Lesson 9: The Product and Quotient Rule

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These rules allow us to differentiate the product or quotient of functions whose derivatives are themselves known.

These rules allow us to differentiate the product or quotient of functions whose derivatives are themselves known.

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  • 1. Section 3.2 The Product and Quotient Rules Math 1a February 22, 2008 Announcements Problem Sessions Sunday, Thursday, 7pm, SC 310 Office hours Tuesday, Wednesday 2–4pm SC 323 Midterm I Friday 2/29 in class (up to §3.2)
  • 2. Outline The Product Rule Derivation of the product rule Examples The Quotient Rule Derivation Examples More on the Power Rule Power Rule for nonnegative integers by induction Power Rule for negative integers
  • 3. 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 ? Is it u v ?
  • 4. Is the derivative of a product the product of the derivatives? NO!
  • 5. Is the derivative of a product the product of the derivatives? NO! Try this with u = x and v = x 2 .
  • 6. Is the derivative of a product the product of the derivatives? NO! Try this with u = x and v = x 2 . Then uv = x 3 , so (uv ) = 3x 2 . But u v = 1(2x) = 2x.
  • 7. Is the derivative of a product the product of the derivatives? NO! Try this with u = x and v = x 2 . Then uv = x 3 , so (uv ) = 3x 2 . But u v = 1(2x) = 2x. So we have to be more careful.
  • 8. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices?
  • 9. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours.
  • 10. 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.
  • 11. 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?
  • 12. 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 = wh + w ∆h + ∆w h + ∆w ∆h − wh = w ∆h + ∆w h + ∆w ∆h
  • 13. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h w ∆w
  • 14. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h w ∆w ∆I = w ∆h + h ∆w + ∆w ∆h
  • 15. Supose wages and hours are changing continuously over time. How does income change? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t
  • 16. Supose wages and hours are changing continuously over time. How does income change? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t So dI ∆I dh dw = lim =w +h +0 dt t→0 ∆t dt dt
  • 17. Supose wages and hours are changing continuously over time. How does income change? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t So dI ∆I dh dw = lim =w +h +0 dt t→0 ∆t dt dt Theorem (The Product Rule) Let u and v be differentiable at x. Then (uv ) (x) = u(x)v (x) + u (x)v (x)
  • 18. Example Apply the product rule to u = x and v = x 2 .
  • 19. Example Apply the product rule to u = x and v = x 2 . Solution (uv ) (x) = u(x)v (x) + u (x)v (x) = x · (2x) + 1 · x 2 = 3x 2 This is what we get the “normal” way.
  • 20. Example Find this derivative two ways: first by FOIL and then by the product rule: d (3 − x 2 )(x 3 − x + 1) dx
  • 21. Example Find this derivative two ways: first by FOIL and then by the product rule: d (3 − x 2 )(x 3 − x + 1) dx Solution (i) by FOIL: d FOILd (3 − x 2 )(x 3 − x + 1) = −x 5 + 4x 3 − x 2 − 3x + 3 dx dx = −5x 4 + 12x 2 − 2x − 3 (ii) by the product rule: dy d d 3 = (3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1) dx dx dx = (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1) = −5x 4 + 12x 2 − 2x − 3
  • 22. One more Example d x Find xe dx
  • 23. One more Example d x Find xe dx Answer y = e x + xe x
  • 24. Mnemonic Let u = “ho” and v = “hi”. Then (uv ) = uv + vu = “ho dee hi plus hi dee ho”
  • 25. Outline The Product Rule Derivation of the product rule Examples The Quotient Rule Derivation Examples More on the Power Rule Power Rule for nonnegative integers by induction Power Rule for negative integers
  • 26. The Quotient Rule What about the derivative of a quotient?
  • 27. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable and let Q = . Then u = Qv . If Q v is differentiable, we have u = (Qv ) = Q v + Qv u − Qv u uv Q = = − 2 v v v u v − uv = v2
  • 28. The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable and let Q = . Then u = Qv . If Q v is differentiable, we have u = (Qv ) = Q v + Qv u − Qv u uv Q = = − 2 v v v u v − uv = v2 This is called the Quotient Rule.
  • 29. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t
  • 30. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers
  • 31. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers 19 1. − (3x − 2)2
  • 32. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers 19 1. − (3x − 2)2 2 x2 + x + 1 2. − (x 2 − 1)2
  • 33. Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x 2 − 1 d t −1 3. 2+t +2 . dt t Answers 19 1. − (3x − 2)2 2 x2 + x + 1 2. − (x 2 − 1)2 −t 2 + 2t + 3 3. (t 2 + t + 2)2
  • 34. Mnemonic Let u = “hi” and v = “lo”. Then u vu − uv = = “lo dee hi minus hi dee lo over lo lo” v v2
  • 35. Outline The Product Rule Derivation of the product rule Examples The Quotient Rule Derivation Examples More on the Power Rule Power Rule for nonnegative integers by induction Power Rule for negative integers
  • 36. Power Rule for nonnegative integers by induction Theorem Let n be a positive integer. Then d n x = nx n−1 dx
  • 37. Power Rule for nonnegative integers by induction Theorem Let n be a positive integer. Then d n x = nx n−1 dx Proof. By induction on n. We have shown it to be true for n = 1. d n Suppose for some n that x = nx n−1 . Then dx d n+1 d x = (x · x n ) dx dx d d n = x xn + x x dx dx = 1 · x n + x · nx n−1 = (n + 1)x n
  • 38. Power Rule for negative integers Use the quotient rule to prove Theorem d −n x = (−n)x −n−1 dx for positive integers n.
  • 39. Power Rule for negative integers 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 x n d d x n · dx 1 − 1 · dx x n = x 2n 0 − nx n−1 = = −nx −n−1 x 2n