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# Lesson 9: The Product and Quotient Rules (slides)

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### Lesson 9: The Product and Quotient Rules (slides)

1. 1. Sec on 2.4 The Product and Quo ent Rules V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 23, 2011.
2. 2. Announcements Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2 Midterm March 7 on all sec ons in class (covers all sec ons up to 2.5)
3. 3. Help! Free resources: Math Tutoring Center (CIWW 524) College Learning Center (schedule on Blackboard) TAs’ oﬃce hours my oﬃce hours each other!
4. 4. Objectives Understand and be able to use the Product Rule for the deriva ve of the product of two func ons. Understand and be able to use the Quo ent Rule for the deriva ve of the quo ent of two func ons.
5. 5. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
6. 6. Recollection and extension We have shown that if u and v are func ons, that (u + v)′ = u′ + v′ (u − v)′ = u′ − v′ What about uv?
7. 7. Is the derivative of a product theproduct of the derivatives? (uv)′ = u′ v′ ? .
8. 8. Is the derivative of a product theproduct of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 .
9. 9. Is the derivative of a product theproduct of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 .
10. 10. Is the derivative of a product theproduct of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x.
11. 11. Is the derivative of a product theproduct of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. So we have to be more careful.
12. 12. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? .
13. 13. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. .
14. 14. 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. .
15. 15. 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? .
16. 16. 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? . ∆I = 5 × \$0.25 = \$1.25?
17. 17. 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? . ∆I = 5 × \$0.25 = \$1.25?
18. 18. 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 me increase of ∆h and a wage increase of ∆w. Income is wages mes hours, so ∆I = (w + ∆w)(h + ∆h) − wh FOIL = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh = w · ∆h + ∆w · h + ∆w · ∆h
19. 19. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h . w ∆w
20. 20. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h . w ∆w ∆I = w ∆h + h ∆w + ∆w ∆h
21. 21. Cash ﬂow Supose wages and hours are changing con nuously over me. Over a me interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t
22. 22. Cash ﬂow Supose wages and hours are changing con nuously over me. Over a me interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t What is the instantaneous rate of change of income? dI ∆I dh dw = lim =w +h +0 dt ∆t→0 ∆t dt dt
23. 23. Eurekamen! We have discovered Theorem (The Product Rule) Let u and v be diﬀeren able at x. Then (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) in Leibniz nota on d du dv (uv) = ·v+u dx dx dx
24. 24. Sanity Check Example Apply the product rule to u = x and v = x2 .
25. 25. Sanity Check Example Apply the product rule to u = x and v = x2 . Solu on (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2 This is what we get the “normal” way.
26. 26. Which is better? Example Find this deriva ve two ways: ﬁrst by direct mul plica on and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx
27. 27. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx
28. 28. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by direct mul plica on: d [ ] FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx
29. 29. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by direct mul plica on: d [ ] FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx = −5x4 + 12x2 − 2x − 3
30. 30. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx
31. 31. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
32. 32. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
33. 33. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
34. 34. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
35. 35. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
36. 36. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) = −5x4 + 12x2 − 2x − 3
37. 37. One more Example d Find x sin x. dx
38. 38. One more Example d Find x sin x. dx Solu on ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx
39. 39. One more Example d Find x sin x. dx Solu on ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x
40. 40. One more Example d Find x sin x. dx Solu on ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x = sin x + x cos x
41. 41. Mnemonic Let u = “hi” and v = “ho”. Then (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
42. 42. Musical interlude jazz bandleader and singer hit song “Minnie the Moocher” featuring “hi de ho” chorus played Cur s in The Blues Brothers Cab Calloway 1907–1994
43. 43. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw.
44. 44. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw. Solu on (uvw)′ .
45. 45. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw. Solu on (uvw)′ = ((uv)w)′ .
46. 46. Iterating the Product Rule Example Apply the product Use the product rule to ﬁnd the deriva ve of a three-fold product uvw. rule to uv and w Solu on (uvw)′ = ((uv)w)′ .
47. 47. Iterating the Product Rule Example Apply the product Use the product rule to ﬁnd the deriva ve of a three-fold product uvw. rule to uv and w Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
48. 48. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw.product Apply the rule to u and v Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
49. 49. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw.product Apply the rule to u and v Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′
50. 50. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw. Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′
51. 51. Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw. Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′ So we write down the product three mes, taking the deriva ve of each factor once.
52. 52. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
53. 53. The Quotient Rule What about the deriva ve of a quo ent?
54. 54. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be diﬀeren able func ons and let Q = . Then v u = Qv
55. 55. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be diﬀeren able func ons and let Q = . Then v u = Qv If Q is diﬀeren able, we have u′ = (Qv)′ = Q′ v + Qv′
56. 56. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be diﬀeren able func ons and let Q = . Then v u = Qv If Q is diﬀeren able, we have u′ = (Qv)′ = Q′ v + Qv′ ′ u′ − Qv′ u′ u v′ =⇒ Q = = − · v v v v
57. 57. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be diﬀeren able func ons and let Q = . Then v u = Qv If Q is diﬀeren able, we have u′ = (Qv)′ = Q′ v + Qv′ ′ u′ − Qv′ u′ u v′ =⇒ Q = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2
58. 58. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be diﬀeren able func ons and let Q = . Then v u = Qv If Q is diﬀeren able, we have u′ = (Qv)′ = Q′ v + Qv′ ′ u′ − Qv′ u′ u v′ =⇒ Q = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2 This is called the Quo ent Rule.
59. 59. The Quotient Rule We have discovered Theorem (The Quo ent Rule) u Let u and v be diﬀeren able at x, and v(x) ̸= 0. Then is v diﬀeren able at x, and ( u )′ u′ (x)v(x) − u(x)v′ (x) (x) = v v(x)2
60. 60. Verifying Example Example ( ) d x2 d Verify the quo ent rule by compu ng and comparing it to (x). dx x dx
61. 61. Verifying Example Example ( ) d x2 d Verify the quo ent rule by compu ng and comparing it to (x). dx x dx Solu on ( 2) ( ) d x x dx x2 − x2 dx (x) x · 2x − x2 · 1 d d = = dx x x2 x2 x2 d = 2 =1= (x) x dx
62. 62. Mnemonic Let u = “hi” and v = “lo”. Then ( u )′ vu′ − uv′ = = “lo dee hi minus hi dee lo over lo lo” v v2
63. 63. Examples Example d 2x + 5 1. dx 3x − 2 d sin x 2. dx x2 d 1 3. 2+t+2 dt t
64. 64. Solution to ﬁrst example Solu on d 2x + 5 dx 3x − 2
65. 65. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
66. 66. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
67. 67. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
68. 68. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
69. 69. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
70. 70. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
71. 71. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
72. 72. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
73. 73. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
74. 74. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
75. 75. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
76. 76. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) = (3x − 2)2
77. 77. Solution to ﬁrst example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) 19 = =− (3x − 2)2 (3x − 2)2
78. 78. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x 2. dx x2 d 1 3. 2+t+2 dt t
79. 79. Solution to second example Solu on d sin x = dx x2
80. 80. Solution to second example Solu on d sin x x2 = dx x2
81. 81. Solution to second example Solu on d d sin x x2 dx sin x = dx x2
82. 82. Solution to second example Solu on d sin x x2 dx sin x − sin x d = dx x2
83. 83. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2
84. 84. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2
85. 85. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 =
86. 86. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 =
87. 87. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x =
88. 88. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x =
89. 89. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x sin x =
90. 90. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x sin x = x4
91. 91. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x sin x = x4 x cos x − 2 sin x = x3
92. 92. Another way to do it Find the deriva ve with the product rule instead. Solu on d sin x d ( ) 2 = sin x · x−2 dx x dx ( ) ( ) d −2 d −2 = sin x · x + sin x · x dx dx = cos x · x−2 + sin x · (−2x−3 ) = x−3 (x cos x − 2 sin x) No ce the technique of factoring out the largest nega ve power, leaving posi ve powers.
93. 93. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x x cos x − 2 sin x 2. 2. dx x2 x3 d 1 3. 2+t+2 dt t
94. 94. Solution to third example Solu on d 1 dt t2 + t + 2
95. 95. Solution to third example Solu on d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2
96. 96. Solution to third example Solu on d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 2t + 1 =− 2 (t + t + 2)2
97. 97. A nice little takeaway Fact 1 Let v be diﬀeren able at x, and v(x) ̸= 0. Then is diﬀeren able at v 0, and ( )′ 1 v′ =− 2 v v Proof. ( ) d 1 v· d dx (1)−1· d dx v v · 0 − 1 · v′ v′ = = =− 2 dx v v2 v2 v
98. 98. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x x cos x − 2 sin x 2. 2. dx x2 x3 d 1 2t + 1 3. 3. − 2 dt t2+t+2 (t + t + 2)2
99. 99. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
100. 100. Derivative of Tangent Example d Find tan x dx
101. 101. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x tan x = dx dx cos x
102. 102. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x
103. 103. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x = cos2 x
104. 104. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = = cos2 x cos2 x
105. 105. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = 2x = 2x = sec2 x cos cos
106. 106. Derivative of Cotangent Example d Find cot x dx
107. 107. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x
108. 108. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x
109. 109. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x = sin2 x
110. 110. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x 1 = =− 2 sin2 x sin x
111. 111. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x 1 = = − 2 = − csc2 x sin2 x sin x
112. 112. Derivative of Secant Example d Find sec x dx
113. 113. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 sec x = dx dx cos x
114. 114. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x
115. 115. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x = cos2 x
116. 116. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · cos2 x cos x cos x
117. 117. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · = sec x tan x cos2 x cos x cos x
118. 118. Derivative of Cosecant Example d Find csc x dx
119. 119. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx
120. 120. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x
121. 121. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x =− 2 sin x
122. 122. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · sin x sin x sin x
123. 123. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · = − csc x cot x sin x sin x sin x
124. 124. Recap: Derivatives oftrigonometric functions y y′ sin x cos x Func ons come in pairs (sin/cos, tan/cot, sec/csc) cos x − sin x Deriva ves of pairs follow tan x sec2 x similar pa erns, with cot x − csc2 x func ons and co-func ons switched sec x sec x tan x and an extra sign. csc x − csc x cot x
125. 125. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
126. 126. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n.
127. 127. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof.
128. 128. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d −n d 1 x = dx dx xn
129. 129. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x x = =− n 2 dx dx xn (x )
130. 130. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x nxn−1 x = = − n 2 = − 2n dx dx xn (x ) x
131. 131. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x nxn−1 x = n = − n 2 = − 2n = −nxn−1−2n dx dx x (x ) x
132. 132. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x nxn−1 x = n = − n 2 = − 2n = −nxn−1−2n = −nx−n−1 dx dx x (x ) x
133. 133. Summary The Product Rule: (uv)′ = u′ v + uv′ ( u )′ vu′ − uv′ The Quo ent Rule: = v v2 Deriva ves of tangent/cotangent, secant/cosecant d d tan x = sec2 x sec x = sec x tan x dx dx d d cot x = − csc2 x csc x = − csc x cot x dx dx The Power Rule is true for all whole number powers, including nega ve powers: d n x = nxn−1 dx