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# Lesson 9: The Product and Quotient Rules (Section 41 slides)

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• 1. Section 2.4 The Product and Quotient Rules V63.0121.041, Calculus I New York University October 4, 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 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! . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 4 / 41
• 5. OutlineDerivative of a Product Derivation ExamplesThe Quotient Rule Derivation ExamplesMore derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and CosecantMore on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 5 / 41
• 6. Recollection and extensionWe have shown that if u and v are functions, that (u + v)′ = u′ + v′ (u − v)′ = u′ − v′What about uv? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 6 / 41
• 7. Is the derivative of a product the product of thederivatives? . uv)′ = u′ v′ ? ( . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 7 / 41
• 8. Is the derivative of a product the product of thederivatives? . uv)′ = u′ v′ ! ( .Try this with u = x and v = x2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 7 / 41
• 9. Is the derivative of a product the product of thederivatives? . uv)′ = u′ v′ ! ( .Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 7 / 41
• 10. Is the derivative of a product the product of thederivatives? . uv)′ = u′ v′ ! ( .Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 7 / 41
• 11. Is the derivative of a product the product of thederivatives? . 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 7 / 41
• 12. Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices? . . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
• 13. Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices? Work longer hours. . . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
• 14. Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices? Work longer hours. Get a raise. . . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
• 15. Mmm...burgersSay 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 inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake? . . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
• 16. Mmm...burgersSay 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 inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake? . . I = 5 × \$0.25 = \$1.25? ∆ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
• 17. Mmm...burgersSay 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 inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake? . . I = 5 × \$0.25 = \$1.25? ∆ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
• 18. Money money money moneyThe answer depends on how much you work already and your currentwage. Suppose you work h hours and are paid w. You get a timeincrease of ∆h and a wage increase of ∆w. Income is wages timeshours, so ∆I = (w + ∆w)(h + ∆h) − wh FOIL = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh = w · ∆h + ∆w · h + ∆w · ∆h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 9 / 41
• 19. A geometric argumentDraw a box: . h ∆ w . ∆h . w ∆h ∆ h . w . h . wh ∆ . w . . w ∆ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 10 / 41
• 20. A geometric argumentDraw a box: . h ∆ w . ∆h . w ∆h ∆ h . w . h . wh ∆ . w . . w ∆ ∆I = w ∆h + h ∆w + ∆w ∆h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 10 / 41
• 21. Cash flowSupose wages and hours are changing continuously over time. Over atime 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 11 / 41
• 22. Cash flowSupose wages and hours are changing continuously over time. Over atime 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 ∆tWhat is the instantaneous rate of change of income? dI ∆I dh dw = lim =w +h +0 dt ∆t→0 ∆t dt dt . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 11 / 41
• 23. Eurekamen!We have discoveredTheorem (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 12 / 41
• 24. Sanity CheckExampleApply the product rule to u = x and v = x2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 13 / 41
• 25. Sanity CheckExampleApply the product rule to u = x and v = x2 .Solution (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2This is what we get the “normal” way. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 13 / 41
• 26. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 27. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby direct multiplication: d [ ] FOIL d [ ] (3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3 dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 28. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby direct multiplication: d [ ] FOIL d [ ] (3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3 dx dx = −5x4 + 12x2 − 2x − 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 29. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 30. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 31. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 32. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 33. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 34. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 35. Which is better?ExampleFind this derivative two ways: first by direct multiplication and then bythe product rule: d [ ] (3 − x2 )(x3 − x + 1) dxSolutionby 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
• 36. One moreExample dFind x sin x. dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 15 / 41
• 37. One moreExample dFind x sin x. dxSolution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 15 / 41
• 38. One moreExample dFind x sin x. dxSolution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 15 / 41
• 39. One moreExample dFind x sin x. dxSolution ( ) ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 15 / 41
• 40. MnemonicLet u = “hi” and v = “ho”. Then (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho” . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 17 / 41
• 42. Iterating the Product RuleExampleUse the product rule to find the derivative of a three-fold product uvw. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 43. Iterating the Product RuleExampleUse the product rule to find the derivative of a three-fold product uvw.Solution (uvw)′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 44. Iterating the Product RuleExampleUse the product rule to find the derivative of a three-fold product uvw.Solution (uvw)′ = ((uv)w)′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 45. Iterating the Product RuleExampleUse the product rule to. find the derivative of a three-fold product uvw. Apply the product ruleSolution to uv and w (uvw)′ = ((uv)w)′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 46. Iterating the Product RuleExampleUse the product rule to. find the derivative of a three-fold product uvw. Apply the product ruleSolution to uv and w (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 47. Iterating the Product RuleExampleUse the product rule to find the derivative of a three-fold product uvw. . Apply the product ruleSolution to u and v (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 48. Iterating the Product RuleExampleUse the product rule to find the derivative of a three-fold product uvw. . Apply the product ruleSolution to u and v (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 49. Iterating the Product RuleExampleUse 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′ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 50. Iterating the Product RuleExampleUse 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 eachfactor once. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
• 51. OutlineDerivative of a Product Derivation ExamplesThe Quotient Rule Derivation ExamplesMore derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and CosecantMore on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 19 / 41
• 52. The Quotient RuleWhat about the derivative of a quotient? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
• 53. The Quotient RuleWhat about the derivative of a quotient? uLet u and v be differentiable functions and let Q = . Then v u = Qv . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
• 54. The Quotient RuleWhat about the derivative of a quotient? uLet u and v be differentiable functions and let Q = . Then v u = QvIf Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
• 55. The Quotient RuleWhat about the derivative of a quotient? uLet u and v be differentiable functions and let Q = . Then v u = QvIf Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
• 56. The Quotient RuleWhat about the derivative of a quotient? uLet u and v be differentiable functions and let Q = . Then v u = QvIf 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
• 57. The Quotient RuleWhat about the derivative of a quotient? uLet u and v be differentiable functions and let Q = . Then v u = QvIf 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 v2This is called the Quotient Rule. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
• 58. The Quotient RuleWe have discoveredTheorem (The Quotient Rule) uLet u and v be differentiable at x, and v(x) ̸= 0. Then is differentiable vat x, and ( u )′ u′ (x)v(x) − u(x)v′ (x) (x) = v v(x)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 21 / 41
• 59. Verifying ExampleExample ( ) d x2Verify the quotient rule by computing and comparing it to dx x d (x).dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 22 / 41
• 60. Verifying ExampleExample ( ) d x2Verify the quotient rule by computing and comparing it to dx x d (x).dxSolution ( ) ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 22 / 41
• 61. MnemonicLet u = “hi” and v = “lo”. Then ( u )′ vu′ − uv′ = = “lo dee hi minus hi dee lo over lo lo” v v2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 23 / 41
• 62. ExamplesExample d 2x + 5 1. dx 3x − 2 d sin x 2. dx x2 d 1 3. 2 dt t + t + 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 24 / 41
• 63. Solution to first exampleSolution d 2x + 5 dx 3x − 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 64. Solution to first exampleSolution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 65. Solution to first exampleSolution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 66. Solution to first exampleSolution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 67. Solution to first exampleSolution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 68. Solution to first exampleSolution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 69. Solution to first exampleSolution d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 70. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 71. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 72. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 73. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 74. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 75. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 76. Solution to first exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
• 77. ExamplesExample 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 26 / 41
• 78. Solution to second exampleSolution d sin x = dx x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 79. Solution to second exampleSolution d sin x x2 = dx x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 80. Solution to second exampleSolution d sin x x2 d sin x = dx dx x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 81. Solution to second exampleSolution d sin x x2 d sin x − sin x = dx dx x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 82. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 83. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 84. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 = . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 85. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 = . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 86. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x = . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 87. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x = . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 88. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x sin x = . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 89. Solution to second exampleSolution d sin x x2 d sin x − sin x dx x2 d = dx dx x2 (x2 )2 x2 cos x − 2x sin x = x4 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 90. Solution to second exampleSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
• 91. Another way to do itSolutionUsing the product rule this time: 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 28 / 41
• 92. ExamplesExample 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 29 / 41
• 93. Solution to third exampleSolution d 1 dt t2 + t + 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 30 / 41
• 94. Solution to third exampleSolution d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 30 / 41
• 95. Solution to third exampleSolution d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 2t + 1 =− 2 (t + t + 2)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 30 / 41
• 96. A nice little takeawayFact 1Let v be differentiable at x, and v(x) ̸= 0. Then is differentiable at 0, vand ( )′ 1 v′ =− 2 v vProof. ( ) d 1 v· d dx (1) −1· d dx v v · 0 − 1 · v′ v′ = = =− 2 dx v v2 v2 v . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 31 / 41
• 97. ExamplesExample 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 32 / 41
• 98. OutlineDerivative of a Product Derivation ExamplesThe Quotient Rule Derivation ExamplesMore derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and CosecantMore on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 33 / 41
• 99. Derivative of TangentExample dFind tan x dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
• 100. Derivative of TangentExample dFind tan x dxSolution ( ) d d sin x tan x = dx dx cos x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
• 101. Derivative of TangentExample dFind tan x dxSolution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
• 102. Derivative of TangentExample dFind tan x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
• 103. Derivative of TangentExample dFind tan x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
• 104. Derivative of TangentExample dFind tan x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
• 105. Derivative of CotangentExample dFind cot x dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
• 106. Derivative of CotangentExample dFind cot x dxAnswer d 1 cot x = − 2 = − csc2 x dx sin x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
• 107. Derivative of CotangentExample dFind cot x dxAnswer d 1 cot x = − 2 = − csc2 x dx sin xSolution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
• 108. Derivative of CotangentExample dFind cot x dxAnswer d 1 cot x = − 2 = − csc2 x dx sin xSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
• 109. Derivative of CotangentExample dFind cot x dxAnswer d 1 cot x = − 2 = − csc2 x dx sin xSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
• 110. Derivative of CotangentExample dFind cot x dxAnswer d 1 cot x = − 2 = − csc2 x dx sin xSolution 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
• 111. Derivative of SecantExample dFind sec x dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
• 112. Derivative of SecantExample dFind sec x dxSolution ( ) d d 1 sec x = dx dx cos x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
• 113. Derivative of SecantExample dFind sec x dxSolution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
• 114. Derivative of SecantExample dFind sec x dxSolution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x = cos2 x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
• 115. Derivative of SecantExample dFind sec x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
• 116. Derivative of SecantExample dFind sec x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
• 117. Derivative of CosecantExample dFind csc x dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
• 118. Derivative of CosecantExample dFind csc x dxAnswer d csc x = − csc x cot x dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
• 119. Derivative of CosecantExample dFind csc x dxAnswer d csc x = − csc x cot x dxSolution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
• 120. Derivative of CosecantExample dFind csc x dxAnswer d csc x = − csc x cot x dxSolution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x =− 2 sin x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
• 121. Derivative of CosecantExample dFind csc x dxAnswer d csc x = − csc x cot x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
• 122. Derivative of CosecantExample dFind csc x dxAnswer d csc x = − csc x cot x dxSolution ( ) 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 38 / 41
• 124. OutlineDerivative of a Product Derivation ExamplesThe Quotient Rule Derivation ExamplesMore derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and CosecantMore on the Power Rule Power Rule for Negative Integers . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 39 / 41
• 125. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor positive integers n. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
• 126. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor positive integers n.Proof. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
• 127. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor positive integers n.Proof. d −n d 1 x = dx dx xn . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
• 128. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor positive integers n.Proof. d n d −n d 1 x x = n = − dxn 2 dx dx x (x ) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
• 129. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor positive integers n.Proof. d n d −n d 1 x x = n = − dxn 2 dx dx x (x ) nxn−1 =− x2n . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
• 130. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
• 131. Power Rule for Negative IntegersWe will use the quotient rule to proveTheorem d −n x = (−n)x−n−1 dxfor 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 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 . . . . . .V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 41 / 41