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# Lesson 7: The Derivative (slides)

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Many functions in nature are described as the rate of change of another function. The concept is called the derivative. Algebraically, the process of finding the derivative involves a limit of difference quotients.

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### Transcript of "Lesson 7: The Derivative (slides)"

1. 1. Sec on 2.1–2.2 The Deriva ve V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 14, 2011. NYUMathematics
2. 2. Announcements Quiz this week on Sec ons 1.1–1.4 No class Monday, February 21
3. 3. ObjectivesThe Derivative Understand and state the deﬁni on of the deriva ve of a func on at a point. Given a func on and a point in its domain, decide if the func on is diﬀeren able at the point and ﬁnd the value of the deriva ve at that point. Understand and give several examples of deriva ves modeling rates of change in science.
4. 4. ObjectivesThe Derivative as a Function Given a func on f, use the deﬁni on of the deriva ve to ﬁnd the deriva ve func on f’. Given a func on, ﬁnd its second deriva ve. Given the graph of a func on, sketch the graph of its deriva ve.
5. 5. Outline Rates of Change Tangent Lines Velocity Popula on growth Marginal costs The deriva ve, deﬁned Deriva ves of (some) power func ons What does f tell you about f′ ? How can a func on fail to be diﬀeren able? Other nota ons The second deriva ve
6. 6. The tangent problemA geometric rate of change Problem Given a curve and a point on the curve, ﬁnd the slope of the line tangent to the curve at that point.
7. 7. A tangent problem Example Find the slope of the line tangent to the curve y = x2 at the point (2, 4).
8. 8. Graphically and numerically y x2 − 22 x m= x−2 4 . x 2
9. 9. Graphically and numerically y x2 − 22 x m= x−2 3 9 4 . x 2 3
10. 10. Graphically and numerically y x2 − 22 x m= x−2 3 5 9 4 . x 2 3
11. 11. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 6.25 4 . x 2 2.5
12. 12. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 6.25 4 . x 2 2.5
13. 13. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.41 4 . x 2.1 2
14. 14. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 4.41 4 . x 2.1 2
15. 15. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.014.0401 4 . x 2.01 2
16. 16. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.014.0401 4 . x 2.01 2
17. 17. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1 . x 1 1 2
18. 18. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1 . x 1 3 1 2
19. 19. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 2.25 . 1.5 x 1 3 1.5 2
20. 20. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 2.25 . 1.5 3.5 x 1 3 1.5 2
21. 21. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 3.61 1.9 . 1.5 3.5 x 1 3 1.9 2
22. 22. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 3.61 1.9 3.9 . 1.5 3.5 x 1 3 1.9 2
23. 23. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 43.9601 1.99 1.9 3.9 . 1.5 3.5 x 1 3 1.99 2
24. 24. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 43.9601 1.99 3.99 1.9 3.9 . 1.5 3.5 x 1 3 1.99 2
25. 25. Graphically and numerically y x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 4 1.99 3.99 1.9 3.9 . 1.5 3.5 x 1 3 2
26. 26. Graphically and numerically y x2 − 22 x m= x−2 3 5 9 2.5 4.5 2.1 4.1 6.25 2.01 4.01 limit 4.414.0401 43.9601 3.61 1.99 3.99 2.25 1.9 3.9 1 1.5 3.5 . x 1 3 1 1.52.1 3 1.99 2.01 1.92.5 2
27. 27. Graphically and numerically y x2 − 22 x m= x−2 3 5 9 2.5 4.5 2.1 4.1 6.25 2.01 4.01 limit 4 4.414.0401 43.9601 3.61 1.99 3.99 2.25 1.9 3.9 1 1.5 3.5 . x 1 3 1 1.52.1 3 1.99 2.01 1.92.5 2
28. 28. The tangent problemA geometric rate of change Problem Given a curve and a point on the curve, ﬁnd the slope of the line tangent to the curve at that point. Solu on If the curve is given by y = f(x), and the point on the curve is (a, f(a)), then the slope of the tangent line is given by f(x) − f(a) mtangent = lim x→a x−a
29. 29. The velocity problemKinematics—Physical rates of change Problem Given the posi on func on of a moving object, ﬁnd the velocity of the object at a certain instant in me.
30. 30. A velocity problemExampleDrop a ball oﬀ the roof of theSilver Center so that its height canbe described by h(t) = 50 − 5t2where t is seconds a er droppingit and h is meters above theground. How fast is it falling onesecond a er we drop it?
31. 31. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15
32. 32. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5
33. 33. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5
34. 34. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1
35. 35. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5
36. 36. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01
37. 37. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05
38. 38. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001
39. 39. Numerical evidence h(t) = 50 − 5t2 Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001 − 10.005
40. 40. A velocity problemExample Solu onDrop a ball oﬀ the roof of the The answer is (50 − 5t2 ) − 45Silver Center so that its height can v = limbe described by t→1 t−1 h(t) = 50 − 5t2where t is seconds a er droppingit and h is meters above theground. How fast is it falling onesecond a er we drop it?
41. 41. A velocity problemExample Solu onDrop a ball oﬀ the roof of the The answer is (50 − 5t2 ) − 45Silver Center so that its height can v = limbe described by t→1 t−1 5 − 5t2 = lim h(t) = 50 − 5t2 t→1 t − 1where t is seconds a er droppingit and h is meters above theground. How fast is it falling onesecond a er we drop it?
42. 42. A velocity problemExample Solu onDrop a ball oﬀ the roof of the The answer is (50 − 5t2 ) − 45Silver Center so that its height can v = limbe described by t→1 t−1 5 − 5t2 = lim h(t) = 50 − 5t2 t→1 t − 1 5(1 − t)(1 + t)where t is seconds a er dropping = lim t→1 t−1it and h is meters above theground. How fast is it falling onesecond a er we drop it?
43. 43. A velocity problemExample Solu onDrop a ball oﬀ the roof of the The answer is (50 − 5t2 ) − 45Silver Center so that its height can v = limbe described by t→1 t−1 5 − 5t2 = lim h(t) = 50 − 5t2 t→1 t − 1 5(1 − t)(1 + t)where t is seconds a er dropping = lim t→1 t−1it and h is meters above the = (−5) lim(1 + t) t→1ground. How fast is it falling onesecond a er we drop it?
44. 44. A velocity problemExample Solu onDrop a ball oﬀ the roof of the The answer is (50 − 5t2 ) − 45Silver Center so that its height can v = limbe described by t→1 t−1 5 − 5t2 = lim h(t) = 50 − 5t2 t→1 t − 1 5(1 − t)(1 + t)where t is seconds a er dropping = lim t→1 t−1it and h is meters above the = (−5) lim(1 + t) t→1ground. How fast is it falling onesecond a er we drop it? = −5 · 2 = −10
45. 45. Velocity in general Upshot y = h(t) If the height func on is given h(t0 ) by h(t), the instantaneous ∆h velocity at me t0 is given by h(t0 + ∆t) h(t) − h(t0 ) v = lim t→t0 t − t0 h(t0 + ∆t) − h(t0 ) = lim ∆t→0 ∆t . ∆t t t0 t
46. 46. Population growthBiological Rates of Change Problem Given the popula on func on of a group of organisms, ﬁnd the rate of growth of the popula on at a par cular instant.
47. 47. Population growth example Example Suppose the popula on of ﬁsh in the East River is given by the func on 3et P(t) = 1 + et where t is in years since 2000 and P is in millions of ﬁsh. Is the ﬁsh popula on growing fastest in 1990, 2000, or 2010? (Es mate numerically)
48. 48. Derivation Solu on Let ∆t be an increment in me and ∆P the corresponding change in popula on: ∆P = P(t + ∆t) − P(t) This depends on ∆t, so ideally we would want ( ) ∆P 1 3et+∆t 3et lim = lim − ∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et
49. 49. Derivation Solu on Let ∆t be an increment in me and ∆P the corresponding change in popula on: ∆P = P(t + ∆t) − P(t) This depends on ∆t, so ideally we would want ( ) ∆P 1 3et+∆t 3et lim = lim − ∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et But rather than compute a complicated limit analy cally, let us approximate numerically. We will try a small ∆t, for instance 0.1.
50. 50. Numerical evidence Solu on (Con nued) To approximate the popula on change in year n, use the diﬀerence P(t + ∆t) − P(t) quo ent , where ∆t = 0.1 and t = n − 2000. ∆t r1990 r2000
51. 51. Numerical evidence Solu on (Con nued) To approximate the popula on change in year n, use the diﬀerence P(t + ∆t) − P(t) quo ent , where ∆t = 0.1 and t = n − 2000. ∆t P(−10 + 0.1) − P(−10) r1990 ≈ 0.1 P(0.1) − P(0) r2000 ≈ 0.1
52. 52. Numerical evidence Solu on (Con nued) To approximate the popula on change in year n, use the diﬀerence P(t + ∆t) − P(t) quo ent , where ∆t = 0.1 and t = n − 2000. ∆t ( ) P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10 r1990 ≈ = − 0.1 0.1 1 + e−9.9 1 + e−10 ( ) P(0.1) − P(0) 1 3e0.1 3e0 r2000 ≈ = − 0.1 0.1 1 + e0.1 1 + e0
53. 53. Numerical evidence Solu on (Con nued) To approximate the popula on change in year n, use the diﬀerence P(t + ∆t) − P(t) quo ent , where ∆t = 0.1 and t = n − 2000. ∆t ( ) P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10 r1990 ≈ = − 0.1 0.1 1 + e−9.9 1 + e−10 = 0.000143229 ( ) P(0.1) − P(0) 1 3e0.1 3e0 r2000 ≈ = − 0.1 0.1 1 + e0.1 1 + e0
54. 54. Numerical evidence Solu on (Con nued) To approximate the popula on change in year n, use the diﬀerence P(t + ∆t) − P(t) quo ent , where ∆t = 0.1 and t = n − 2000. ∆t ( ) P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10 r1990 ≈ = − 0.1 0.1 1 + e−9.9 1 + e−10 = 0.000143229 ( ) P(0.1) − P(0) 1 3e0.1 3e0 r2000 ≈ = − 0.1 0.1 1 + e0.1 1 + e0 = 0.749376
55. 55. Solu on (Con nued) r2010
56. 56. Solu on (Con nued) P(10 + 0.1) − P(10) r2010 ≈ 0.1
57. 57. Solu on (Con nued) ( ) P(10 + 0.1) − P(10) 1 3e10.1 3e10 r2010 ≈ = − 0.1 0.1 1 + e10.1 1 + e10
58. 58. Solu on (Con nued) ( ) P(10 + 0.1) − P(10) 1 3e10.1 3e10 r2010 ≈ = − 0.1 0.1 1 + e10.1 1 + e10 = 0.0001296
59. 59. Population growth example Example Suppose the popula on of ﬁsh in the East River is given by the func on 3et P(t) = 1 + et where t is in years since 2000 and P is in millions of ﬁsh. Is the ﬁsh popula on growing fastest in 1990, 2000, or 2010? (Es mate numerically) Answer We es mate the rates of growth to be 0.000143229, 0.749376, and 0.0001296. So the popula on is growing fastest in 2000.
60. 60. Population growthBiological Rates of Change Problem Given the popula on func on of a group of organisms, ﬁnd the rate of growth of the popula on at a par cular instant. Solu on The instantaneous popula on growth is given by P(t + ∆t) − P(t) lim ∆t→0 ∆t
61. 61. Marginal costsRates of change in economics Problem Given the produc on cost of a good, ﬁnd the marginal cost of produc on a er having produced a certain quan ty.
62. 62. Marginal Cost Example Example Suppose the cost of producing q tons of rice on our paddy in a year is C(q) = q3 − 12q2 + 60q We are currently producing 5 tons a year. Should we change that?
63. 63. Marginal Cost Example Example Suppose the cost of producing q tons of rice on our paddy in a year is C(q) = q3 − 12q2 + 60q We are currently producing 5 tons a year. Should we change that? Answer If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should produce more to lower average costs.
64. 64. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) 4 5 6
65. 65. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) 4 112 5 6
66. 66. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) 4 112 5 125 6
67. 67. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) 4 112 5 125 6 144
68. 68. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q 4 112 5 125 6 144
69. 69. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q 4 112 28 5 125 6 144
70. 70. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q 4 112 28 5 125 25 6 144
71. 71. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q 4 112 28 5 125 25 6 144 24
72. 72. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q) 4 112 28 5 125 25 6 144 24
73. 73. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q) 4 112 28 13 5 125 25 6 144 24
74. 74. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q) 4 112 28 13 5 125 25 19 6 144 24
75. 75. Comparisons Solu on C(q) = q3 − 12q2 + 60q Fill in the table: q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q) 4 112 28 13 5 125 25 19 6 144 24 31
76. 76. Marginal Cost Example Example Suppose the cost of producing q tons of rice on our paddy in a year is C(q) = q3 − 12q2 + 60q We are currently producing 5 tons a year. Should we change that? Answer If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should produce more to lower average costs.
77. 77. Marginal costsRates of change in economics Problem Given the produc on cost of a good, ﬁnd the marginal cost of produc on a er having produced a certain quan ty. Solu on The marginal cost a er producing q is given by C(q + ∆q) − C(q) MC = lim ∆q→0 ∆q
78. 78. Outline Rates of Change Tangent Lines Velocity Popula on growth Marginal costs The deriva ve, deﬁned Deriva ves of (some) power func ons What does f tell you about f′ ? How can a func on fail to be diﬀeren able? Other nota ons The second deriva ve
79. 79. The deﬁnition All of these rates of change are found the same way!
80. 80. The deﬁnition All of these rates of change are found the same way! Deﬁni on Let f be a func on and a a point in the domain of f. If the limit f(a + h) − f(a) f(x) − f(a) f′ (a) = lim = lim h→0 h x→a x−a exists, the func on is said to be diﬀeren able at a and f′ (a) is the deriva ve of f at a.
81. 81. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a).
82. 82. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a). Solu on f(a + h) − f(a) f′ (a) = lim h→0 h
83. 83. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a). Solu on ′ f(a + h) − f(a) (a + h)2 − a2 f (a) = lim = lim h→0 h h→0 h
84. 84. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a). Solu on ′ f(a + h) − f(a) (a + h)2 − a2 f (a) = lim = lim h→0 h h→0 h (a + 2ah + h ) − a 2 2 2 = lim h→0 h
85. 85. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a). Solu on ′ f(a + h) − f(a) (a + h)2 − a2 f (a) = lim = lim h→0 h h→0 h (a + 2ah + h ) − a 2 2 2 2ah + h2 = lim = lim h→0 h h→0 h
86. 86. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a). Solu on ′ f(a + h) − f(a) (a + h)2 − a2 f (a) = lim = lim h→0 h h→0 h (a + 2ah + h ) − a 2 2 2 2ah + h2 = lim = lim h→0 h h→0 h = lim (2a + h) h→0
87. 87. Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (a). Solu on ′ f(a + h) − f(a) (a + h)2 − a2 f (a) = lim = lim h→0 h h→0 h (a + 2ah + h ) − a 2 2 2 2ah + h2 = lim = lim h→0 h h→0 h = lim (2a + h) = 2a h→0
88. 88. Derivative of the reciprocal Example 1 Suppose f(x) = . Use the deﬁni on of the deriva ve to ﬁnd f′ (2). x
89. 89. Derivative of the reciprocal Example 1 Suppose f(x) = . Use the deﬁni on of the deriva ve to ﬁnd f′ (2). x Solu on y 1/x − 1/2 f′ (2) = lim x→2 x−2 . x
90. 90. Derivative of the reciprocal Example 1 Suppose f(x) = . Use the deﬁni on of the deriva ve to ﬁnd f′ (2). x Solu on y 1/x − 1/2 2−x f′ (2) = lim = lim x→2 x−2 x→2 2x(x − 2) . x
91. 91. Derivative of the reciprocal Example 1 Suppose f(x) = . Use the deﬁni on of the deriva ve to ﬁnd f′ (2). x Solu on y 1/x − 1/2 2−x f′ (2) = lim = lim x→2 x−2 x→2 2x(x − 2) −1 = lim x→2 2x . x
92. 92. Derivative of the reciprocal Example 1 Suppose f(x) = . Use the deﬁni on of the deriva ve to ﬁnd f′ (2). x Solu on y 1/x − 1/2 2−x f′ (2) = lim = lim x→2 x−2 x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 . x
93. 93. “Can you do it the other way?”Same limit, diﬀerent form Solu on ′ f(2 + h) − f(2) 1 2+h− 1 2 f (2) = lim = lim h→0 h h→0 h
94. 94. “Can you do it the other way?”Same limit, diﬀerent form Solu on ′ f(2 + h) − f(2) 1 2+h− 1 2 f (2) = lim = lim h→0 h h→0 h 2 − (2 + h) = lim h→0 2h(2 + h)
95. 95. “Can you do it the other way?”Same limit, diﬀerent form Solu on 2+h − 2 1 1 ′ f(2 + h) − f(2) f (2) = lim = lim h→0 h h→0 h 2 − (2 + h) −h = lim = lim h→0 2h(2 + h) h→0 2h(2 + h)
96. 96. “Can you do it the other way?”Same limit, diﬀerent form Solu on 2+h − 2 1 1 ′ f(2 + h) − f(2) f (2) = lim = lim h→0 h h→0 h 2 − (2 + h) −h = lim = lim h→0 2h(2 + h) h→0 2h(2 + h) −1 = lim h→0 2(2 + h)
97. 97. “Can you do it the other way?”Same limit, diﬀerent form Solu on 2+h − 2 1 1 ′ f(2 + h) − f(2) f (2) = lim = lim h→0 h h→0 h 2 − (2 + h) −h = lim = lim h→0 2h(2 + h) h→0 2h(2 + h) −1 1 = lim =− h→0 2(2 + h) 4
98. 98. “How did you get that?”The Sure-Fire Sally Rule (SFSR) for adding fractions Fact a c ad ± bc ± = b d bd 1 1 2−x − x 2 = 2x = 2 − x x−2 x−2 2x(x − 2)
99. 99. “How did you get that?”The Sure-Fire Sally Rule (SFSR) for adding fractions Fact a c ad ± bc ± = b d bd 1 1 2−x − x 2 = 2x = 2 − x x−2 x−2 2x(x − 2) Paul Sally
100. 100. What does f tell you about f′? If f is a func on, we can compute the deriva ve f′ (x) at each point x where f is diﬀeren able, and come up with another func on, the deriva ve func on. What can we say about this func on f′ ?
101. 101. What does f tell you about f′? If f is a func on, we can compute the deriva ve f′ (x) at each point x where f is diﬀeren able, and come up with another func on, the deriva ve func on. What can we say about this func on f′ ? If f is decreasing on an interval, f′ is nega ve (technically, nonposi ve) on that interval
102. 102. Derivative of the reciprocal Example 1 Suppose f(x) = . Use the deﬁni on of the deriva ve to ﬁnd f′ (2). x Solu on y 1/x − 1/2 2−x f′ (2) = lim = lim x→2 x−2 x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 . x
103. 103. What does f tell you about f′? If f is a func on, we can compute the deriva ve f′ (x) at each point x where f is diﬀeren able, and come up with another func on, the deriva ve func on. What can we say about this func on f′ ? If f is decreasing on an interval, f′ is nega ve (technically, nonposi ve) on that interval If f is increasing on an interval, f′ is posi ve (technically, nonnega ve) on that interval
104. 104. Graphically and numerically y x2 − 22 x m= x−2 3 5 9 2.5 4.5 2.1 4.1 6.25 2.01 4.01 limit 4 4.414.0401 43.9601 3.61 1.99 3.99 2.25 1.9 3.9 1 1.5 3.5 . x 1 3 1 1.52.1 3 1.99 2.01 1.92.5 2
105. 105. What does f tell you about f′? Fact If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).
106. 106. What does f tell you about f′? Fact If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b). Picture Proof. If f is decreasing, then all secant lines point downward, hence have y nega ve slope. The deriva ve is a limit of slopes of secant lines, which are all nega ve, so the limit must be ≤ 0. . x
107. 107. What does f tell you about f′? Fact If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b). The Real Proof. If ∆x > 0, then f(x + ∆x) − f(x) f(x + ∆x) < f(x) =⇒ <0 ∆x
108. 108. What does f tell you about f′? Fact If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b). The Real Proof. If ∆x > 0, then f(x + ∆x) − f(x) f(x + ∆x) < f(x) =⇒ <0 ∆x If ∆x < 0, then x + ∆x < x, and f(x + ∆x) − f(x) f(x + ∆x) > f(x) =⇒ <0 ∆x
109. 109. What does f tell you about f′? Fact If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b). The Real Proof. f(x + ∆x) − f(x) Either way, < 0, so ∆x f(x + ∆x) − f(x) f′ (x) = lim ≤0 ∆x→0 ∆x
110. 110. Going the Other Way? Ques on If a func on has a nega ve deriva ve on an interval, must it be decreasing on that interval?
111. 111. Going the Other Way? Ques on If a func on has a nega ve deriva ve on an interval, must it be decreasing on that interval? Answer Maybe.
112. 112. Outline Rates of Change Tangent Lines Velocity Popula on growth Marginal costs The deriva ve, deﬁned Deriva ves of (some) power func ons What does f tell you about f′ ? How can a func on fail to be diﬀeren able? Other nota ons The second deriva ve
113. 113. Diﬀerentiability is super-continuity Theorem If f is diﬀeren able at a, then f is con nuous at a.
114. 114. Diﬀerentiability is super-continuity Theorem If f is diﬀeren able at a, then f is con nuous at a. Proof. We have f(x) − f(a) lim (f(x) − f(a)) = lim · (x − a) x→a x→a x−a f(x) − f(a) = lim · lim (x − a) x→a x−a x→a
115. 115. Diﬀerentiability is super-continuity Theorem If f is diﬀeren able at a, then f is con nuous at a. Proof. We have f(x) − f(a) lim (f(x) − f(a)) = lim · (x − a) x→a x→a x−a f(x) − f(a) = lim · lim (x − a) x→a x−a x→a ′ = f (a) · 0
116. 116. Diﬀerentiability is super-continuity Theorem If f is diﬀeren able at a, then f is con nuous at a. Proof. We have f(x) − f(a) lim (f(x) − f(a)) = lim · (x − a) x→a x→a x−a f(x) − f(a) = lim · lim (x − a) x→a x−a x→a ′ = f (a) · 0 = 0
117. 117. Diﬀerentiability FAILKinks Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) . x
118. 118. Diﬀerentiability FAILKinks Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
119. 119. Diﬀerentiability FAILKinks Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
120. 120. Diﬀerentiability FAILCusps Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) . x
121. 121. Diﬀerentiability FAILCusps Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
122. 122. Diﬀerentiability FAILCusps Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
123. 123. Diﬀerentiability FAILCusps Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
124. 124. Diﬀerentiability FAILVertical Tangents Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) . x
125. 125. Diﬀerentiability FAILVertical Tangents Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
126. 126. Diﬀerentiability FAILVertical Tangents Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
127. 127. Diﬀerentiability FAILVertical Tangents Example Let f have the graph on the le -hand side below. Sketch the graph of the deriva ve f′ . f(x) f′ (x) . x . x
128. 128. Diﬀerentiability FAILWeird, Wild, Stuﬀ Example f(x) . x This func on is diﬀeren able at 0.
129. 129. Diﬀerentiability FAILWeird, Wild, Stuﬀ Example f(x) f′ (x) . x . x This func on is diﬀeren able at 0.
130. 130. Diﬀerentiability FAILWeird, Wild, Stuﬀ Example f(x) f′ (x) . x . x This func on is diﬀeren able at 0.
131. 131. Diﬀerentiability FAILWeird, Wild, Stuﬀ Example f(x) f′ (x) . x . x This func on is diﬀeren able But the deriva ve is not at 0. con nuous at 0!
132. 132. Outline Rates of Change Tangent Lines Velocity Popula on growth Marginal costs The deriva ve, deﬁned Deriva ves of (some) power func ons What does f tell you about f′ ? How can a func on fail to be diﬀeren able? Other nota ons The second deriva ve
133. 133. Notation Newtonian nota on f′ (x) y′ (x) y′ Leibnizian nota on dy d df f(x) dx dx dx These all mean the same thing.
134. 134. Meet the MathematicianIsaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathema ca published 1687
135. 135. Meet the MathematicianGottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathema cian Contemporarily disgraced by the calculus priority dispute
136. 136. Outline Rates of Change Tangent Lines Velocity Popula on growth Marginal costs The deriva ve, deﬁned Deriva ves of (some) power func ons What does f tell you about f′ ? How can a func on fail to be diﬀeren able? Other nota ons The second deriva ve
137. 137. The second derivative If f is a func on, so is f′ , and we can seek its deriva ve. f′′ = (f′ )′ It measures the rate of change of the rate of change!
138. 138. The second derivative If f is a func on, so is f′ , and we can seek its deriva ve. f′′ = (f′ )′ It measures the rate of change of the rate of change! Leibnizian nota on: d2 y d2 d2 f f(x) dx2 dx2 dx2
139. 139. Function, derivative, second derivative y f(x) = x2 f′ (x) = 2x f′′ (x) = 2 . x
140. 140. SummaryWhat have we learned today? The deriva ve measures instantaneous rate of change The deriva ve has many interpreta ons: slope of the tangent line, velocity, marginal quan es, etc. The deriva ve reﬂects the monotonicity (increasing-ness or decreasing-ness) of the graph
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