ) = 3x 2 . But u v = 1(2x) = 2x. So we have to be more careful.
8. 8. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices?
9. 9. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours.
10. 10. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise.
11. 11. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make?
12. 12. Money money money money The answer depends on how much you work already and your current wage. Suppose you work h hours and are paid w . You get a time increase of ∆h and a wage increase of ∆w . Income is wages times hours, so ∆I = (w + ∆w )(h + ∆h) − wh FOIL = wh + w ∆h + ∆w h + ∆w ∆h − wh = w ∆h + ∆w h + ∆w ∆h
13. 13. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h w ∆w
14. 14. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h w ∆w ∆I = w ∆h + h ∆w + ∆w ∆h
15. 15. Supose wages and hours are changing continuously over time. How does income change? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t
16. 16. Supose wages and hours are changing continuously over time. How does income change? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t So dI ∆I dh dw = lim =w +h +0 dt t→0 ∆t dt dt
17. 17. Supose wages and hours are changing continuously over time. How does income change? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t So dI ∆I dh dw = lim =w +h +0 dt t→0 ∆t dt dt Theorem (The Product Rule) Let u and v be diﬀerentiable at x. Then (uv ) (x) = u(x)v (x) + u (x)v (x)
18. 18. Example Apply the product rule to u = x and v = x 2 .
19. 19. Example Apply the product rule to u = x and v = x 2 . Solution (uv ) (x) = u(x)v (x) + u (x)v (x) = x · (2x) + 1 · x 2 = 3x 2 This is what we get the “normal” way.
20. 20. Example Find this derivative two ways: ﬁrst by FOIL and then by the product rule: d (3 − x 2 )(x 3 − x + 1) dx
21. 21. Example Find this derivative two ways: ﬁrst by FOIL and then by the product rule: d (3 − x 2 )(x 3 − x + 1) dx Solution (i) by FOIL: d FOILd (3 − x 2 )(x 3 − x + 1) = −x 5 + 4x 3 − x 2 − 3x + 3 dx dx = −5x 4 + 12x 2 − 2x − 3 (ii) by the product rule: dy d d 3 = (3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1) dx dx dx = (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1) = −5x 4 + 12x 2 − 2x − 3
22. 22. One more Example d x Find xe dx
23. 23. One more Example d x Find xe dx Answer y = e x + xe x
24. 24. Mnemonic Let u = “ho” and v = “hi”. Then (uv ) = uv + vu = “ho dee hi plus hi dee ho”
25. 25. Outline The Product Rule Derivation of the product rule Examples The Quotient Rule Derivation Examples More on the Power Rule Power Rule for nonnegative integers by induction Power Rule for negative integers
26. 26. The Quotient Rule What about the derivative of a quotient?
27. 27. The Quotient Rule What about the derivative of a quotient? u Let u and v be diﬀerentiable and let Q = . Then u = Qv . If Q v is diﬀerentiable, we have u = (Qv ) = Q v + Qv u − Qv u uv Q = = − 2 v v v u v − uv = v2
28. 28. The Quotient Rule What about the derivative of a quotient? u Let u and v be diﬀerentiable and let Q = . Then u = Qv . If Q v is diﬀerentiable, we have u = (Qv ) = Q v + Qv u − Qv u uv Q = = − 2 v v v u v − uv = v2 This is called the Quotient Rule.
29. 29. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t
30. 30. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers
31. 31. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers 19 1. − (3x − 2)2
32. 32. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers 19 1. − (3x − 2)2 2 x2 + x + 1 2. − (x 2 − 1)2
33. 33. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers 19 1. − (3x − 2)2 2 x2 + x + 1 2. − (x 2 − 1)2 −t 2 + 2t + 3 3. (t 2 + t + 2)2
34. 34. Mnemonic Let u = “hi” and v = “lo”. Then u vu − uv = = “lo dee hi minus hi dee lo over lo lo” v v2
35. 35. Outline The Product Rule Derivation of the product rule Examples The Quotient Rule Derivation Examples More on the Power Rule Power Rule for nonnegative integers by induction Power Rule for negative integers
36. 36. Power Rule for nonnegative integers by induction Theorem Let n be a positive integer. Then d n x = nx n−1 dx
37. 37. Power Rule for nonnegative integers by induction Theorem Let n be a positive integer. Then d n x = nx n−1 dx Proof. By induction on n. We have shown it to be true for n = 1. d n Suppose for some n that x = nx n−1 . Then dx d n+1 d x = (x · x n ) dx dx d d n = x xn + x x dx dx = 1 · x n + x · nx n−1 = (n + 1)x n
38. 38. Power Rule for negative integers Use the quotient rule to prove Theorem d −n x = (−n)x −n−1 dx for positive integers n.
39. 39. Power Rule for negative integers Use the quotient rule to prove Theorem d −n x = (−n)x −n−1 dx for positive integers n. Proof. d −n d 1 x = dx dx x n d d x n · dx 1 − 1 · dx x n = x 2n 0 − nx n−1 = = −nx −n−1 x 2n