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 V63.0121.027, Calculus I September 24, 2009 Announcements WebAssignments due Tuesday. Office Hours today 3-4. See Section Calendar for up-to-date OH. . . . . . .
  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 y . x m . . 4 . . . x . 2 . . . . . . .
  9. 9. Graphically and numerically y . x m 3 5 . . 9 . . . 4 . . . . x . 2 . 3 . . . . . . .
  10. 10. Graphically and numerically y . x m 3 5 2.5 4.25 . .25 . 6 . . . 4 . . . . x . 22 . . .5 . . . . . .
  11. 11. Graphically and numerically y . x m 3 5 2.5 4.25 2.1 4.1 . .41 . 4 . . . 4 . . .. x . .. .1 22 . . . . . .
  12. 12. Graphically and numerically y . x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 . .0401 . 4 4 . . . . x . 22 . ..01 . . . . . .
  13. 13. Graphically and numerically y . x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 . . 4 . 1 3 . . 1 . . . . x . 1 . 2 . . . . . . .
  14. 14. Graphically and numerically y . x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 . . 4 . 1.5 3.5 . .25 . 2 . 1 3 . . . x . 1 2 . .5 . . . . . . .
  15. 15. Graphically and numerically y . x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 . . 4 . 1.9 3.9 . .61 . 3 . 1.5 3.5 1 3 . .. x . 12 . ..9 . . . . . .
  16. 16. Graphically and numerically y . x m 3 5 2.5 4.25 2.1 4.1 2.01 4.01 1.99 3.99 . .9601 . 3 4 . . 1.9 3.9 1.5 3.5 1 3 . . x . 12 . ..99 . . . . . .
  17. 17. Graphically and numerically y . x m 3 5 . . 9 . 2.5 4.25 2.1 4.1 2.01 4.01 . .25 . 6 . limit 4 1.99 3.99 . .41 . 4 . . .0401 . 4 9601 . 3 . .61 3 4 . .. 1.9 3.9 1.5 3.5 . .25 . 2 . 1 3 . . 1 . . . . ... . . x . 1 1 . .2.1 3 . . .5 ..99 .5 . 12 . 2 9201 . . . . . .
  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 f(x) − f(a) mtangent = lim x→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 h(t) − h(1) t vave = t−1 2 − 15 . . . . . .
  21. 21. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 . . . . . .
  22. 22. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 . . . . . .
  23. 23. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 . . . . . .
  24. 24. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 . . . . . .
  25. 25. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 . . . . . .
  26. 26. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 . . . . . .
  27. 27. Numerical evidence h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 − 10.05 1.001 . . . . . .
  28. 28. Numerical evidence 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 . . . . . .
  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 (50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t) v = lim = lim = lim t→1 t−1 t→1 t − 1 t→1 t−1 = (−5) lim(1 + t) = −5 · 2 = −10 t→1 . . . . . .
  30. 30. y . = h (t ) Upshot . If the height function is given by h(t), the instantaneous velocity at time t0 is given by . h(t) − h(t0 ) v = lim t→t0 t − t0 h(t0 + ∆t) − h(t0 ) = lim ∆t→0 ∆t . . . t . ∆ t . t .0 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 3et P(t) = 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 ( ) ∆P 1 3et+∆t 3et lim = lim − ∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 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 ( ) ∆P 1 3et+∆t 3et lim = lim − ∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et Too hard! Try a small ∆t to approximate. . . . . . .
  35. 35. Numerical evidence P(−10 + 0.1) − P(−10) r1990 ≈ ≈ 0.1 . . . . . .
  36. 36. Numerical evidence P(−10 + 0.1) − P(−10) r1990 ≈ ≈ 0.000136 0.1 . . . . . .
  37. 37. Numerical evidence P(−10 + 0.1) − P(−10) r1990 ≈ ≈ 0.000136 0.1 P(0.1) − P(0) r2000 ≈ ≈ 0.1 . . . . . .
  38. 38. Numerical evidence P(−10 + 0.1) − P(−10) r1990 ≈ ≈ 0.000136 0.1 P(0.1) − P(0) r2000 ≈ ≈ 0.75 0.1 . . . . . .
  39. 39. Numerical evidence P(−10 + 0.1) − P(−10) r1990 ≈ ≈ 0.000136 0.1 P(0.1) − P(0) r2000 ≈ ≈ 0.75 0.1 P(10 + 0.1) − P(10) r2010 ≈ ≈ 0.1 . . . . . .
  40. 40. Numerical evidence P(−10 + 0.1) − P(−10) r1990 ≈ ≈ 0.000136 0.1 P(0.1) − P(0) r2000 ≈ ≈ 0.75 0.1 P(10 + 0.1) − P(10) r2010 ≈ ≈ 0.000136 0.1 . . . . . .
  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 3et P(t) = 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 P(t + ∆t) − P(t) lim ∆t→0 ∆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 C(q + ∆q) − C(q) MC = lim ∆q→0 ∆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 + h) − f(a) f′ (a) = lim h→0 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 + h) − f(a) (a + h)2 − a2 f′ (a) = lim = lim h→0 h h→0 h (a 2 + 2ah + h2 ) − a2 2ah + h2 = lim = lim h→0 h h→0 h = lim (2a + h) = 2a. h→0 . . . . . .
  65. 65. Derivative of the reciprocal function Example 1 Suppose f(x) = . Use the x definition of the derivative to find f′ (2). . . . . . .
  66. 66. Derivative of the reciprocal function Example 1 Suppose f(x) = . Use the x x . definition of the derivative to find f′ (2). Solution 1/x − 1/2 2−x . f′ (2) = lim = lim . x . x→2 x−2 x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 . . . . . .
  67. 67. The Sure-Fire Sally Rule (SFSR) for adding Fractions In anticipation of the question, “How did you get that?” a c ad ± bc ± = b d bd So 1 1 2−x − x 2 = 2x x−2 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 c ad ± bc ± = b d bd So 1 1 2−x − x 2 = 2x x−2 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 1 Suppose f(x) = . Use the x x . definition of the derivative to find f′ (2). Solution 1/x − 1/2 2−x . f′ (2) = lim = lim . x . x→2 x−2 x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 . . . . . .
  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 y . x m 3 5 . . 9 . 2.5 4.25 2.1 4.1 2.01 4.01 . .25 . 6 . limit 4 1.99 3.99 . .41 . 4 . . .0401 . 4 9601 . 3 . .61 3 4 . .. 1.9 3.9 1.5 3.5 . .25 . 2 . 1 3 . . 1 . . . . ... . . x . 1 1 . .2.1 3 . . .5 ..99 .5 . 12 . 2 9201 . . . . . .
  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) =⇒ <0 ∆x . . . . . .
  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) =⇒ <0 ∆x But if ∆x < 0, then x + ∆x < x, and f(x + ∆x) − f(x) f(x + ∆x) > f(x) =⇒ <0 ∆x 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) =⇒ <0 ∆x But if ∆x < 0, then x + ∆x < x, and f(x + ∆x) − f(x) f(x + ∆x) > f(x) =⇒ <0 ∆x f(x + ∆x) − f(x) still! Either way, < 0, so ∆x f(x + ∆x) − f(x) f′ (x) = lim ≤0 ∆x→0 ∆x . . . . . .
  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 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 . . . . . .
  80. 80. Differentiability is super-continuity Theorem If f is differentiable at a, then f is continuous 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 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 f .(x) . x . . . . . . .
  82. 82. How can a function fail to be differentiable? Kinks f .(x) . ′ (x ) f . x . . x . . . . . . .
  83. 83. How can a function fail to be differentiable? Kinks f .(x) . ′ (x ) f . . x . . x . . . . . . .
  84. 84. How can a function fail to be differentiable? Cusps f .(x) . x . . . . . . .
  85. 85. How can a function fail to be differentiable? Cusps f .(x) . ′ (x ) f . x . . x . . . . . . .
  86. 86. How can a function fail to be differentiable? Cusps f .(x) . ′ (x ) f . x . . x . . . . . . .
  87. 87. How can a function fail to be differentiable? Vertical Tangents f .(x) . x . . . . . . .
  88. 88. How can a function fail to be differentiable? Vertical Tangents f .(x) . ′ (x ) f . x . . x . . . . . . .
  89. 89. How can a function fail to be differentiable? Vertical Tangents f .(x) . ′ (x ) f . x . . x . . . . . . .
  90. 90. How can a function fail to be differentiable? Weird, Wild, Stuff f .(x) . x . This function is differentiable at 0. . . . . . .
  91. 91. How can a function fail to be differentiable? Weird, Wild, Stuff f .(x) . ′ (x ) f . x . . x . This function is differentiable But the derivative is not at 0. 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 d df f(x) dx dx 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 d2 d2 f f(x) dx2 dx2 dx2 . . . . . .
  99. 99. function, derivative, second derivative y . . (x ) = x 2 f .′ (x) = 2x f .′′ (x) = 2 f . x . . . . . . .

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