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# Lesson 7: The Derivative

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The derivative is a major tool for investigating the behavior of a function. Since functions are ubiquitous, so are their derivatives. Velocity, growth rates, marginal costs, and material strain are all examples of derivatives. We motivate and define the derivative and compute a few examples, then discuss how features of a function are manifested in its derivative.

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### Lesson 7: The Derivative

1. 1. . . . . . . Section 2.1 The Derivative and Rates of Change V63.0121.034, Calculus I September 23, 2009 Announcements WebAssignments due Monday. Email me if you need an extension for Yom Kippur.
2. 2. . . . . . . Regarding WebAssign We feel your pain
3. 3. . . . . . . Explanations From the syllabus: Graders will be expecting you to express your ideas clearly, legibly, and completely, often requiring complete English sentences rather than merely just a long string of equations or unconnected mathematical expressions. This means you could lose points for unexplained answers.
4. 4. . . . . . . Rubric Points Description of Work 3 Work is completely accurate and essentially perfect. Work is thoroughly developed, neat, and easy to read. Complete sentences are used. 2 Work is good, but incompletely developed, hard to read, unexplained, or jumbled. Answers which are not explained, even if correct, will generally receive 2 points. Work contains “right idea” but is ﬂawed. 1 Work is sketchy. There is some correct work, but most of work is incorrect. 0 Work minimal or non-existent. Solution is completely incorrect.
5. 5. . . . . . . Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, deﬁned Derivatives of (some) power functions What does f tell you about f′ ? How can a function fail to be differentiable? Other notations The second derivative
6. 6. . . . . . . The tangent problem 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. . . . . . . The tangent problem Problem Given a curve and a point on the curve, ﬁnd the slope of the line tangent to the curve at that point. Example Find the slope of the line tangent to the curve y = x2 at the point (2, 4).
8. 8. . . . . . . Graphically and numerically . .x .y . .2 ..4 . x m
9. 9. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .3 ..9 x m 3 5
10. 10. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .2.5 ..6.25 x m 3 5 2.5 4.25
11. 11. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .2.1 ..4.41 x m 3 5 2.5 4.25 2.1 4.1
12. 12. . . . . . . Graphically and numerically . .x .y . .2 ..4 .. . .2.01 ..4.0401 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01
13. 13. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .1 ..1 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 1 3
14. 14. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .1.5 ..2.25 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 1.5 3.5 1 3
15. 15. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .1.9 ..3.61 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 1.9 3.9 1.5 3.5 1 3
16. 16. . . . . . . Graphically and numerically . .x .y . .2 ..4 .. . .1.99 ..3.9601 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 1.99 3.99 1.9 3.9 1.5 3.5 1 3
17. 17. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .3 ..9 . . .2.5 ..6.25 . . .2.1 ..4.41 . . .2.01 ..4.0401 . . .1 ..1 . . .1.5 ..2.25 . . .1.9 ..3.61 . . .1.99 ..3.9601 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 limit 4 1.99 3.99 1.9 3.9 1.5 3.5 1 3
18. 18. . . . . . . The tangent problem Problem Given a curve and a point on the curve, ﬁnd the slope of the line tangent to the curve at that point. Example Find the slope of the line tangent to the curve y = x2 at the point (2, 4). Upshot 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 mtangent = lim x→a f(x) − f(a) x − a
19. 19. . . . . . . Velocity Problem Given the position function of a moving object, ﬁnd the velocity of the object at a certain instant in time. Example Drop a ball off the roof of the Silver Center so that its height can be described by h(t) = 50 − 5t2 where t is seconds after dropping it and h is meters above the ground. How fast is it falling one second after we drop it?
20. 20. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15
21. 21. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5
22. 22. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5
23. 23. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5 1.1
24. 24. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5 1.1 − 10.5
25. 25. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01
26. 26. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05
27. 27. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001
28. 28. . . . . . . Numerical evidence t vave = h(t) − h(1) t − 1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001 − 10.005
29. 29. . . . . . . Velocity Problem Given the position function of a moving object, ﬁnd the velocity of the object at a certain instant in time. Example Drop a ball off the roof of the Silver Center so that its height can be described by h(t) = 50 − 5t2 where t is seconds after dropping it and h is meters above the ground. How fast is it falling one second after we drop it? Solution The answer is v = lim t→1 (50 − 5t2) − 45 t − 1 = lim t→1 5 − 5t2 t − 1 = lim t→1 5(1 − t)(1 + t) t − 1 = (−5) lim t→1 (1 + t) = −5 · 2 = −10
30. 30. . . . . . . Upshot If the height function is given by h(t), the instantaneous velocity at time t0 is given by v = lim t→t0 h(t) − h(t0) t − t0 = lim ∆t→0 h(t0 + ∆t) − h(t0) ∆t . .t .y = h(t) . . . .t0 . .t .∆t
31. 31. . . . . . . Population growth Problem Given the population function of a group of organisms, ﬁnd the rate of growth of the population at a particular instant.
32. 32. . . . . . . Population growth Problem Given the population function of a group of organisms, ﬁnd the rate of growth of the population at a particular instant. Example Suppose the population of ﬁsh in the East River is given by the function P(t) = 3et 1 + et where t is in years since 2000 and P is in millions of ﬁsh. Is the ﬁsh population growing fastest in 1990, 2000, or 2010? (Estimate numerically)?
33. 33. . . . . . . Derivation Let ∆t be an increment in time and ∆P the corresponding change in population: ∆P = P(t + ∆t) − P(t) This depends on ∆t, so we want lim ∆t→0 ∆P ∆t = lim ∆t→0 1 ∆t ( 3et+∆t 1 + et+∆t − 3et 1 + et )
34. 34. . . . . . . Derivation Let ∆t be an increment in time and ∆P the corresponding change in population: ∆P = P(t + ∆t) − P(t) This depends on ∆t, so we want lim ∆t→0 ∆P ∆t = lim ∆t→0 1 ∆t ( 3et+∆t 1 + et+∆t − 3et 1 + et ) Too hard! Try a small ∆t to approximate.
35. 35. . . . . . . Numerical evidence r1990 ≈ P(−10 + 0.1) − P(−10) 0.1 ≈
36. 36. . . . . . . Numerical evidence r1990 ≈ P(−10 + 0.1) − P(−10) 0.1 ≈ 0.000136
37. 37. . . . . . . Numerical evidence r1990 ≈ P(−10 + 0.1) − P(−10) 0.1 ≈ 0.000136 r2000 ≈ P(0.1) − P(0) 0.1 ≈
38. 38. . . . . . . Numerical evidence r1990 ≈ P(−10 + 0.1) − P(−10) 0.1 ≈ 0.000136 r2000 ≈ P(0.1) − P(0) 0.1 ≈ 0.75
39. 39. . . . . . . Numerical evidence r1990 ≈ P(−10 + 0.1) − P(−10) 0.1 ≈ 0.000136 r2000 ≈ P(0.1) − P(0) 0.1 ≈ 0.75 r2010 ≈ P(10 + 0.1) − P(10) 0.1 ≈
40. 40. . . . . . . Numerical evidence r1990 ≈ P(−10 + 0.1) − P(−10) 0.1 ≈ 0.000136 r2000 ≈ P(0.1) − P(0) 0.1 ≈ 0.75 r2010 ≈ P(10 + 0.1) − P(10) 0.1 ≈ 0.000136
41. 41. . . . . . . Population growth Problem Given the population function of a group of organisms, ﬁnd the rate of growth of the population at a particular instant. Example Suppose the population of ﬁsh in the East River is given by the function P(t) = 3et 1 + et where t is in years since 2000 and P is in millions of ﬁsh. Is the ﬁsh population growing fastest in 1990, 2000, or 2010? (Estimate numerically)? Solution The estimated rates of growth are 0.000136, 0.75, and 0.000136.
42. 42. . . . . . . Upshot The instantaneous population growth is given by lim ∆t→0 P(t + ∆t) − P(t) ∆t
43. 43. . . . . . . Marginal costs Problem Given the production cost of a good, ﬁnd the marginal cost of production after having produced a certain quantity.
44. 44. . . . . . . Marginal costs Problem Given the production cost of a good, ﬁnd the marginal cost of production after having produced a certain quantity. 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?
45. 45. . . . . . . Comparisons q C(q) 4 5 6
46. 46. . . . . . . Comparisons q C(q) 4 112 5 6
47. 47. . . . . . . Comparisons q C(q) 4 112 5 125 6
48. 48. . . . . . . Comparisons q C(q) 4 112 5 125 6 144
49. 49. . . . . . . Comparisons q C(q) AC(q) = C(q)/q 4 112 5 125 6 144
50. 50. . . . . . . Comparisons q C(q) AC(q) = C(q)/q 4 112 28 5 125 6 144
51. 51. . . . . . . Comparisons q C(q) AC(q) = C(q)/q 4 112 28 5 125 25 6 144
52. 52. . . . . . . Comparisons q C(q) AC(q) = C(q)/q 4 112 28 5 125 25 6 144 24
53. 53. . . . . . . Comparisons q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q) 4 112 28 5 125 25 6 144 24
54. 54. . . . . . . Comparisons q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q) 4 112 28 13 5 125 25 6 144 24
55. 55. . . . . . . Comparisons 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
56. 56. . . . . . . Comparisons 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
57. 57. . . . . . . Marginal costs Problem Given the production cost of a good, ﬁnd the marginal cost of production after having produced a certain quantity. 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? Example If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should produce more to lower average costs.
58. 58. . . . . . . Upshot The incremental cost ∆C = C(q + 1) − C(q) is useful, but depends on units.
59. 59. . . . . . . Upshot The incremental cost ∆C = C(q + 1) − C(q) is useful, but depends on units. The marginal cost after producing q given by MC = lim ∆q→0 C(q + ∆q) − C(q) ∆q is more useful since it’s unit-independent.
60. 60. . . . . . . Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, deﬁned Derivatives of (some) power functions What does f tell you about f′ ? How can a function fail to be differentiable? Other notations The second derivative
61. 61. . . . . . . The deﬁnition All of these rates of change are found the same way!
62. 62. . . . . . . The deﬁnition All of these rates of change are found the same way! Deﬁnition Let f be a function and a a point in the domain of f. If the limit f′ (a) = lim h→0 f(a + h) − f(a) h exists, the function is said to be differentiable at a and f′ (a) is the derivative of f at a.
63. 63. . . . . . . Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁnition of derivative to ﬁnd f′ (a).
64. 64. . . . . . . Derivative of the squaring function Example Suppose f(x) = x2 . Use the deﬁnition of derivative to ﬁnd f′ (a). Solution f′ (a) = lim h→0 f(a + h) − f(a) h = lim h→0 (a + h)2 − a2 h = lim h→0 (a2 + 2ah + h2 ) − a2 h = lim h→0 2ah + h2 h = lim h→0 (2a + h) = 2a.
65. 65. . . . . . . Derivative of the reciprocal function Example Suppose f(x) = 1 x . Use the deﬁnition of the derivative to ﬁnd f′ (2).
66. 66. . . . . . . Derivative of the reciprocal function Example Suppose f(x) = 1 x . Use the deﬁnition of the derivative to ﬁnd f′ (2). Solution f′ (2) = lim x→2 1/x − 1/2 x − 2 = lim x→2 2 − x 2x(x − 2) = lim x→2 −1 2x = − 1 4 . .x .x .
67. 67. . . . . . . The Sure-Fire Sally Rule (SFSR) for adding Fractions In anticipation of the question, “How did you get that?” a b ± c d = ad ± bc bd So 1 x − 1 2 x − 2 = 2 − x 2x x − 2 = 2 − x 2x(x − 2)
68. 68. . . . . . . The Sure-Fire Sally Rule (SFSR) for adding Fractions In anticipation of the question, “How did you get that?” a b ± c d = ad ± bc bd So 1 x − 1 2 x − 2 = 2 − x 2x x − 2 = 2 − x 2x(x − 2)
69. 69. . . . . . . What does f tell you about f′ ? If f is a function, we can compute the derivative f′ (x) at each point x where f is differentiable, and come up with another function, the derivative function. What can we say about this function f′ ?
70. 70. . . . . . . What does f tell you about f′ ? If f is a function, we can compute the derivative f′ (x) at each point x where f is differentiable, and come up with another function, the derivative function. What can we say about this function f′ ? If f is decreasing on an interval, f′ is negative (well, nonpositive) on that interval
71. 71. . . . . . . Derivative of the reciprocal function Example Suppose f(x) = 1 x . Use the deﬁnition of the derivative to ﬁnd f′ (2). Solution f′ (2) = lim x→2 1/x − 1/2 x − 2 = lim x→2 2 − x 2x(x − 2) = lim x→2 −1 2x = − 1 4 . .x .x .
72. 72. . . . . . . What does f tell you about f′ ? If f is a function, we can compute the derivative f′ (x) at each point x where f is differentiable, and come up with another function, the derivative function. What can we say about this function f′ ? If f is decreasing on an interval, f′ is negative (well, nonpositive) on that interval If f is increasing on an interval, f′ is positive (well, nonnegative) on that interval
73. 73. . . . . . . Graphically and numerically . .x .y . .2 ..4 . . . .3 ..9 . . .2.5 ..6.25 . . .2.1 ..4.41 . . .2.01 ..4.0401 . . .1 ..1 . . .1.5 ..2.25 . . .1.9 ..3.61 . . .1.99 ..3.9601 x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 limit 4 1.99 3.99 1.9 3.9 1.5 3.5 1 3
74. 74. . . . . . . What does f tell you about f′ ? Fact If f is decreasing on (a, b), then f′ ≤ 0 on (a, b). Proof. If f is decreasing on (a, b), and ∆x > 0, then f(x + ∆x) < f(x) =⇒ f(x + ∆x) − f(x) ∆x < 0
75. 75. . . . . . . What does f tell you about f′ ? Fact If f is decreasing on (a, b), then f′ ≤ 0 on (a, b). Proof. If f is decreasing on (a, b), and ∆x > 0, then f(x + ∆x) < f(x) =⇒ f(x + ∆x) − f(x) ∆x < 0 But if ∆x < 0, then x + ∆x < x, and f(x + ∆x) > f(x) =⇒ f(x + ∆x) − f(x) ∆x < 0 still!
76. 76. . . . . . . What does f tell you about f′ ? Fact If f is decreasing on (a, b), then f′ ≤ 0 on (a, b). Proof. If f is decreasing on (a, b), and ∆x > 0, then f(x + ∆x) < f(x) =⇒ f(x + ∆x) − f(x) ∆x < 0 But if ∆x < 0, then x + ∆x < x, and f(x + ∆x) > f(x) =⇒ f(x + ∆x) − f(x) ∆x < 0 still! Either way, f(x + ∆x) − f(x) ∆x < 0, so f′ (x) = lim ∆x→0 f(x + ∆x) − f(x) ∆x ≤ 0
77. 77. . . . . . . Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, deﬁned Derivatives of (some) power functions What does f tell you about f′ ? How can a function fail to be differentiable? Other notations The second derivative
78. 78. . . . . . . Differentiability is super-continuity Theorem If f is differentiable at a, then f is continuous at a.
79. 79. . . . . . . Differentiability is super-continuity Theorem If f is differentiable at a, then f is continuous at a. Proof. We have lim x→a (f(x) − f(a)) = lim x→a f(x) − f(a) x − a · (x − a) = lim x→a f(x) − f(a) x − a · lim x→a (x − a) = f′ (a) · 0 = 0
80. 80. . . . . . . Differentiability is super-continuity Theorem If f is differentiable at a, then f is continuous at a. Proof. We have lim x→a (f(x) − f(a)) = lim x→a f(x) − f(a) x − a · (x − a) = lim x→a f(x) − f(a) x − a · lim x→a (x − a) = f′ (a) · 0 = 0 Note the proper use of the limit law: if the factors each have a limit at a, the limit of the product is the product of the limits.
81. 81. . . . . . . How can a function fail to be differentiable? Kinks . .x .f(x)
82. 82. . . . . . . How can a function fail to be differentiable? Kinks . .x .f(x) . .x .f′ (x) .
83. 83. . . . . . . How can a function fail to be differentiable? Kinks . .x .f(x) . .x .f′ (x) . .
84. 84. . . . . . . How can a function fail to be differentiable? Cusps . .x .f(x)
85. 85. . . . . . . How can a function fail to be differentiable? Cusps . .x .f(x) . .x .f′ (x)
86. 86. . . . . . . How can a function fail to be differentiable? Cusps . .x .f(x) . .x .f′ (x)
87. 87. . . . . . . How can a function fail to be differentiable? Vertical Tangents . .x .f(x)
88. 88. . . . . . . How can a function fail to be differentiable? Vertical Tangents . .x .f(x) . .x .f′ (x)
89. 89. . . . . . . How can a function fail to be differentiable? Vertical Tangents . .x .f(x) . .x .f′ (x)
90. 90. . . . . . . How can a function fail to be differentiable? Weird, Wild, Stuff . .x .f(x) This function is differentiable at 0.
91. 91. . . . . . . How can a function fail to be differentiable? Weird, Wild, Stuff . .x .f(x) This function is differentiable at 0. . .x .f′ (x) But the derivative is not continuous at 0!
92. 92. . . . . . . Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, deﬁned Derivatives of (some) power functions What does f tell you about f′ ? How can a function fail to be differentiable? Other notations The second derivative
93. 93. . . . . . . Notation Newtonian notation f′ (x) y′ (x) y′ Leibnizian notation dy dx d dx f(x) df dx These all mean the same thing.
94. 94. . . . . . . Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687
95. 95. . . . . . . Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute
96. 96. . . . . . . Outline Rates of Change Tangent Lines Velocity Population growth Marginal costs The derivative, deﬁned Derivatives of (some) power functions What does f tell you about f′ ? How can a function fail to be differentiable? Other notations The second derivative
97. 97. . . . . . . The second derivative If f is a function, so is f′ , and we can seek its derivative. f′′ = (f′ )′ It measures the rate of change of the rate of change!
98. 98. . . . . . . The second derivative If f is a function, so is f′ , and we can seek its derivative. f′′ = (f′ )′ It measures the rate of change of the rate of change! Leibnizian notation: d2 y dx2 d2 dx2 f(x) d2 f dx2
99. 99. . . . . . . function, derivative, second derivative . .x .y .f(x) = x2 .f′ (x) = 2x .f′′ (x) = 2