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- 1. Section 2.3 Basic Differentiation Rules V63.0121.002.2010Su, Calculus I New York University May 24, 2010 Announcements Homework 1 due Tuesday Quiz 2 Thursday in class on Sections 1.5–2.5 . . . . . .
- 2. Announcements Homework 1 due Tuesday Quiz 2 Thursday in class on Sections 1.5–2.5 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 2 / 36
- 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 3 / 36
- 4. Outline 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 4 / 36
- 5. Derivative of the squaring function Example Suppose f(x) = x2 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 5 / 36
- 6. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 5 / 36
- 7. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 5 / 36
- 8. 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 + 2xh + h − x2 x2 = lim h→0 h . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 5 / 36
- 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) (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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 5 / 36
- 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 2 + 2xh + h − x2 x2 2x + h¡ h 2 = lim = lim h→0 h h→0 h = lim (2x + h) = 2x. h→0 So f′ (x) = 2x. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 5 / 36
- 11. Derivative of the cubing function Example Suppose f(x) = x3 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 6 / 36
- 12. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 6 / 36
- 13. 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 + 3x h + 3xh + h − x3 2 x3 = lim h→0 h . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 6 / 36
- 14. 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 + 3x2 h 2 + 3xh + h − 3 x3 3x2 h ¡ ! 2 ! ¡ 3 + 3xh + h = lim = lim h→0 h h→0 h . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 6 / 36
- 15. 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 + 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→0 So f′ (x) = 3x2 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 6 / 36
- 16. The cubing function and its derivatives y . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 7 / 36
- 17. The cubing function and its derivatives y . f . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 7 / 36
- 18. The cubing function and its derivatives y . Notice that f is increasing, .′ f and f′ 0 except f′ (0) = 0 f . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 7 / 36
- 19. The cubing function and its derivatives y . Notice that f is increasing, .′′ .′ f f and f′ 0 except f′ (0) = 0 f . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 7 / 36
- 20. 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) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 7 / 36
- 21. Derivative of the square root function Example √ Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 8 / 36
- 22. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 8 / 36
- 23. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 8 / 36
- 24. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 8 / 36
- 25. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 8 / 36
- 26. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 8 / 36
- 27. The square root function and its derivatives y . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 9 / 36
- 28. The square root function and its derivatives y . f . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 9 / 36
- 29. The square root function and its derivatives y . f . Here lim+ f′ (x) = ∞ and f x→0 . .′ f is not differentiable at 0 x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 9 / 36
- 30. 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→∞ . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 9 / 36
- 31. Derivative of the cube root function Example √ Suppose f(x) = 3 x = x1/3 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36
- 32. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36
- 33. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36
- 34. 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) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36
- 35. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36
- 36. 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 3x2/3 So f′ (x) = 1 x−2/3 . . . . . . . 3 V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36
- 37. The cube root function and its derivatives y . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 11 / 36
- 38. The cube root function and its derivatives y . f . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 11 / 36
- 39. The cube root function and its derivatives y . Here lim f′ (x) = ∞ and f is f . x→0 not differentiable at 0 . .′ f x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 11 / 36
- 40. 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→±∞ . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 11 / 36
- 41. One more Example Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 12 / 36
- 42. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 12 / 36
- 43. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 12 / 36
- 44. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 12 / 36
- 45. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 12 / 36
- 46. The function x → x2/3 and its derivative y . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 13 / 36
- 47. The function x → x2/3 and its derivative y . f . . x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 13 / 36
- 48. The function x → x2/3 and its derivative y . f is not differentiable at 0 f . and lim f′ (x) = ±∞ x→0± . .′ f x . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 13 / 36
- 49. 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→±∞ . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 13 / 36
- 50. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 51. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 52. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 53. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 54. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 55. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 56. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 57. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 58. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 59. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 60. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 61. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 62. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 14 / 36
- 63. 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 15 / 36
- 64. The other Tower of Power . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 16 / 36
- 65. Outline 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 17 / 36
- 66. 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) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 67. 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−k n copies k=0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 68. 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 69. 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 70. 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 71. 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 72. 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) = (x + h) · (x + h) · (x + h) · · · (x + h) = n ck xk hn−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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 18 / 36
- 73. 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 . ... ... . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 19 / 36
- 74. 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 . ... ... . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 19 / 36
- 75. 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 . ... ... . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 19 / 36
- 76. 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 . ... ... . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 19 / 36
- 77. Theorem (The Power Rule) Let r be a positive whole number. Then d r x = rxr−1 dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 20 / 36
- 78. 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 n−1 = nx + (stuff with at least one h in it) and this tends to nxn−1 as h → 0. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 20 / 36
- 79. The Power Rule for constants Theorem Let c be a constant. Then d c=0 dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 21 / 36
- 80. The Power Rule for constants Theorem d 0 l .ike x = 0x−1 Let c be a constant. Then dx d c=0. dx (although x → 0x−1 is not defined at zero.) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 21 / 36
- 81. The Power Rule for constants Theorem d 0 l .ike x = 0x−1 Let 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 h So f′ (x) = lim 0 = 0. h→0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 21 / 36
- 82. 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. . . . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 22 / 36
- 83. 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′ . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 23 / 36
- 84. Proof. Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 24 / 36
- 85. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 24 / 36
- 86. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 24 / 36
- 87. 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 24 / 36
- 88. 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′ . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 25 / 36
- 89. Proof. Again, follow your nose. (cf)(x + h) − (cf)(x) (cf)′ (x) = lim h→0 h . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 26 / 36
- 90. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 26 / 36
- 91. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 26 / 36
- 92. 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) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 26 / 36
- 93. Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 27 / 36
- 94. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 27 / 36
- 95. 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 3 d 4 d 12 = 2 x + x − 17 x + 0 dx dx dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 27 / 36
- 96. 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 3 d 4 d 12 = 2 x + x − 17 x + 0 dx dx dx = 2 · 3x + 4x − 17 · 12x11 2 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 27 / 36
- 97. 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 3 d 4 d 12 = 2 x + x − 17 x + 0 dx dx dx = 2 · 3x + 4x − 17 · 12x11 2 3 = 6x2 + 4x3 − 204x11 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 27 / 36
- 98. Outline 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 28 / 36
- 99. Derivatives of Sine and Cosine Fact d sin x = ??? dx Proof. From the definition: . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 29 / 36
- 100. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 29 / 36
- 101. Angle addition formulas See Appendix A . sin(A + B) = sin . cos B + cos A sin B A cos(A + B) = cos A cos B − sin A sin B . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 30 / 36
- 102. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 31 / 36
- 103. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 31 / 36
- 104. Two important trigonometric limits See 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 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 32 / 36
- 105. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 33 / 36
- 106. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 33 / 36
- 107. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 33 / 36
- 108. Illustration of Sine and Cosine y . . x . . π −2 . π 0 . .π . π 2 s . in x . . . . . .
- 109. 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 34 / 36
- 110. 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? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 34 / 36
- 111. Derivatives of Sine and Cosine Fact d d sin x = cos x cos x = − sin x dx dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 35 / 36
- 112. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 35 / 36
- 113. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 35 / 36
- 114. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 35 / 36
- 115. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 35 / 36
- 116. Summary d r Power Rule: x = rxr−1 dx Sum, Difference, Constant, and Constant Multiple Rules Derivative of sin is cos. Derivative of cos is − sin. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 36 / 36

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