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Splash Screen. Find an equation for the slope of the graph of y = x 3 + 3 x at any point. A. m = 3 x 2 B. m = 3 x 2 + 3 x C. m = 3 x 2 + 3 D. m = 3 x 4 + 3 x 2. 5–Minute Check 2. You calculated the slope of tangent lines to find the instantaneous rate of change. (Lesson 12-3).

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  1. Splash Screen

  2. Find an equation for the slope of the graph of y = x3+ 3x at any point. A.m = 3x2 B.m = 3x2 + 3x C.m = 3x2 + 3 D.m = 3x4 + 3x2 5–Minute Check 2

  3. You calculated the slope of tangent lines to find the instantaneous rate of change. (Lesson 12-3) • Find instantaneous rates of change by calculating derivatives. • Use the Product and Quotient Rules to calculate derivatives. Then/Now

  4. derivative • differentiation • differential equation • differential operator Vocabulary

  5. Definition of Derivative f(x + h) = 2(x + h)3 + 2(x + h) – 7(x + h) + 12 Derivative of a Function at Any Point Find the derivative of f(x) = 2x3 + 2x2 – 7x + 12. Then evaluate the derivative at x = 1 and x = 4. Example 1

  6. Factor. Divide by h. Expand and simplify. Derivative of a Function at Any Point Example 1

  7. Sum and Difference Properties of Limits and Limits of Constant and Identity Functions Original equation x = 1 Simplify. Derivative of a Function at Any Point The derivative of f(x) is f ′(x) = 6x2 + 4x – 7. Evaluate f ′(x) for x = 1 and x = 4. Example 1

  8. Original equation x = 1 and 4 Simplify. Derivative of a Function at Any Point Answer: f ′(x) = 6x2 + 4x – 7; f ′(1) = 3, f ′(4) = 105 Example 1

  9. Find the derivative of f(x) = –2x4 + 3x2 – 5x. Then evaluate it at x = –1. A. f΄(x) = –8x3 + 6x; f΄(–1) = –14 B. f΄(x) = –8x3 + 6x; f΄(–1) = 2 C. f΄(x) = –8x3 + 6x – 5; f΄(–1) = –19 D. f΄(x) = –8x3 + 6x – 5; f΄(–1) = –3 Example 1

  10. Key Concept 2

  11. Power Rule for Derivatives A. Find the derivative of the function f(x) = x5. f(x) = x5 Original equation f′(x) = 5x5 – 1 Power Rule = 5x4 Simplify. Answer:f′(x) = 5x4 Example 2

  12. B. Find the derivative of the function . Original equation Rewrite using a rational exponent. Power Rule Simplify. Answer:g ′(x) = Power Rule for Derivatives Example 2

  13. C. Find the derivative of the function . Original equation Rewrite using a negative exponent. Power Rule Simplify. Answer:h ′(x) = Power Rule for Derivatives Example 2

  14. Find the derivative of . A. B. C. D. Example 2

  15. Key Concept 3

  16. Derivative Rules A. Find the derivative of the function f(x) = 6x2 – 3. Answer:f ′(x) = 12x Example 3

  17. Derivative Rules B. Find the derivative of the functiong(x) = 2x3 (5x – 3). Answer:g′(x) = 40x3 – 18x2 Example 3

  18. C. Find the derivative of the function . Original equation Divide each term in the numerator by x. Divide out the common factor. Derivative Rules Example 3

  19. Constant, Constant Multiple of a Power, and Sum and Difference Rules Simplify. Derivative Rules Answer:h′(x) = 6x – 2 Example 3

  20. Find the derivative of . A. B. C. D. Example 3

  21. Instantaneous Velocity PARTICLES The distance a particle moves along a path is defined by s(t) = 6t – 2t3 + 4, where t is given in seconds and the distance of the particle is given in millimeters. Find the expression for the instantaneous velocity v(t) of the particle. The instantaneous velocity v(t) is equivalent to s′(t). s(t) = 6t – 2t3 + 4 Original equation s′(t) = 6 • 1t1 – 1 – 2 •3t3 – 1 + 0 Constant, Constant Multiple of a Power, and Sum and Difference Rules Example 4

  22. Instantaneous Velocity = 6 – 6t2 Simplify. The instantaneous velocity is v(t) = 6 – 6t2. Answer:v(t) = 6 – 6t2 Example 4

  23. PARTICLES The distance a particle moves along a path is defined by s(t) = –t3 – 4t2 – 7, where t is given in seconds and the distance of the particle is given in meters. Find the expression for the instantaneous velocity v(t) of the particle. A. v(t) = –3t2 – 8t – 1 B. v(t) = –3t2 – 8t C. v(t) = 3t2 + 8t D. v(t) = 3t2 + 8t + 1 Example 4

  24. Key Concept 3

  25. Maximum and Minimum of a Graph TRAMPOLINE The height h in feet of a person jumping off a trampoline can be defined by h(t) = 4 + 5t – 2t2 on the interval [0, 3], where time t is given in seconds. Find the maximum and minimum heights of the jump. Find the derivative of h(t). h(t) = 4 + 5t – 2t2 Original equation h′(t) = 0 + 5 •1t1 – 1 – 2 • 2t2 – 1 Constant, Constant Multiple of a Power, and Sum and Difference Rules Example 5

  26. Original equation h′(t) = 5 – 4t Subtract 5 from both sides and divide both sides by –4. A critical point for this function occurs at . Evaluate h(t) for t = 0, , and t = 3. Maximum and Minimum of a Graph = 5 – 4t Simplify. Solve h′(t) = 0 to find the critical points. Example 5

  27. maximum minimum Maximum and Minimum of a Graph The person jumping off a trampoline will achieve a maximum height of 7.125 feet at 1.25 seconds and a minimum height of 1 foot at 3 seconds. Answer:maximum of 7.125 ft at 1.25 seconds; minimum of 1 ft at 3 seconds Example 5

  28. Maximum and Minimum of a Graph CHECK The graph of h(t) = 4 + 5t – 2t2supports the result that at 1.25 seconds the maximum height is 7.125 feet and at 3 seconds the minimum height is 1 foot.  Example 5

  29. BALL The height of a bouncing ball can be defined by h(t) = 2 + 7t – 3t2 on the interval [0,2], where t is given in seconds and h is given feet. Find the maximum and minimum heights of the ball. A. maximum of 4 feet at 2 seconds; minimum of 2 feet at 0 seconds B. maximum of approximately 6.1 feet at approximately 1.2 seconds; minimum of 4 feet at 2 seconds C. maximum of approximately 6.1 feet at approximately 1.2 seconds; minimum of 2 feet at 0 seconds D. maximum of 4 feet at 2 seconds; there is no minimum height on the interval [0, 2] Example 5

  30. Key Concept 6

  31. Product Rule A. Find the derivative of the functionh(x) = (x2 – 2x + 3) •(x3 – 4). Let f(x) = x2 – 2x + 3 and g(x) = x3 – 4 Example 6

  32. Product Rule Use f(x), f′(x), g(x), and g′(x) to find the derivative of h(x). = 5x4 – 8x3 + 9x2 – 8x + 8 Simplify. Answer:h′(x) = 5x4 – 8x3 + 9x2 – 8x + 8 Example 6

  33. Product Rule B. Find the derivative of the functionh(x) = (x4 – x2 + 2) • (x3 – x + 1). Let f(x) = x4 – x2 + 2 and g(x) = x3 – x + 1 Example 6

  34. Product Rule Use f(x), f′(x), g(x), and g′(x) to find the derivative of h(x). Example 6

  35. Find the derivative of . A. B. C. D. Example 6

  36. Key Concept 7

  37. A. Find the derivative of the function . Let f(x) = 4x3 and g(x) = x2 – 2. So, . Quotient Rule Example 7

  38. Quotient Rule Substitution Distributive Property Simplify. Quotient Rule Use f(x), f′(x), g(x), and g′(x) to find the derivative of h(x). Example 7

  39. Answer: Quotient Rule Example 7

  40. B. Find the derivative of the function . Let f(x) = x3 – 4 and g(x) = x2 + 2. So, . Quotient Rule f(x) = x3 – 4 Original equation f′(x) = 3x2 Power, Constant, and Difference Rules g(x) = x2 + 1 Original equation g′(x) = 2x Power, Constant, and Sum Rules Example 7

  41. Quotient Rule Substitution Expand and simplify. Quotient Rule Use f(x), f′(x), g(x), and g′(x) to find the derivative of h(x). Example 7

  42. Answer: Quotient Rule Example 7

  43. Find the derivative of . A. B. C. D. Example 7

  44. derivative • differentiation • differential equation • differential operator Vocabulary

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