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# Lesson 8: Basic Differentiation Rules (Section 41 slides)

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### Lesson 8: Basic Differentiation Rules (Section 41 slides)

1. 1. Section 2.3 Basic Differentiation Rules V63.0121.041, Calculus I New York University September 29, 2010 Announcements Last chance for extra credit on Quiz 1: Do the get-to-know you survey and photo by October 1.. . . . . . .
2. 2. Announcements Last chance for extra credit on Quiz 1: Do the get-to-know you survey and photo by October 1. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 2 / 42
3. 3. Objectives Understand and use these differentiation rules: the derivative of a constant function (zero); the Constant Multiple Rule; the Sum Rule; the Difference Rule; the derivatives of sine and cosine. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 3 / 42
4. 4. Recall: the derivativeDefinitionLet f be a function and a a point in the domain of f. If the limit f(a + h) − f(a) f(x) − f(a) f′ (a) = lim = lim h→0 h x→a x−aexists, the function is said to be differentiable at a and f′ (a) is thederivative of f at a.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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 4 / 42
5. 5. Notation Newtonian notation Leibnizian notation dy d df f′ (x) y′ (x) y′ f(x) dx dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 5 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 6 / 42
7. 7. OutlineDerivatives so far Derivatives of power functions by hand The Power RuleDerivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple RuleDerivatives of sine and cosine . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 7 / 42
8. 8. Derivative of the squaring functionExampleSuppose f(x) = x2 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
9. 9. Derivative of the squaring functionExampleSuppose f(x) = x2 . Use the definition of derivative to find f′ (x).Solution f(x + h) − f(x) f′ (x) = lim h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
10. 10. Derivative of the squaring functionExampleSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
11. 11. Derivative of the squaring functionExampleSuppose 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     + 2xh + h −   x2 x2 = lim h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
12. 12. Derivative of the squaring functionExampleSuppose 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     + 2xh + h −   x2 x2 2x + h¡ h 2 = lim = lim h→0 h h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
13. 13. Derivative of the squaring functionExampleSuppose 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     + 2xh + h −   x2 x2 2x + h¡ h 2 = lim = lim h→0 h h→0 h = lim (2x + h) = 2x. h→0So f′ (x) = 2x. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
14. 14. The second derivativeIf 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! . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 9 / 42
15. 15. The second derivativeIf 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! Leibniziannotation: d2 y d2 d2 f f(x) dx2 dx2 dx2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 9 / 42
16. 16. The squaring function and its derivatives y . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42
17. 17. The squaring function and its derivatives y . f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42
20. 20. Derivative of the cubing functionExampleSuppose f(x) = x3 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 42
21. 21. Derivative of the cubing functionExampleSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 42
22. 22. Derivative of the cubing functionExampleSuppose 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     + 3x h + 3xh + h −   x3 2 x3 = lim h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 42
23. 23. Derivative of the cubing functionExampleSuppose 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 + 3x2 h 2 + 3xh + h −   3   x3 3x2 h ¡ ! 2 ! ¡ 3   + 3xh + h = lim = lim h→0 h h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 42
24. 24. Derivative of the cubing functionExampleSuppose 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   + 3xh + h −   x3 + 3x2 h 2 3   x3 3x2 h ¡ ! 2 ! ¡ 3   + 3xh + h = lim = lim h→0 ( h ) h→0 h 2 2 2 = lim 3x + 3xh + h = 3x . h→0So f′ (x) = 3x2 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 42
25. 25. The cubing function and its derivatives y . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42
26. 26. The cubing function and its derivatives y . f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42
27. 27. The cubing function and its derivatives y . Notice that f is increasing, .′ f and f′ 0 except f′ (0) = 0 f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42
28. 28. The cubing function and its derivatives y . Notice that f is increasing, .′′ .′ f f and f′ 0 except f′ (0) = 0 f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42
29. 29. The cubing function and its derivatives y . Notice that f is increasing, .′′ .′ f f and f′ 0 except f′ (0) = 0 Notice also that the f . tangent line to the graph of . x . f at (0, 0) crosses the graph (contrary to a popular “definition” of the tangent line) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42
30. 30. Derivative of the square root function.Example √Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x).. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
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 (x + h) − x ¡ ¡ = lim (√ √ ) h→0 h x+h+ x. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
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 (x + h) − x ¡ ¡ h = lim (√ √ ) = lim (√ √ ) h→0 h x+h+ x h→0 h x+h+ x. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
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 (x + h) − 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
36. 36. The square root function and its derivatives y . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42
37. 37. The square root function and its derivatives y . f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42
38. 38. The square root function and its derivatives y . f . Here lim+ f′ (x) = ∞ and f x→0 . .′ f is not differentiable at 0 x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42
39. 39. The square root function and its derivatives y . f . Here lim+ f′ (x) = ∞ and f x→0 . .′ f is not differentiable at 0 x . Notice also lim f′ (x) = 0 x→∞ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42
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).. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
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 (x + h) − x ¡ ¡ = lim ( 2/3 + (x + h)1/3 x1/3 + x2/3 ) h→0 h (x + h). . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
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 (x + h) − 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
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 (x + h) − 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 3x2/3So f′ (x) = 1 x−2/3 . 3. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
46. 46. The cube root function and its derivatives y . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42
47. 47. The cube root function and its derivatives y . f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42
48. 48. The cube root function and its derivatives y . Here lim f′ (x) = ∞ and f is f . x→0 not differentiable at 0 . .′ f x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42
49. 49. The cube root function and its derivatives y . Here lim f′ (x) = ∞ and f is f . x→0 not differentiable at 0 .′ f . x . Notice also lim f′ (x) = 0 x→±∞ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42
50. 50. One moreExampleSuppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 42
51. 51. One moreExampleSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 42
52. 52. One moreExampleSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 42
53. 53. One moreExampleSuppose 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 42
54. 54. One moreExampleSuppose 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 3So f′ (x) = 2 x−1/3 . 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 42
55. 55. The function x → x2/3 and its derivative y . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 42
56. 56. The function x → x2/3 and its derivative y . f . . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 42
57. 57. The function x → x2/3 and its derivative y . f is not differentiable at 0 f . and lim f′ (x) = ±∞ x→0± . .′ f x . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 42
58. 58. The function x → x2/3 and its derivative y . f is not differentiable at 0 f . and lim f′ (x) = ±∞ x→0± . .′ f x . Notice also lim f′ (x) = 0 x→±∞ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
65. 65. Recap: The Tower of Power y y′ x2 2x1 The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 x 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
66. 66. Recap: The Tower of Power y y′ x2 2x The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 x 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
67. 67. Recap: The Tower of Power y y′ x2 2x The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 x 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
68. 68. Recap: The Tower of Power y y′ x2 2x The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 x 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
69. 69. Recap: The Tower of Power y y′ x2 2x The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 x 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
70. 70. Recap: The Tower of Power y y′ x2 2x The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 x 2x 1 −2/3 x1/3 3x 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
71. 71. Recap: The Tower of Power y y′ x2 2x The power goes down by x 3 3x 2 one in each derivative 1/2 1 −1/2 The coefficient in the x 2x derivative is the power of 1 −2/3 x1/3 3x the original function 2 −1/3 x2/3 3x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
72. 72. The Power RuleThere is mounting evidence forTheorem (The Power Rule)Let r be a real number and f(x) = xr . Then f′ (x) = rxr−1as 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 20 / 42
73. 73. The other Tower of Power . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 21 / 42
74. 74. OutlineDerivatives so far Derivatives of power functions by hand The Power RuleDerivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple RuleDerivatives of sine and cosine . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 22 / 42
75. 75. Remember your algebraFactLet n be a positive whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
76. 76. Remember your algebraFactLet 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
77. 77. Remember your algebraFactLet 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0The coefficient of xn is 1 because we have to choose x from eachbinomial, and there’s only one way to do that. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
78. 78. Remember your algebraFactLet 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0The coefficient of xn is 1 because we have to choose x from eachbinomial, and there’s only one way to do that. The coefficient of xn−1 his the number of ways we can choose x n − 1 times, which is the sameas the number of different hs we can pick, which is n. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
79. 79. Remember your algebraFactLet 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0The coefficient of xn is 1 because we have to choose x from eachbinomial, and there’s only one way to do that. The coefficient of xn−1 his the number of ways we can choose x n − 1 times, which is the sameas the number of different hs we can pick, which is n. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
80. 80. Remember your algebraFactLet 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0The coefficient of xn is 1 because we have to choose x from eachbinomial, and there’s only one way to do that. The coefficient of xn−1 his the number of ways we can choose x n − 1 times, which is the sameas the number of different hs we can pick, which is n. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
81. 81. Remember your algebraFactLet 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0The coefficient of xn is 1 because we have to choose x from eachbinomial, and there’s only one way to do that. The coefficient of xn−1 his the number of ways we can choose x n − 1 times, which is the sameas the number of different hs we can pick, which is n. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
82. 82. Pascals 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 . ... ... . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 42
83. 83. Pascals 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 . ... ... . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 42
84. 84. Pascals 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 . ... ... . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 42
85. 85. Pascals 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 . ... ... . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 42
86. 86. Proving the Power Rule.Theorem (The Power Rule)Let n be a positive whole number. Then d n x = nxn−1 dx. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 25 / 42
87. 87. Proving the Power Rule.Theorem (The Power Rule)Let n be a positive whole number. Then d n x = nxn−1 dxProof.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.. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 25 / 42
88. 88. The Power Rule for constantsTheoremLet c be a constant. Then d c=0 dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42
89. 89. The Power Rule for constantsTheorem d 0 l .ike x = 0x−1Let c be a constant. Then dx d c=0. dx(although x → 0x−1 is not defined at zero.) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42
90. 90. The Power Rule for constantsTheorem d 0 l .ike x = 0x−1Let c be a constant. Then 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 hSo f′ (x) = lim 0 = 0. h→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42
91. 91. Calculus . . . . . .
92. 92. Recall the Limit LawsFactSuppose 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. . . . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 28 / 42
93. 93. Adding functionsTheorem (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′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 29 / 42
94. 94. Proof of the Sum RuleProof.Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 42
95. 95. Proof of the Sum RuleProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 42
96. 96. Proof of the Sum RuleProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 42
97. 97. Proof of the Sum RuleProof.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 thedifference quotients for for f and g exist, the limit of the sum is the sumof the limits. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 42
98. 98. Scaling functionsTheorem (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′ . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 31 / 42
99. 99. Proof of the Constant Multiple RuleProof.Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 42
100. 100. Proof of the Constant Multiple RuleProof.Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h cf(x + h) − cf(x) = lim h→0 h . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 42
101. 101. Proof of the Constant Multiple RuleProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 42
102. 102. Proof of the Constant Multiple RuleProof.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) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 42
103. 103. Derivatives of polynomialsExample d ( 3 )Find 2x + x4 − 17x12 + 37 dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42
104. 104. Derivatives of polynomialsExample d ( 3 )Find 2x + x4 − 17x12 + 37 dxSolution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42
105. 105. Derivatives of polynomialsExample d ( 3 )Find 2x + x4 − 17x12 + 37 dxSolution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx d 3 d 4 d 12 = 2 x + x − 17 x + 0 dx dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42
106. 106. Derivatives of polynomialsExample d ( 3 )Find 2x + x4 − 17x12 + 37 dxSolution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx d 3 d 4 d 12 = 2 x + x − 17 x + 0 dx dx dx = 2 · 3x + 4x − 17 · 12x11 2 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42
107. 107. Derivatives of polynomialsExample d ( 3 )Find 2x + x4 − 17x12 + 37 dxSolution d ( 3 ) 2x + x4 − 17x12 + 37 dx d ( 3) d d ( ) d = 2x + x4 + −17x12 + (37) dx dx dx dx d 3 d 4 d 12 = 2 x + x − 17 x + 0 dx dx dx = 2 · 3x + 4x − 17 · 12x11 2 3 = 6x2 + 4x3 − 204x11 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42
108. 108. OutlineDerivatives so far Derivatives of power functions by hand The Power RuleDerivatives of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Multiple RuleDerivatives of sine and cosine . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 34 / 42
109. 109. Derivatives of Sine and Cosine.Fact d sin x = ??? dxProof.From the definition:. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 35 / 42
110. 110. Derivatives of Sine and Cosine.Fact d sin x = ??? dxProof.From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 35 / 42
111. 111. Angle addition formulasSee Appendix A . sin(A + B) = sin A cos B + cos A sin B . cos(A + B) = cos A cos B − sin A sin B . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 36 / 42
112. 112. Derivatives of Sine and Cosine.Fact d sin x = ??? dxProof.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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 37 / 42
113. 113. Derivatives of Sine and Cosine.Fact d sin x = ??? dxProof.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. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 37 / 42
114. 114. Two important trigonometric limitsSee Section 1.4 . sin θ lim . =1 θ→0 θ . in θ . s θ cos θ − 1 lim =0 . θ θ→0 θ . . − cos θ 1 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 38 / 42
115. 115. Derivatives of Sine and CosineFact d sin x = ??? dxProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 39 / 42
116. 116. Derivatives of Sine and CosineFact d sin x = ??? dxProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 39 / 42
117. 117. Derivatives of Sine and CosineFact d sin x = cos x dxProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 39 / 42
118. 118. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 s . in x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42
119. 119. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 c . os x s . in x f(x) = sin x has horizontal tangents where f′ = cos(x) is zero. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42
120. 120. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 c . os x s . in x f(x) = sin x has horizontal tangents where f′ = cos(x) is zero. what happens at the horizontal tangents of cos? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42
121. 121. Derivatives of Sine and Cosine.Fact d d sin x = cos x cos x = − sin x dx dx. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42
122. 122. Derivatives of Sine and Cosine.Fact d d sin x = cos x cos x = − sin x dx dxProof.We already did the first. The second is similar (mutatis mutandis): d cos(x + h) − cos x cos x = lim dx h→0 h. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42