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- 1. Section 2.4 The Product and Quotient Rules V63.0121.021, Calculus I New York University October 5, 2010 Announcements Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2 Midterm in class (covers all sections up to 2.5) . . . . . . .
- 2. Announcements Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2 Midterm in class (covers all sections up to 2.5) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 2 / 41
- 3. Help! Free resources: Math Tutoring Center (CIWW 524) College Learning Center (schedule on Blackboard) TAs’ office hours my office hours each other! . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 3 / 41
- 4. Objectives Understand and be able to use the Product Rule for the derivative of the product of two functions. Understand and be able to use the Quotient Rule for the derivative of the quotient of two functions. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 4 / 41
- 5. Outline Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 5 / 41
- 6. Recollection and extension We have shown that if u and v are functions, that (u + v)′ = u′ + v′ (u − v)′ = u′ − v′ What about uv? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 6 / 41
- 7. Is the derivative of a product the product of the derivatives? . uv)′ = u′ v′ ? ( . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41
- 8. Is the derivative of a product the product of the derivatives? . uv)′ = u′ v′ ! ( . Try this with u = x and v = x2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41
- 9. Is the derivative of a product the product of the derivatives? . uv)′ = u′ v′ ! ( . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41
- 10. Is the derivative of a product the product of the derivatives? . uv)′ = u′ v′ ! ( . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41
- 11. Is the derivative of a product the product of the derivatives? . uv)′ = u′ v′ ! ( . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. So we have to be more careful. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41
- 12. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41
- 13. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41
- 14. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41
- 15. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41
- 16. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . . I = 5 × $0.25 = $1.25? ∆ . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41
- 17. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . . I = 5 × $0.25 = $1.25? ∆ . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41
- 18. Money money money money The answer depends on how much you work already and your current wage. Suppose you work h hours and are paid w. You get a time increase of ∆h and a wage increase of ∆w. Income is wages times hours, so ∆I = (w + ∆w)(h + ∆h) − wh FOIL = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh = w · ∆h + ∆w · h + ∆w · ∆h . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 9 / 41
- 19. A geometric argument Draw a box: . h ∆ w . ∆h . w ∆h ∆ h . w . h . wh ∆ . w . . w ∆ . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 10 / 41
- 20. A geometric argument Draw a box: . h ∆ w . ∆h . w ∆h ∆ h . w . h . wh ∆ . w . . w ∆ ∆I = w ∆h + h ∆w + ∆w ∆h . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 10 / 41
- 21. Cash flow Supose wages and hours are changing continuously over time. Over a time interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 11 / 41
- 22. Cash flow Supose wages and hours are changing continuously over time. Over a time interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t What is the instantaneous rate of change of income? dI ∆I dh dw = lim =w +h +0 dt ∆t→0 ∆t dt dt . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 11 / 41
- 23. Eurekamen! We have discovered Theorem (The Product Rule) Let u and v be differentiable at x. Then (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) in Leibniz notation d du dv (uv) = ·v+u dx dx dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 12 / 41
- 24. Sanity Check Example Apply the product rule to u = x and v = x2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 13 / 41
- 25. Sanity Check Example Apply the product rule to u = x and v = x2 . Solution (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2 This is what we get the “normal” way. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 13 / 41
- 26. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 27. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by direct multiplication: d [ ] FOIL d [ ] (3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3 dx dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 28. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by direct multiplication: d [ ] FOIL d [ ] (3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3 dx dx = −5x4 + 12x2 − 2x − 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 29. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 30. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 31. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 32. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 33. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 34. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 35. Which is better? Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) = −5x4 + 12x2 − 2x − 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41
- 36. One more Example d Find x sin x. dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41
- 37. One more Example d Find x sin x. dx Solution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41
- 38. One more Example d Find x sin x. dx Solution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41
- 39. One more Example d Find x sin x. dx Solution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x = sin x + x cos x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41
- 40. Mnemonic Let u = “hi” and v = “ho”. Then (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho” . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 16 / 41
- 41. Musical interlude jazz bandleader and singer hit song “Minnie the Moocher” featuring “hi de ho” chorus played Curtis in The Blues Brothers Cab Calloway 1907–1994 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 17 / 41
- 42. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 43. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 44. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ = ((uv)w)′ . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 45. Iterating the Product Rule Example Use the product rule to. find the derivative of a three-fold product uvw. Apply the product rule Solution to uv and w (uvw)′ = ((uv)w)′ . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 46. Iterating the Product Rule Example Use the product rule to. find the derivative of a three-fold product uvw. Apply the product rule Solution to uv and w (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 47. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . Apply the product rule Solution to u and v (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 48. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . Apply the product rule Solution to u and v (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 49. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′ . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 50. Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′ So we write down the product three times, taking the derivative of each factor once. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41
- 51. Outline Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 19 / 41
- 52. The Quotient Rule What about the derivative of a quotient? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41
- 53. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41
- 54. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41
- 55. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41
- 56. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41
- 57. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2 This is called the Quotient Rule. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41
- 58. The Quotient Rule We have discovered Theorem (The Quotient Rule) u Let u and v be differentiable at x, and v(x) ̸= 0. Then is differentiable v at x, and ( u )′ u′ (x)v(x) − u(x)v′ (x) (x) = v v(x)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 21 / 41
- 59. Verifying Example Example ( ) d x2 Verify the quotient rule by computing and comparing it to dx x d (x). dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 22 / 41
- 60. Verifying Example Example ( ) d x2 Verify the quotient rule by computing and comparing it to dx x d (x). dx Solution ( ) ( ) d x2 x dx x2 − x2 dx (x) d d = dx x x2 x · 2x − x2 · 1 = x2 x2 d = 2 =1= (x) x dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 22 / 41
- 61. Mnemonic Let u = “hi” and v = “lo”. Then ( u )′ vu′ − uv′ = = “lo dee hi minus hi dee lo over lo lo” v v2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 23 / 41
- 62. Examples Example d 2x + 5 1. dx 3x − 2 d sin x 2. dx x2 d 1 3. 2 dt t + t + 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 24 / 41
- 63. Solution to first example Solution d 2x + 5 dx 3x − 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 64. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 65. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 66. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 67. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 68. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 69. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 70. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 71. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 72. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 73. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 74. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 75. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) = (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 76. Solution to first example Solution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) 19 = =− (3x − 2) 2 (3x − 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41
- 77. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x 2. dx x2 d 1 3. 2 dt t + t + 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 26 / 41
- 78. Solution to second example Solution d sin x = dx x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 79. Solution to second example Solution d sin x x2 = dx x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 80. Solution to second example Solution d sin x x2 d sin x = dx dx x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 81. Solution to second example Solution d sin x x2 d sin x − sin x = dx dx x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 82. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 83. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 84. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 = . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 85. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 = . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 86. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x = . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 87. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x = . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 88. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x sin x = . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 89. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x sin x = x4 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 90. Solution to second example Solution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x sin x = x4 x cos x − 2 sin x = x3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
- 91. Another way to do it Find the derivative with the product rule instead. Solution d sin x d ( ) = sin x · x−2 dx x2 dx ( ) ( ) d −2 d −2 = sin x · x + sin x · x dx dx = cos x · x−2 + sin x · (−2x−3 ) = x−3 (x cos x − 2 sin x) Notice the technique of factoring out the largest negative power, leaving positive powers. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 28 / 41
- 92. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x x cos x − 2 sin x 2. 2. dx x2 x3 d 1 3. 2 dt t + t + 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 29 / 41
- 93. Solution to third example Solution d 1 dt t2 + t + 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 30 / 41
- 94. Solution to third example Solution d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 30 / 41
- 95. Solution to third example Solution d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 2t + 1 =− 2 (t + t + 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 30 / 41
- 96. A nice little takeaway Fact 1 Let v be differentiable at x, and v(x) ̸= 0. Then is differentiable at 0, v and ( )′ 1 v′ =− 2 v v Proof. ( ) d 1 v· d dx (1) −1· d dx v v · 0 − 1 · v′ v′ = = =− 2 dx v v2 v2 v . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 31 / 41
- 97. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x x cos x − 2 sin x 2. 2. dx x2 x3 d 1 2t + 1 3. 2 3. − 2 dt t + t + 2 (t + t + 2)2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 32 / 41
- 98. Outline Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 33 / 41
- 99. Derivative of Tangent Example d Find tan x dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41
- 100. Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x tan x = dx dx cos x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41
- 101. Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41
- 102. Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x = cos2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41
- 103. Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = 2x = cos cos2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41
- 104. Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = 2x = = sec2 x cos cos2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41
- 105. Derivative of Cotangent Example d Find cot x dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41
- 106. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41
- 107. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41
- 108. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x 2 − sin x − cos2 x = sin2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41
- 109. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x 2 − sin x − cos2 x 1 = 2 =− 2 sin x sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41
- 110. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x 2 − sin x − cos2 x 1 = 2 = − 2 = − csc2 x sin x sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41
- 111. Derivative of Secant Example d Find sec x dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41
- 112. Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 sec x = dx dx cos x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41
- 113. Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41
- 114. Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x = cos2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41
- 115. Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · cos2 x cos x cos x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41
- 116. Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · = sec x tan x cos2 x cos x cos x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41
- 117. Derivative of Cosecant Example d Find csc x dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41
- 118. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41
- 119. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41
- 120. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x =− 2 sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41
- 121. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · sin x sin x sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41
- 122. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · = − csc x cot x sin x sin x sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41
- 123. Recap: Derivatives of trigonometric functions y y′ sin x cos x Functions come in pairs cos x − sin x (sin/cos, tan/cot, sec/csc) tan x sec2 x Derivatives of pairs follow similar patterns, with cot x − csc2 x functions and co-functions sec x sec x tan x switched and an extra sign. csc x − csc x cot x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 38 / 41
- 124. Outline Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 39 / 41
- 125. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 126. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 127. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d −n d 1 x = dx dx xn . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 128. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d n d −n d 1 x x = n = − dxn 2 dx dx x (x ) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 129. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d n d −n d 1 x x = n = − dxn 2 dx dx x (x ) nxn−1 =− x2n . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 130. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d n d −n d 1 x x = n = − dxn 2 dx dx x (x ) nxn−1 =− = −nxn−1−2n x2n . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 131. Power Rule for Negative Integers We will use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d n d −n d 1 x x = n = − dxn 2 dx dx x (x ) nxn−1 =− = −nxn−1−2n = −nx−n−1 x2n . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41
- 132. Summary The Product Rule: (uv)′ = u′ v + uv′ ( u )′ vu′ − uv′ The Quotient Rule: = v v2 Derivatives of tangent/cotangent, secant/cosecant d d tan x = sec2 x sec x = sec x tan x dx dx d d cot x = − csc2 x csc x = − csc x cot x dx dx The Power Rule is true for all whole number powers, including negative powers: d n x = nxn−1 dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 41 / 41

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