Lesson 8: Basic Differentiation Rules

1,483 views
1,376 views

Published on

After defining the limit and calculating a few, we introduced the limit laws. Today we do the same for the derivative. We calculate a few and introduce laws which allow us to computer more. The Power Rule shows us how to compute derivatives of polynomials, and we can also find directly the derivative of sine and cosine.

Published in: Education, Technology
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
1,483
On SlideShare
0
From Embeds
0
Number of Embeds
7
Actions
Shares
0
Downloads
45
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide

Lesson 8: Basic Differentiation Rules

  1. 1. Section 2.3 Basic Differentiation Rules V63.0121.027, Calculus I September 24, 2009 Announcements Quiz next week (up to Section 2.1) OH today 3-4 See website for up-to-date events. . . . . . .
  2. 2. Outline Recall Derivatives so far Derivatives of power functions by hand The Power Rule Derivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple Rule Derivatives of sine and cosine . . . . . .
  3. 3. Derivative . . . . . .
  4. 4. Recall: the derivative Definition Let 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−a exists, the function is said to be differentiable at a and f′ (a) is the derivative of f at a. The derivative … …measures the slope of the line through (a, f(a)) tangent to the curve y = f(x); …represents the instantaneous rate of change of f at a …produces the best possible linear approximation to f near a. . . . . . .
  5. 5. Notation Newtonian notation Leibnizian notation dy d df f′ (x) y′ (x) y′ f(x) dx dx dx . . . . . .
  6. 6. Link between the notations f(x + ∆x) − f(x) ∆y dy f′ (x) = lim = lim = ∆x→0 ∆x ∆x→0 ∆x dx dy Leibniz thought of as a quotient of “infinitesimals” dx dy We think of as representing a limit of (finite) difference dx quotients, not as an actual fraction itself. The notation suggests things which are true even though they don’t follow from the notation per se . . . . . .
  7. 7. Outline Recall Derivatives so far Derivatives of power functions by hand The Power Rule Derivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple Rule Derivatives of sine and cosine . . . . . .
  8. 8. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). . . . . . .
  9. 9. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) f′ (x) = lim h→0 h . . . . . .
  10. 10. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2 − x2 f′ (x) = lim = lim h→0 h h→0 h . . . . . .
  11. 11. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2 − x2 f′ (x) = lim = lim h→0 h h→0 h 2 x2 x2   + 2xh + h −       = lim h→0 h . . . . . .
  12. 12. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2 − x2 f′ (x) = lim = lim h→0 h h→0 h 2 2 x2 x2   + 2xh + h −       2xh + h¡ ¡ = lim = lim h→0 h h→0 h ¡ . . . . . .
  13. 13. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2 − x2 f′ (x) = lim = lim h→0 h h→0 h 2 2 x2 x2   + 2xh + h −       2xh + h¡ ¡ = lim = lim h→0 h h→0 h ¡ = lim (2x + h) = 2x. h→0 So f′ (x) = 2x. . . . . . .
  14. 14. The second derivative If f is a function, so is f′ , and we can seek its derivative. f′′ = (f′ )′ It measures the rate of change of the rate of change! . . . . . .
  15. 15. The second derivative If f is a function, so is f′ , and we can seek its derivative. f′′ = (f′ )′ It measures the rate of change of the rate of change! Leibnizian notation: d2 y d2 d2 f f(x) dx2 dx2 dx2 . . . . . .
  16. 16. The squaring function and its derivatives y . . x . . . . . . .
  17. 17. The squaring function and its derivatives y . . f . x . . . . . . .
  18. 18. The squaring function and its derivatives y . .′ f f increasing =⇒ f′ ≥ 0 f . f decreasing =⇒ f′ ≤ 0 . x . horizontal tangent at 0 =⇒ f′ (0) = 0 . . . . . .
  19. 19. The squaring function and its derivatives y . .′ f .′′ f f increasing =⇒ f′ ≥ 0 f . f decreasing =⇒ f′ ≤ 0 . x . horizontal tangent at 0 =⇒ f′ (0) = 0 . . . . . .
  20. 20. Derivative of the cubing function Example Suppose f(x) = x3 . Use the definition of derivative to find f′ (x). . . . . . .
  21. 21. Derivative of the cubing function Example Suppose f(x) = x3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)3 − x3 f′ (x) = lim = lim h→0 h h→0 h . . . . . .
  22. 22. Derivative of the cubing function Example Suppose f(x) = x3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)3 − x3 f′ (x) = lim = lim h→0 h h→0 h 2 3 x3 2 x3   + 3x h + 3xh + h −       = lim h→0 h . . . . . .
  23. 23. Derivative of the cubing function Example Suppose f(x) = x3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)3 − x3 f′ (x) = lim = lim h→0 h h→0 h 1 2 x3 2 2 3 x3 3x2 h ¡ ! 2 ! ¡ 3   + 3x h + 3xh + h −       ¡ + 3xh + h = lim = lim h→0 h h→0 h ¡ . . . . . .
  24. 24. Derivative of the cubing function Example Suppose f(x) = x3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)3 − x3 f′ (x) = lim = lim h→0 h h→0 h 1 2 x3 2 2 3 x3 3x2 h ¡ ! 2 ! ¡ 3   + 3x h + 3xh + h −       ¡ + 3xh + h = lim = lim h→0 h h→0 h ¡ ( ) = lim 3x2 + 3xh + h2 = 3x2 . h→0 So f′ (x) = 3x2 . . . . . . .
  25. 25. The cubing function and its derivatives y . . x . . . . . . .
  26. 26. The cubing function and its derivatives y . f . . x . . . . . . .
  27. 27. The cubing function and its derivatives Notice that f is y . f . ′ increasing, and f′ > 0 except f′ (0) = 0 f . . x . . . . . . .
  28. 28. The cubing function and its derivatives Notice that f is y . f . ′′ f . ′ increasing, and f′ > 0 except f′ (0) = 0 f . . x . . . . . . .
  29. 29. The cubing function and its derivatives Notice that f is y . f . ′′ f . ′ increasing, and f′ > 0 except f′ (0) = 0 Notice also that the f . tangent line to the graph . x . of f at (0, 0) crosses the graph (contrary to a popular “definition” of the tangent line) . . . . . .
  30. 30. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). . . . . . .
  31. 31. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). Solution √ √ ′ f(x + h) − f(x) x+h− x f (x) = lim = lim h→0 h h→0 h . . . . . .
  32. 32. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). Solution √ √ ′ f(x + h) − f(x) x+h− x f (x) = lim = lim h→0 h h→0 h √ √ √ √ x+h− x x+h+ x = lim ·√ √ h→0 h x+h+ x . . . . . .
  33. 33. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). Solution √ √ ′ f(x + h) − f(x) x+h− x f (x) = lim = lim h→0 h h→0 h √ √ √ √ x+h− x x+h+ x = lim ·√ √ h→0 h x+h+ x (¡ + h) − ¡ x x = lim (√ √ ) h→0 h x+h+ x . . . . . .
  34. 34. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). Solution √ √ ′ f(x + h) − f(x) x+h− x f (x) = lim = lim h→0 h h→0 h √ √ √ √ x+h− x x+h+ x = lim ·√ √ h→0 h x+h+ x (¡ + h) − ¡ x x h ¡ = lim (√ √ ) = lim (√ √ ) h→0 h x+h+ x h→0 h ¡ x+h+ x . . . . . .
  35. 35. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). Solution √ √ ′ f(x + h) − f(x) x+h− x f (x) = lim = lim h→0 h h→0 h √ √ √ √ x+h− x x+h+ x = lim ·√ √ h→0 h x+h+ x (¡ + h) − ¡ x x h ¡ = lim (√ √ ) = lim (√ √ ) h→0 h x+h+ x h→0 h ¡ x+h+ x 1 = √ 2 x √ So f′ (x) = x = 1 x−1/2 . 2 . . . . . .
  36. 36. The square root function and its derivatives y . . x . . . . . . .
  37. 37. The square root function and its derivatives y . f . . x . . . . . . .
  38. 38. The square root function and its derivatives y . Here lim f′ (x) = ∞ and f . x→0+ f is not differentiable at 0 . .′ f x . . . . . . .
  39. 39. The square root function and its derivatives y . Here lim f′ (x) = ∞ and f . x→0+ f is not differentiable at 0 . .′ f x . Notice also lim f′ (x) = 0 x→∞ . . . . . .
  40. 40. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). . . . . . .
  41. 41. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)1/3 − x1/3 f′ (x) = lim = lim h→0 h h→0 h . . . . . .
  42. 42. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)1/3 − x1/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 = lim · h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 . . . . . .
  43. 43. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)1/3 − x1/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 = lim · h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 (¡ + h) − ¡ x x = lim ( 2/3 + (x + h)1/3 x1/3 + x2/3 ) h→0 h (x + h) . . . . . .
  44. 44. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)1/3 − x1/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 = lim · h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 (¡ + h) − ¡ x x = lim ( 2/3 + (x + h)1/3 x1/3 + x2/3 ) h→0 h (x + h) h ¡ = lim ( ) h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 ¡ . . . . . .
  45. 45. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)1/3 − x1/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 = lim · h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 (¡ + h) − ¡ x x = lim ( 2/3 + (x + h)1/3 x1/3 + x2/3 ) h→0 h (x + h) h ¡ 1 = lim ( )= h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 ¡ 3x2/3 So f′ (x) = 1 x−2/3 . 3 . . . . . .
  46. 46. The cube root function and its derivatives y . . x . . . . . . .
  47. 47. The cube root function and its derivatives y . f . . x . . . . . . .
  48. 48. The cube root function and its derivatives y . Here lim f′ (x) = ∞ and f x→0 f . is not differentiable at 0 . .′ f x . . . . . . .
  49. 49. The cube root function and its derivatives y . Here lim f′ (x) = ∞ and f x→0 f . is not differentiable at 0 . .′ f Notice also x . lim f′ (x) = 0 x→±∞ . . . . . .
  50. 50. One more Example Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). . . . . . .
  51. 51. One more Example Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2/3 − x2/3 f′ (x) = lim = lim h→0 h h→0 h . . . . . .
  52. 52. One more Example Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2/3 − x2/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 ( ) = lim · (x + h)1/3 + x1/3 h→0 h . . . . . .
  53. 53. One more Example Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2/3 − x2/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 ( ) = lim · (x + h)1/3 + x1/3 h→0 h ( ) 1 −2/3 1 /3 = 3x 2x . . . . . .
  54. 54. One more Example Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). Solution f(x + h) − f(x) (x + h)2/3 − x2/3 f′ (x) = lim = lim h→0 h h→0 h (x + h) 1/3 − x1/3 ( ) = lim · (x + h)1/3 + x1/3 h→0 h ( ) 1 −2/3 = 3x 2x 1 /3 = 2 x−1/3 3 So f′ (x) = 2 x−1/3 . 3 . . . . . .
  55. 55. The function x → x2/3 and its derivative y . . x . . . . . . .
  56. 56. The function x → x2/3 and its derivative y . f . . x . . . . . . .
  57. 57. The function x → x2/3 and its derivative y . f is not differentiable at f . 0 and lim f′ (x) = ±∞ x→0± ′ . f . x . . . . . . .
  58. 58. The function x → x2/3 and its derivative y . f is not differentiable at f . 0 and lim f′ (x) = ±∞ x→0± ′ . f . Notice also x . lim f′ (x) = 0 x→±∞ . . . . . .
  59. 59. Recap y y′ x2 2x x3 3x2 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  60. 60. Recap y y′ x2 2x x3 3x2 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  61. 61. Recap y y′ x2 2x x3 3x2 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  62. 62. Recap y y′ x2 2x x3 3x2 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  63. 63. Recap y y′ x2 2x x3 3x2 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  64. 64. Recap y y′ x2 2x1 x3 3x2 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  65. 65. Recap: The Tower of Power y y′ x2 2x1 The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  66. 66. Recap: The Tower of Power y y′ x2 2x The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  67. 67. Recap: The Tower of Power y y′ x2 2x The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  68. 68. Recap: The Tower of Power y y′ x2 2x The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  69. 69. Recap: The Tower of Power y y′ x2 2x The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  70. 70. Recap: The Tower of Power y y′ x2 2x The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . .
  71. 71. Recap: The Tower of Power y y′ x2 2x The power goes down by one in each x3 3x2 derivative 1 −1/2 x1/2 2x The coefficient in the 1 −2/3 derivative is the power x1/3 3x of the original function 2 −1/3 x2/3 3x . . . . . .
  72. 72. The Power Rule There is mounting evidence for Theorem (The Power Rule) Let r be a real number and f(x) = xr . Then f′ (x) = rxr−1 as long as the expression on the right-hand side is defined. Perhaps the most famous rule in calculus We will assume it as of today We will prove it many ways for many different r. . . . . . .
  73. 73. The other Tower of Power . . . . . .
  74. 74. Outline Recall Derivatives so far Derivatives of power functions by hand The Power Rule Derivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple Rule Derivatives of sine and cosine . . . . . .
  75. 75. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) . . . . . .
  76. 76. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have n ∑ (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k n copies k=0 . . . . . .
  77. 77. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have n ∑ (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k n copies k=0 The coefficient of xn is 1 because we have to choose x from each binomial, and there’s only one way to do that. . . . . . .
  78. 78. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have n ∑ (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k n copies k=0 The coefficient of xn is 1 because we have to choose x from each binomial, and there’s only one way to do that. The coefficient of xn−1 h is the number of ways we can choose x n − 1 times, which is the same as the number of different hs we can pick, which is n. . . . . . .
  79. 79. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have n ∑ (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k n copies k=0 The coefficient of xn is 1 because we have to choose x from each binomial, and there’s only one way to do that. The coefficient of xn−1 h is the number of ways we can choose x n − 1 times, which is the same as the number of different hs we can pick, which is n. . . . . . .
  80. 80. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have n ∑ (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k n copies k=0 The coefficient of xn is 1 because we have to choose x from each binomial, and there’s only one way to do that. The coefficient of xn−1 h is the number of ways we can choose x n − 1 times, which is the same as the number of different hs we can pick, which is n. . . . . . .
  81. 81. Remember your algebra Fact Let n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have n ∑ (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k n copies k=0 The coefficient of xn is 1 because we have to choose x from each binomial, and there’s only one way to do that. The coefficient of xn−1 h is the number of ways we can choose x n − 1 times, which is the same as the number of different hs we can pick, which is n. . . . . . .
  82. 82. Pascal’s Triangle .. 1 1 . 1 . 1 . 2 . 1 . 1 . 3 . 3 . 1 . (x + h)0 = 1 1 . 4 . 6 . 4 . 1 . (x + h)1 = 1x + 1h (x + h)2 = 1x2 + 2xh + 1h2 1 . 5 1 1 5 . .0 .0 . 1 . (x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3 1 . 6 1 2 1 6 . .5 .0 .5 . 1 . ... ... . . . . . .
  83. 83. Pascal’s Triangle .. 1 1 . 1 . 1 . 2 . 1 . 1 . 3 . 3 . 1 . (x + h)0 = 1 1 . 4 . 6 . 4 . 1 . (x + h)1 = 1x + 1h (x + h)2 = 1x2 + 2xh + 1h2 1 . 5 1 1 5 . .0 .0 . 1 . (x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3 1 . 6 1 2 1 6 . .5 .0 .5 . 1 . ... ... . . . . . .
  84. 84. Pascal’s Triangle .. 1 1 . 1 . 1 . 2 . 1 . 1 . 3 . 3 . 1 . (x + h)0 = 1 1 . 4 . 6 . 4 . 1 . (x + h)1 = 1x + 1h (x + h)2 = 1x2 + 2xh + 1h2 1 . 5 1 1 5 . .0 .0 . 1 . (x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3 1 . 6 1 2 1 6 . .5 .0 .5 . 1 . ... ... . . . . . .
  85. 85. Pascal’s Triangle .. 1 1 . 1 . 1 . 2 . 1 . 1 . 3 . 3 . 1 . (x + h)0 = 1 1 . 4 . 6 . 4 . 1 . (x + h)1 = 1x + 1h (x + h)2 = 1x2 + 2xh + 1h2 1 . 5 1 1 5 . .0 .0 . 1 . (x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3 1 . 6 1 2 1 6 . .5 .0 .5 . 1 . ... ... . . . . . .
  86. 86. Theorem (The Power Rule) Let r be a positive whole number. Then d r x = rxr−1 dx . . . . . .
  87. 87. Theorem (The Power Rule) Let r be a positive whole number. Then d r x = rxr−1 dx Proof. As we showed above, (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) So (x + h)n − xn nxn−1 h + (stuff with at least two hs in it) = h h = nxn−1 + (stuff with at least one h in it) and this tends to nxn−1 as h → 0. . . . . . .
  88. 88. The Power Rule for constants Theorem Let c be a constant. Then d c=0 dx . . . . . .
  89. 89. The Power Rule for constants Theorem d 0 Let c be a constant. Then l .ike x = 0x−1 dx d c=0. dx (although x → 0x−1 is not defined at zero.) . . . . . .
  90. 90. The Power Rule for constants Theorem d 0 Let c be a constant. Then l .ike x = 0x−1 dx d c=0. dx (although x → 0x−1 is not defined at zero.) Proof. Let f(x) = c. Then f(x + h) − f(x) c−c = =0 h h So f′ (x) = lim 0 = 0. h→0 . . . . . .
  91. 91. New derivatives from old This is where the calculus starts to get really powerful! . . . . . .
  92. 92. Calculus . . . . . .
  93. 93. Recall the Limit Laws Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M x→a 2. lim [f(x) − g(x)] = L − M x→a 3. lim [cf(x)] = cL x→a 4. . . . . . . . . .
  94. 94. Adding functions Theorem (The Sum Rule) Let f and g be functions and define (f + g)(x) = f(x) + g(x) Then if f and g are differentiable at x, then so is f + g and (f + g)′ (x) = f′ (x) + g′ (x). Succinctly, (f + g)′ = f′ + g′ . . . . . . .
  95. 95. Proof. Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h . . . . . .
  96. 96. Proof. Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h f(x + h) + g(x + h) − [f(x) + g(x)] = lim h→0 h . . . . . .
  97. 97. Proof. Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h f(x + h) + g(x + h) − [f(x) + g(x)] = lim h→0 h f(x + h) − f(x) g(x + h) − g(x) = lim + lim h→0 h h→0 h . . . . . .
  98. 98. Proof. Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h f(x + h) + g(x + h) − [f(x) + g(x)] = lim h→0 h f(x + h) − f(x) g(x + h) − g(x) = lim + lim h→0 h h→0 h ′ ′ = f (x) + g (x) Note the use of the Sum Rule for limits. Since the limits of the difference quotients for for f and g exist, the limit of the sum is the sum of the limits. . . . . . .
  99. 99. Scaling functions Theorem (The Constant Multiple Rule) Let f be a function and c a constant. Define (cf)(x) = cf(x) Then if f is differentiable at x, so is cf and (cf)′ (x) = c · f′ (x) Succinctly, (cf)′ = cf′ . . . . . . .
  100. 100. Proof. Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h . . . . . .
  101. 101. Proof. Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h cf(x + h) − cf(x) = lim h→0 h . . . . . .
  102. 102. Proof. Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h cf(x + h) − cf(x) = lim h→0 h f(x + h) − f(x) = c lim h→0 h . . . . . .
  103. 103. Proof. Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h cf(x + h) − cf(x) = lim h→0 h f(x + h) − f(x) = c lim h→0 h ′ = c · f (x) . . . . . .
  104. 104. Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx . . . . . .
  105. 105. Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx Solution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx . . . . . .
  106. 106. Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx Solution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx d d d = 2 x3 + x4 − 17 x12 + 0 dx dx dx . . . . . .
  107. 107. Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx Solution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx d d d = 2 x3 + x4 − 17 x12 + 0 dx dx dx = 2 · 3x2 + 4x3 − 17 · 12x11 . . . . . .
  108. 108. Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx Solution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx d d d = 2 x3 + x4 − 17 x12 + 0 dx dx dx = 2 · 3x2 + 4x3 − 17 · 12x11 = 6x2 + 4x3 − 204x11 . . . . . .
  109. 109. Outline Recall Derivatives so far Derivatives of power functions by hand The Power Rule Derivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple Rule Derivatives of sine and cosine . . . . . .
  110. 110. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: . . . . . .
  111. 111. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h . . . . . .
  112. 112. Angle addition formulas See Appendix A . . sin(A + B) = sin A cos B + cos A sin B cos(A + B) = cos A cos B − sin A sin B . . . . . .
  113. 113. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h ( sin x cos h + cos x sin h) − sin x = lim h→0 h . . . . . .
  114. 114. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h ( sin x cos h + cos x sin h) − sin x = lim h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h . . . . . .
  115. 115. Two important trigonometric limits See Section 1.4 . sin θ lim . =1 θ→0 θ s . in θ . cos θ − 1 θ lim =0 . θ θ→0 θ . . − cos θ 1 1 . . . . . . .
  116. 116. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h ( sin x cos h + cos x sin h) − sin x = lim h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h = sin x · 0 + cos x · 1 . . . . . .
  117. 117. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h ( sin x cos h + cos x sin h) − sin x = lim h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h = sin x · 0 + cos x · 1 . . . . . .
  118. 118. Derivatives of Sine and Cosine Fact d sin x = cos x dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h ( sin x cos h + cos x sin h) − sin x = lim h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h = sin x · 0 + cos x · 1 = cos x . . . . . .
  119. 119. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 . in x s . . . . . .
  120. 120. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 . os x c . in x s f(x) = sin x has horizontal tangents where f′ = cos(x) is zero. . . . . . .
  121. 121. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 . os x c . in x s f(x) = sin x has horizontal tangents where f′ = cos(x) is zero. what happens at the horizontal tangents of cos? . . . . . .
  122. 122. Derivatives of Sine and Cosine Fact d d sin x = cos x cos x = − sin x dx dx . . . . . .
  123. 123. Derivatives of Sine and Cosine Fact d d sin x = cos x cos x = − sin x dx dx Proof. We already did the first. The second is similar (mutatis mutandis): d cos(x + h) − cos x cos x = lim dx h→0 h . . . . . .
  124. 124. Derivatives of Sine and Cosine Fact d d sin x = cos x cos x = − sin x dx dx Proof. We already did the first. The second is similar (mutatis mutandis): d cos(x + h) − cos x cos x = lim dx h→0 h (cos x cos h − sin x sin h) − cos x = lim h→0 h . . . . . .
  125. 125. Derivatives of Sine and Cosine Fact d d sin x = cos x cos x = − sin x dx dx Proof. We already did the first. The second is similar (mutatis mutandis): d cos(x + h) − cos x cos x = lim dx h→0 h (cos x cos h − sin x sin h) − cos x = lim h→0 h cos h − 1 sin h = cos x · lim − sin x · lim h→0 h h→0 h . . . . . .
  126. 126. Derivatives of Sine and Cosine Fact d d sin x = cos x cos x = − sin x dx dx Proof. We already did the first. The second is similar (mutatis mutandis): d cos(x + h) − cos x cos x = lim dx h→0 h (cos x cos h − sin x sin h) − cos x = lim h→0 h cos h − 1 sin h = cos x · lim − sin x · lim h→0 h h→0 h = cos x · 0 − sin x · 1 = − sin x . . . . . .
  127. 127. What have we learned today? The Power Rule . . . . . .
  128. 128. What have we learned today? The Power Rule The derivative of a sum is the sum of the derivatives The derivative of a constant multiple of a function is that constant multiple of the derivative . . . . . .
  129. 129. What have we learned today? The Power Rule The derivative of a sum is the sum of the derivatives The derivative of a constant multiple of a function is that constant multiple of the derivative The derivative of sine is cosine The derivative of cosine is the opposite of sine. . . . . . .

×