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# Lesson 7: The Derivative (Section 21 slide)

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The derivative measure the instantaneous rate of change of a function.

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### Lesson 7: The Derivative (Section 21 slide)

1. 1. Section 2.1–2.2 The Derivative and Rates of Change The Derivative as a Function V63.0121.021, Calculus I New York University September 28, 2010Announcements Quiz this week in recitation on §§1.1–1.4 Get-to-know-you/photo due Friday October 1 . . . . . .
2. 2. Announcements Quiz this week in recitation on §§1.1–1.4 Get-to-know-you/photo due Friday October 1 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 2 / 49
3. 3. Format of written workPlease: Use scratch paper and copy your final work onto fresh paper. Use loose-leaf paper (not torn from a notebook). Write your name, lecture section, assignment number, recitation, and date at the top. Staple your homework together.See the website for more information. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 3 / 49
4. 4. Objectives for Section 2.1 Understand and state the definition of the derivative of a function at a point. Given a function and a point in its domain, decide if the function is differentiable at the point and find the value of the derivative at that point. Understand and give several examples of derivatives modeling rates of change in science. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 4 / 49
5. 5. Objectives for Section 2.2 Given a function f, use the definition of the derivative to find the derivative function f’. Given a function, find its second derivative. Given the graph of a function, sketch the graph of its derivative. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 5 / 49
6. 6. OutlineRates of Change Tangent Lines Velocity Population growth Marginal costsThe derivative, defined Derivatives of (some) power functions What does f tell you about f′ ?How can a function fail to be differentiable?Other notationsThe second derivative . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 6 / 49
7. 7. The tangent problemProblemGiven a curve and a point on the curve, find the slope of the linetangent to the curve at that point. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 7 / 49
8. 8. The tangent problemProblemGiven a curve and a point on the curve, find the slope of the linetangent to the curve at that point.ExampleFind the slope of the line tangent to the curve y = x2 at the point (2, 4). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 7 / 49
9. 9. Graphically and numerically y . x2 − 22 x m= x−2 . . 4 . . . x . 2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
10. 10. Graphically and numerically y . x2 − 22 x m= x−2 3 . . 9 . . . 4 . . . . x . 2 . 3 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
11. 11. Graphically and numerically y . x2 − 22 x m= x−2 3 5 . . 9 . . . 4 . . . . x . 2 . 3 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
12. 12. Graphically and numerically y . x2 − 22 x m= x−2 3 5 2.5 . .25 . 6 . . . 4 . . . . x . 22 . . .5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
13. 13. Graphically and numerically y . x2 − 22 x m= x−2 3 5 2.5 4.5 . .25 . 6 . . . 4 . . . . x . 22 . . .5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
14. 14. Graphically and numerically y . x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 . .41 . 4 . . . 4 . . .. x . 2 .. .1 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
15. 15. Graphically and numerically y . x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 . .41 . 4 . . . 4 . . .. x . 2 .. .1 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
16. 16. Graphically and numerically y . x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01. .0401 .4 4 . . . . x . 2. . .01 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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. .0401 .4 4 . . . . x . 2. . .01 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 . . 1 . . . . x . 1 . 2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 . 1 3 . . 1 . . . . x . 1 . 2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 . 1.5 . .25 . 2 . 1 3 . . . x . 1 2 . .5 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 . 1.5 3.5 . .25 . 2 . 1 3 . . . x . 1 2 . .5 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 . . .61 . 3 . 1.9 1.5 3.5 1 3 . .. x . 1. . .9 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 . . 4 . . .61 . 3 . 1.9 3.9 1.5 3.5 1 3 . .. x . 1. . .9 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 1.99. .9601 .3 4 . . 1.9 3.9 1.5 3.5 1 3 . . x . 1. . .99 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
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 1.99 3.99. .9601 .3 4 . . 1.9 3.9 1.5 3.5 1 3 . . x . 1. . .99 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
26. 26. Graphically and numerically y . x2 − 22 x m= x−2 3 5 2.5 4.5 2.1 4.1 2.01 4.01 1.99 3.99 . . 4 . 1.9 3.9 1.5 3.5 1 3 . . x . 2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
27. 27. Graphically and numerically y . x2 − 22 x m= x−2 3 5 . . 9 . 2.5 4.5 2.1 4.1 2.01 4.01 . .25 . 6 . limit . .41 . 4 . 1.99 3.99. .0401 .4.9601 .3 . .61 3 4 . .. 1.9 3.9 1.5 3.5 . .25 . 2 . 1 3 . . 1 . . . . ... . . x . 1 1 . .. .1 3 . . .5 .99 .5 . 12 . 2.9 2 2 .01 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
28. 28. Graphically and numerically y . x2 − 22 x m= x−2 3 5 . . 9 . 2.5 4.5 2.1 4.1 2.01 4.01 . .25 . 6 . limit 4 . .41 . 4 . 1.99 3.99. .0401 .4.9601 .3 . .61 3 4 . .. 1.9 3.9 1.5 3.5 . .25 . 2 . 1 3 . . 1 . . . . ... . . x . 1 1 . .. .1 3 . . .5 .99 .5 . 12 . 2.9 2 2 .01 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
29. 29. The tangent problemProblemGiven a curve and a point on the curve, find the slope of the linetangent to the curve at that point.ExampleFind the slope of the line tangent to the curve y = x2 at the point (2, 4).UpshotIf 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 9 / 49
30. 30. Velocity.ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.ExampleDrop a ball off the roof of the Silver Center so that its height can be describedby h(t) = 50 − 5t2where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 10 / 49
31. 31. Numerical evidence h(t) = 50 − 5t2Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
32. 32. Numerical evidence h(t) = 50 − 5t2Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
33. 33. Numerical evidence h(t) = 50 − 5t2Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
34. 34. Numerical evidence h(t) = 50 − 5t2Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
35. 35. Numerical evidence h(t) = 50 − 5t2Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
36. 36. Numerical evidence h(t) = 50 − 5t2Fill in the table: h(t) − h(1) t vave = t−1 2 − 15 1.5 − 12.5 1.1 − 10.5 1.01 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
37. 37. Numerical evidence h(t) = 50 − 5t2Fill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
38. 38. Numerical evidence h(t) = 50 − 5t2Fill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
39. 39. Numerical evidence h(t) = 50 − 5t2Fill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
40. 40. Velocity.ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.ExampleDrop a ball off the roof of the Silver Center so that its height can be describedby h(t) = 50 − 5t2where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?SolutionThe answer is (50 − 5t2 ) − 45 v = lim t→1 t−1. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
41. 41. Velocity.ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.ExampleDrop a ball off the roof of the Silver Center so that its height can be describedby h(t) = 50 − 5t2where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?SolutionThe answer is (50 − 5t2 ) − 45 5 − 5t2 v = lim = lim t→1 t−1 t→1 t − 1. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
42. 42. Velocity.ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.ExampleDrop a ball off the roof of the Silver Center so that its height can be describedby h(t) = 50 − 5t2where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?SolutionThe 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
43. 43. Velocity.ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.ExampleDrop a ball off the roof of the Silver Center so that its height can be describedby h(t) = 50 − 5t2where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?SolutionThe 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) t→1. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
44. 44. Velocity.ProblemGiven the position function of a moving object, find the velocity of the object ata certain instant in time.ExampleDrop a ball off the roof of the Silver Center so that its height can be describedby h(t) = 50 − 5t2where t is seconds after dropping it and h is meters above the ground. Howfast is it falling one second after we drop it?SolutionThe 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
45. 45. Velocity in general . . = h(t) yUpshot . (t0 ) . h .If the height function is given byh(t), the instantaneous velocity . h ∆at time t0 is given by . (t0 + ∆t) . h . h(t) − h(t0 ) v = lim t→t0 t − t0 h(t0 + ∆t) − h(t0 ) = lim ∆t→0 ∆t . . . t . ∆ .. t t .0 t . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 13 / 49
46. 46. Population growthProblemGiven the population function of a group of organisms, find the rate ofgrowth of the population at a particular instant. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 14 / 49
47. 47. Population growthProblemGiven the population function of a group of organisms, find the rate ofgrowth of the population at a particular instant.ExampleSuppose the population of fish in the East River is given by the function 3et P(t) = 1 + etwhere t is in years since 2000 and P is in millions of fish. Is the fishpopulation growing fastest in 1990, 2000, or 2010? (Estimatenumerically) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 14 / 49
48. 48. DerivationSolutionLet ∆t be an increment in time and ∆P the corresponding change inpopulation: ∆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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 15 / 49
49. 49. DerivationSolutionLet ∆t be an increment in time and ∆P the corresponding change inpopulation: ∆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 + etBut rather than compute a complicated limit analytically, let usapproximate numerically. We will try a small ∆t, for instance 0.1. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 15 / 49
50. 50. Numerical evidenceSolution (Continued)To approximate the population change in year n, use the difference P(t + ∆t) − P(t)quotient , where ∆t = 0.1 and t = n − 2000. ∆t r1990 r2000 r2010 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
51. 51. Numerical evidenceSolution (Continued)To approximate the population change in year n, use the difference P(t + ∆t) − P(t)quotient , 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 P(10 + 0.1) − P(10) r2010 ≈ 0.1 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
52. 52. Numerical evidenceSolution (Continued)To approximate the population change in year n, use the difference P(t + ∆t) − P(t)quotient , 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 ( ) P(10 + 0.1) − P(10) 1 3e10.1 3e10 r2010 ≈ = − 0.1 0.1 1 + e10.1 1 + e10 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
53. 53. Numerical evidenceSolution (Continued)To approximate the population change in year n, use the difference P(t + ∆t) − P(t)quotient , 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 ( ) P(10 + 0.1) − P(10) 1 3e10.1 3e10 r2010 ≈ = − 0.1 0.1 1 + e10.1 1 + e10 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
54. 54. Numerical evidenceSolution (Continued)To approximate the population change in year n, use the difference P(t + ∆t) − P(t)quotient , 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 ( ) P(10 + 0.1) − P(10) 1 3e10.1 3e10 r2010 ≈ = − 0.1 0.1 1 + e10.1 1 + e10 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
55. 55. Numerical evidenceSolution (Continued)To approximate the population change in year n, use the difference P(t + ∆t) − P(t)quotient , 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 ( ) P(10 + 0.1) − P(10) 1 3e10.1 3e10 r2010 ≈ = − 0.1 0.1 1 + e10.1 1 + e10 = 0.0001296 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
56. 56. Population growth.ProblemGiven the population function of a group of organisms, find the rate of growthof the population at a particular instant.ExampleSuppose the population of fish in the East River is given by the function 3et P(t) = 1 + etwhere t is in years since 2000 and P is in millions of fish. Is the fish populationgrowing fastest in 1990, 2000, or 2010? (Estimate numerically)AnswerWe estimate the rates of growth to be 0.000143229, 0.749376, and 0.0001296.So the population is growing fastest in 2000.. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 17 / 49
57. 57. Population growth in generalUpshotThe instantaneous population growth is given by P(t + ∆t) − P(t) lim ∆t→0 ∆t . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 18 / 49
58. 58. Marginal costsProblemGiven the production cost of a good, find the marginal cost ofproduction after having produced a certain quantity. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 19 / 49
59. 59. Marginal costsProblemGiven the production cost of a good, find the marginal cost ofproduction after having produced a certain quantity.ExampleSuppose the cost of producing q tons of rice on our paddy in a year is C(q) = q3 − 12q2 + 60qWe are currently producing 5 tons a year. Should we change that? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 19 / 49
60. 60. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) 4 5 6 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
61. 61. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) 4 112 5 6 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
62. 62. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) 4 112 5 125 6 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
63. 63. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) 4 112 5 125 6 144 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
64. 64. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) AC(q) = C(q)/q 4 112 5 125 6 144 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
65. 65. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) AC(q) = C(q)/q 4 112 28 5 125 6 144 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
66. 66. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) AC(q) = C(q)/q 4 112 28 5 125 25 6 144 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
67. 67. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill in the table: q C(q) AC(q) = C(q)/q 4 112 28 5 125 25 6 144 24 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
68. 68. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
69. 69. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
70. 70. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
71. 71. ComparisonsSolution C(q) = q3 − 12q2 + 60qFill 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
72. 72. Marginal costsProblemGiven the production cost of a good, find the marginal cost ofproduction after having produced a certain quantity.ExampleSuppose the cost of producing q tons of rice on our paddy in a year is C(q) = q3 − 12q2 + 60qWe are currently producing 5 tons a year. Should we change that?AnswerIf q = 5, then C = 125, ∆C = 19, while AC = 25. So we shouldproduce more to lower average costs. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 21 / 49
73. 73. Marginal Cost in GeneralUpshot The incremental cost ∆C = C(q + 1) − C(q) is useful, but is still only an average rate of change. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 22 / 49
74. 74. Marginal Cost in GeneralUpshot The incremental cost ∆C = C(q + 1) − C(q) is useful, but is still only an average rate of change. The marginal cost after producing q given by C(q + ∆q) − C(q) MC = lim ∆q→0 ∆q is more useful since it’s an instantaneous rate of change. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 22 / 49
75. 75. OutlineRates of Change Tangent Lines Velocity Population growth Marginal costsThe derivative, defined Derivatives of (some) power functions What does f tell you about f′ ?How can a function fail to be differentiable?Other notationsThe second derivative . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 23 / 49
76. 76. The definitionAll of these rates of change are found the same way! . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 24 / 49
77. 77. The definitionAll of these rates of change are found the same way!DefinitionLet f be a function 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−aexists, the function is said to be differentiable at a and f′ (a) is thederivative of f at a. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 24 / 49
78. 78. Derivative of the squaring functionExampleSuppose f(x) = x2 . Use the definition of derivative to find f′ (a). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
79. 79. Derivative of the squaring functionExampleSuppose f(x) = x2 . Use the definition of derivative to find f′ (a).Solution f(a + h) − f(a) f′ (a) = lim h→0 h . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
80. 80. Derivative of the squaring functionExampleSuppose 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
81. 81. Derivative of the squaring functionExampleSuppose 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 (a2 + 2ah + h2 ) − a2 = lim h→0 h . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
82. 82. Derivative of the squaring functionExampleSuppose 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 (a2 + 2ah + h2 ) − a2 2ah + h2 = lim = lim h→0 h h→0 h . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
83. 83. Derivative of the squaring functionExampleSuppose 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 (a2 + 2ah + h2 ) − a2 2ah + h2 = lim = lim h→0 h h→0 h = lim (2a + h) h→0 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
84. 84. Derivative of the squaring functionExampleSuppose 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 (a2 + 2ah + h2 ) − a2 2ah + h2 = lim = lim h→0 h h→0 h = lim (2a + h) = 2a h→0 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
85. 85. Derivative of the reciprocal functionExample 1Suppose f(x) = . Use the xdefinition of the derivative tofind f′ (2). x . . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
86. 86. Derivative of the reciprocal functionExample 1Suppose f(x) = . Use the xdefinition of the derivative tofind f′ (2). x .Solution 1/x − 1/2 f′ (2) = lim . x→2 x−2 . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
87. 87. Derivative of the reciprocal functionExample 1Suppose f(x) = . Use the xdefinition of the derivative tofind f′ (2). x .Solution 1/x − 1/2 f′ (2) = lim . x→2 x−2 . x . 2−x = lim x→2 2x(x − 2) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
88. 88. Derivative of the reciprocal functionExample 1Suppose f(x) = . Use the xdefinition of the derivative tofind f′ (2). x .Solution 1/x − 1/2 f′ (2) = lim . x→2 x−2 . x . 2−x = lim x→2 2x(x − 2) −1 = lim x→2 2x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
89. 89. Derivative of the reciprocal functionExample 1Suppose f(x) = . Use the xdefinition of the derivative tofind f′ (2). x .Solution 1/x − 1/2 f′ (2) = lim . x→2 x−2 . x . 2−x = lim x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
90. 90. “Can you do it the other way?"Same limit, different formSolution f(2 + h) − f(2) f′ (2) = lim h→0 h 1 −1 = lim 2+h 2 h→0 h . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
91. 91. “Can you do it the other way?"Same limit, different formSolution f(2 + h) − f(2) f′ (2) = lim h→0 h 1 −1 = lim 2+h 2 h→0 h 2 − (2 + h) = lim h→0 2h(2 + h) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
92. 92. “Can you do it the other way?"Same limit, different formSolution f(2 + h) − f(2) f′ (2) = lim h→0 h 1 −1 = lim 2+h 2 h→0 h 2 − (2 + h) −h = lim = lim h→0 2h(2 + h) h→0 2h(2 + h) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
93. 93. “Can you do it the other way?"Same limit, different formSolution f(2 + h) − f(2) f′ (2) = lim h→0 h 1 −1 = lim 2+h 2 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
94. 94. “Can you do it the other way?"Same limit, different formSolution f(2 + h) − f(2) f′ (2) = lim h→0 h 1 −1 = lim 2+h 2 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
95. 95. “How did you get that?"The Sure-Fire Sally Rule (SFSR) for adding FractionsFact a c ad ± bc ± = b d bdSo 1 1 2−x − x 2 = 2x x−2 x−2 2−x = 2x(x − 2) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 28 / 49
96. 96. “How did you get that?"The Sure-Fire Sally Rule (SFSR) for adding FractionsFact a c ad ± bc ± = b d bdSo 1 1 2−x − x 2 = 2x x−2 x−2 2−x = 2x(x − 2) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 28 / 49
97. 97. 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′ ? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 29 / 49
98. 98. 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 (technically, nonpositive) on that interval . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 29 / 49
99. 99. Derivative of the reciprocal functionExample 1Suppose f(x) = . Use the xdefinition of the derivative tofind f′ (2). x .Solution 1/x − 1/2 f′ (2) = lim . x→2 x−2 . x . 2−x = lim x→2 2x(x − 2) −1 1 = lim =− x→2 2x 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 30 / 49
100. 100. 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 (technically, nonpositive) on that interval If f is increasing on an interval, f′ is positive (technically, nonnegative) on that interval . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 31 / 49
101. 101. Graphically and numerically y . x2 − 22 x m= x−2 3 5 . . 9 . 2.5 4.5 2.1 4.1 2.01 4.01 . .25 . 6 . limit 4 . .41 . 4 . 1.99 3.99. .0401 .4.9601 .3 . .61 3 4 . .. 1.9 3.9 1.5 3.5 . .25 . 2 . 1 3 . . 1 . . . . ... . . x . 1 1 . .. .1 3 . . .5 .99 .5 . 12 . 2.9 2 2 .01 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 32 / 49
102. 102. What does f tell you about f′ ?FactIf f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 33 / 49
103. 103. What does f tell you about f′ ?FactIf f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).Picture Proof. . y .If f is decreasing, then allsecant lines point downward,hence have negative slope.The derivative is a limit ofslopes of secant lines, whichare all negative, so the limit .must be ≤ 0. . . . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 33 / 49
104. 104. What does f tell you about f′ ?.FactIf f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).The Real Proof.If f is decreasing on (a, b), and ∆x > 0, then f(x + ∆x) − f(x) f(x + ∆x) < f(x) =⇒ <0 ∆x. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
105. 105. What does f tell you about f′ ?.FactIf f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).The Real Proof.If f is decreasing on (a, b), and ∆x > 0, then f(x + ∆x) − f(x) f(x + ∆x) < f(x) =⇒ <0 ∆xBut if ∆x < 0, then x + ∆x < x, and f(x + ∆x) − f(x) f(x + ∆x) > f(x) =⇒ <0 ∆xstill!. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
106. 106. What does f tell you about f′ ?.FactIf f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).The Real Proof.If f is decreasing on (a, b), and ∆x > 0, then f(x + ∆x) − f(x) f(x + ∆x) < f(x) =⇒ <0 ∆xBut 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
107. 107. Going the Other Way?QuestionIf a function has a negative derivative on an interval, must it bedecreasing on that interval? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 35 / 49
108. 108. Going the Other Way?QuestionIf a function has a negative derivative on an interval, must it bedecreasing on that interval?AnswerMaybe. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 35 / 49
109. 109. OutlineRates of Change Tangent Lines Velocity Population growth Marginal costsThe derivative, defined Derivatives of (some) power functions What does f tell you about f′ ?How can a function fail to be differentiable?Other notationsThe second derivative . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 36 / 49
110. 110. Differentiability is super-continuityTheoremIf f is differentiable at a, then f is continuous at a. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
111. 111. Differentiability is super-continuityTheoremIf 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
112. 112. Differentiability is super-continuityTheoremIf 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 = 0Note the proper use of the limit law: if the factors each have a limit ata, the limit of the product is the product of the limits. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
113. 113. Differentiability FAILKinksExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
114. 114. Differentiability FAILKinksExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) .′ (x) f . x . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
115. 115. Differentiability FAILKinksExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) .′ (x) f . . x . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
116. 116. Differentiability FAILCuspsExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
117. 117. Differentiability FAILCuspsExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) .′ (x) f . x . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
118. 118. Differentiability FAILCuspsExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) .′ (x) f . x . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
119. 119. Differentiability FAILCuspsExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) .′ (x) f . x . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
120. 120. Differentiability FAILVertical TangentsExampleLet f have the graph on the left-hand side below. Sketch the graph ofthe derivative f′ . f .(x) . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
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