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  • 1. Slopes and Derivatives
  • 2. (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run Geometry of Slope Δy = y2 – y1 = the difference in the heights of the points. Δx = x2 – x1 = the difference in the run of the points. Slopes and Derivatives
  • 3. (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run Geometry of Slope Δy = y2 – y1 = the difference in the heights of the points. Δx = x2 – x1 = the difference in the run of the points. Δy Δx=The slope m is the ratio of the “rise” to the “run”. Slopes and Derivatives
  • 4. (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run Geometry of Slope Δy = y2 – y1 = the difference in the heights of the points. Δx = x2 – x1 = the difference in the run of the points. Δy Δx=The slope m is the ratio of the “rise” to the “run”. The slope measures the tilt of a line in relation to the horizon, that is, the steepness in relation to the x axis. Slopes and Derivatives
  • 5. (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run Geometry of Slope Δy = y2 – y1 = the difference in the heights of the points. Δx = x2 – x1 = the difference in the run of the points. Δy Δx=The slope m is the ratio of the “rise” to the “run”. The slope measures the tilt of a line in relation to the horizon, that is, the steepness in relation to the x axis. Therefore horizontal lines have its steepness or slope = 0 . Slopes and Derivatives
  • 6. (x1, y1) (x2, y2) Δy=y2–y1=rise Δx=x2–x1=run Geometry of Slope Δy = y2 – y1 = the difference in the heights of the points. Δx = x2 – x1 = the difference in the run of the points. Δy Δx=The slope m is the ratio of the “rise” to the “run”. * http://www.mathwarehouse.com/algebra/linear_equation/interactive- slope.php The slope measures the tilt of a line in relation to the horizon, that is, the steepness in relation to the x axis. Therefore horizontal lines have its steepness or slope = 0 . Steeper lines have “larger” slopes*. Slopes and Derivatives
  • 7. Algebra of Slope Slopes and Derivatives
  • 8. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Slopes and Derivatives
  • 9. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Δx = x2 – x1 = the difference in the inputs x Slopes and Derivatives
  • 10. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Δx = x2 – x1 = the difference in the inputs x Algebraically the slope m = Δy/Δx = Δy : Δx is the ratio the difference in the outputs the difference in the inputs: Slopes and Derivatives
  • 11. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Δx = x2 – x1 = the difference in the inputs x Algebraically the slope m = Δy/Δx = Δy : Δx is the ratio the difference in the outputs the difference in the inputs: The units of this ratio are (units of y) / (units of x). Slopes and Derivatives
  • 12. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Δx = x2 – x1 = the difference in the inputs x Algebraically the slope m = Δy/Δx = Δy : Δx is the ratio the difference in the outputs the difference in the inputs: The units of this ratio are (units of y) / (units of x). This is also the amount of change in y for each unit change in x. Slopes and Derivatives
  • 13. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Δx = x2 – x1 = the difference in the inputs x Algebraically the slope m = Δy/Δx = Δy : Δx is the ratio the difference in the outputs the difference in the inputs: The units of this ratio are (units of y) / (units of x). This is also the amount of change in y for each unit change in x. The ratio “2 eggs : 3 cakes” is the same as “2/3 egg per cake”. Slopes and Derivatives
  • 14. Algebra of Slope Δy = y2 – y1 = the difference in the outputs y Δx = x2 – x1 = the difference in the inputs x Algebraically the slope m = Δy/Δx = Δy : Δx is the ratio the difference in the outputs the difference in the inputs: The units of this ratio are (units of y) / (units of x). This is also the amount of change in y for each unit change in x. The ratio “2 eggs : 3 cakes” is the same as “2/3 egg per cake”. The slope m is “the amount of change of y if x changes by one unit. Slopes and Derivatives
  • 15. Example A. We had 6 gallons of gas at the start of a trip and the odometer was registered at 75,000 miles. Two hours later, the gas gauge indicated there were 3 gallons left and the odometer was at 75,300 miles. Let x = the gas–tank indicator reading y = the odometer reading t = the time in hours. Find the following slopes. Slopes and Derivatives
  • 16. Example A. We had 6 gallons of gas at the start of a trip and the odometer was registered at 75,000 miles. Two hours later, the gas gauge indicated there were 3 gallons left and the odometer was at 75,300 miles. Let x = the gas–tank indicator reading y = the odometer reading t = the time in hours. Find the following slopes. a. Compare the measurements of the fuel amount x versus the distance y traveled, or (x, y), Slopes and Derivatives
  • 17. Example A. We had 6 gallons of gas at the start of a trip and the odometer was registered at 75,000 miles. Two hours later, the gas gauge indicated there were 3 gallons left and the odometer was at 75,300 miles. Let x = the gas–tank indicator reading y = the odometer reading t = the time in hours. Find the following slopes. a. Compare the measurements of the fuel amount x versus the distance y traveled, or (x, y). We have (6, 75000), (3, 75300). Slopes and Derivatives
  • 18. The slope m = (75,300 – 75,000) / (3 – 6) = –100 mpg. Example A. We had 6 gallons of gas at the start of a trip and the odometer was registered at 75,000 miles. Two hours later, the gas gauge indicated there were 3 gallons left and the odometer was at 75,300 miles. Let x = the gas–tank indicator reading y = the odometer reading t = the time in hours. Find the following slopes. a. Compare the measurements of the fuel amount x versus the distance y traveled, or (x, y). We have (6, 75000), (3, 75300). Slopes and Derivatives
  • 19. The slope m = (75,300 – 75,000) / (3 – 6) = –100 mpg. So the distance–to–fuel rate or the fuel efficiency is 100 miles per gallon. Example A. We had 6 gallons of gas at the start of a trip and the odometer was registered at 75,000 miles. Two hours later, the gas gauge indicated there were 3 gallons left and the odometer was at 75,300 miles. Let x = the gas–tank indicator reading y = the odometer reading t = the time in hours. Find the following slopes. a. Compare the measurements of the fuel amount x versus the distance y traveled, or (x, y). We have (6, 75000), (3, 75300). Slopes and Derivatives
  • 20. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). Slopes and Derivatives
  • 21. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). we have (0, 75000), (2, 75300). Slopes and Derivatives
  • 22. The slope n = (75,300 – 70,000) / (2 – 0) = 150 mph. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). we have (0, 75000), (2, 75300). Slopes and Derivatives
  • 23. So the distance–to–time rate (or the velocity) per hour is 150 miles per hour. The slope n = (75,300 – 70,000) / (2 – 0) = 150 mph. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). we have (0, 75000), (2, 75300). Slopes and Derivatives
  • 24. So the distance–to–time rate (or the velocity) per hour is 150 miles per hour. The slope n = (75,300 – 70,000) / (2 – 0) = 150 mph. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). we have (0, 75000), (2, 75300). c. Compare the measurements of the time t versus the amount of fuel x, or (t, x), Slopes and Derivatives
  • 25. So the distance–to–time rate (or the velocity) per hour is 150 miles per hour. The slope n = (75,300 – 70,000) / (2 – 0) = 150 mph. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). we have (0, 75000), (2, 75300). the slope n = (6 – 3) / (0 – 2) = –1.5 gph. c. Compare the measurements of the time t versus the amount of fuel x, or (t, x). We have (0, 6), (2, 3). The rate of change is Slopes and Derivatives
  • 26. So the distance–to–time rate (or the velocity) per hour is 150 miles per hour. The slope n = (75,300 – 70,000) / (2 – 0) = 150 mph. b. Compare the measurements of the time t versus the distance y traveled, or (t, y). we have (0, 75000), (2, 75300). So the fuel–to–time rate of the fuel consumption per hour is 1.50 gallon per hour. c. Compare the measurements of the time t versus the amount of fuel x, or (t, x). We have (0, 6), (2, 3). The rate of change is Slopes and Derivatives the slope n = (6 – 3) / (0 – 2) = –1.5 gph.
  • 27. Question: What are the reciprocals of the above rates and what do they measure? Slopes and Derivatives
  • 28. Question: What are the reciprocals of the above rates and what do they measure? Slopes measure steepness of straight lines. Slopes and Derivatives
  • 29. Question: What are the reciprocals of the above rates and what do they measure? Slopes measure steepness of straight lines. We want to extend this system of geometric measurement to measuring “steepness” of curves. Slopes and Derivatives
  • 30. Question: What are the reciprocals of the above rates and what do they measure? Slopes measure steepness of straight lines. We want to extend this system of geometric measurement to measuring “steepness” of curves. A curve has “ups” and “downs” so a curve has different “slopes” at different points. Slopes and Derivatives
  • 31. Question: What are the reciprocals of the above rates and what do they measure? x P y= f(x) Slopes measure steepness of straight lines. We want to extend this system of geometric measurement to measuring “steepness” of curves. Q A curve has “ups” and “downs” so a curve has different “slopes” at different points. This may be seen in the figure shown at points P and Q. Slopes and Derivatives
  • 32. Question: What are the reciprocals of the above rates and what do they measure? x P y= f(x) Slopes measure steepness of straight lines. We want to extend this system of geometric measurement to measuring “steepness” of curves. Q A curve has “ups” and “downs” so a curve has different “slopes” at different points. This may be seen in the figure shown at points P and Q. Obviously the “slope” at the point P should be positive, and the “slope” at the point Q should be negative. Slopes and Derivatives
  • 33. Question: What are the reciprocals of the above rates and what do they measure? x P y= f(x) Slopes measure steepness of straight lines. We want to extend this system of geometric measurement to measuring “steepness” of curves. Q A curve has “ups” and “downs” so a curve has different “slopes” at different points. This may be seen in the figure shown at points P and Q. We define the “slope at a point P on the curve y = f(x)” to be “the slope of the tangent line to y = f(x) at P” . Obviously the “slope” at the point P should be positive, and the “slope” at the point Q should be negative. Slopes and Derivatives
  • 34. Slopes and Derivatives Derivatives
  • 35. Slopes and Derivatives Derivatives If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x).
  • 36. Slopes and Derivatives x P y= f(x) Derivatives If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x).
  • 37. Slopes and Derivatives x P y= f(x) Derivatives If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x). The slope at a point P is also called the derivative at P.
  • 38. Slopes and Derivatives x P y= f(x) Derivatives If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x). The slope at a point P is also called the derivative at P. By “well defined” we mean that the geometric notion of “the tangent line at P” is intuitive and unambiguous so its slope is unambiguous.
  • 39. Slopes and Derivatives x P y= f(x) Derivatives By “well defined” we mean that the geometric notion of “the tangent line at P” is intuitive and unambiguous so its slope is unambiguous. We will examine the notion of the “tangent line” in the next section. For now, we accept the tangent line intuitively as a straight line that leans against y = f(x) at P as shown. If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x). The slope at a point P is also called the derivative at P.
  • 40. Slopes and Derivatives x P y= f(x) Derivatives By “well defined” we mean that the geometric notion of “the tangent line at P” is intuitive and unambiguous so its slope is unambiguous. Furthermore, it is well defined because we are able to compute the slopes of tangents at different locations algebraically. If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x). The slope at a point P is also called the derivative at P.
  • 41. Slopes and Derivatives x P y= f(x) Derivatives By “well defined” we mean that the geometric notion of “the tangent line at P” is intuitive and unambiguous so its slope is unambiguous. Furthermore, it is well defined because we are able to compute the slopes of tangents at different locations algebraically. We will carry out these computations for the 2nd degree example from the last section. If f(x) is an elementary function, then the “slope” of the tangent is well defined for “most” of the points on its graph y = f(x). The slope at a point P is also called the derivative at P.
  • 42. Example B. Given f(x) = x2 – 2x + 2 a. Find the slope of the cord connecting the points (x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2. Slopes and Derivatives
  • 43. Example B. Given f(x) = x2 – 2x + 2 a. Find the slope of the cord connecting the points (x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2. Substitute the values of x and h, we are find the slopes of the cord connecting (2, f(2)=2) and (2.2, f(2.2)). (2.2, f(2.2)) (2, 2) 2 2.2 0.2 Slopes and Derivatives
  • 44. Example B. Given f(x) = x2 – 2x + 2 a. Find the slope of the cord connecting the points (x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2. f(x+h) – f(x) h Substitute the values of x and h, we are find the slopes of the cord connecting (2, f(2)=2) and (2.2, f(2.2)). Its slope is f(2.2) – f(2) 0.2 = (2, 2) 2 2.2 0.2 (2.2, f(2.2)) Slopes and Derivatives
  • 45. Example B. Given f(x) = x2 – 2x + 2 a. Find the slope of the cord connecting the points (x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2. f(x+h) – f(x) h Substitute the values of x and h, we are find the slopes of the cord connecting (2, f(2)=2) and (2.2, f(2.2)). Its slope is f(2.2) – f(2) 0.2 = = 2.44 – 2 0.2 = 2.2 (2, 2) 2 2.2= 0.44 0.2 0.44 0.2 slope m = 2.2 (2.2, f(2.2)) Slopes and Derivatives
  • 46. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). Slopes and Derivatives
  • 47. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) Slopes and Derivatives
  • 48. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). We are to simplify the difference quotient formula with f(x) = x2 – 2x + 2 at x = 2. (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) Slopes and Derivatives
  • 49. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). f(2+h) – f(2) h We are to simplify the difference quotient formula with f(x) = x2 – 2x + 2 at x = 2. (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) Slopes and Derivatives
  • 50. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). f(2+h) – f(2) h We are to simplify the difference quotient formula with f(x) = x2 – 2x + 2 at x = 2. = [(2+h)2 – 2(2+h) + 2] – (2) h (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) Slopes and Derivatives
  • 51. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). f(2+h) – f(2) h We are to simplify the difference quotient formula with f(x) = x2 – 2x + 2 at x = 2. = [(2+h)2 – 2(2+h) + 2] – (2) h h2 + 2h h = (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) Slopes and Derivatives
  • 52. b. Given f(x) = x2 – 2x + 2, simplify the formula for the slope of the cord connecting the points (2, 2) and (2+h, f(2+h)). f(2+h) – f(2) h We are to simplify the difference quotient formula with f(x) = x2 – 2x + 2 at x = 2. = [(2+h)2 – 2(2+h) + 2] – (2) h h2 + 2h h = h + 2 = (2+h, f(2+h) (2, 2) 2 2 + h h slope = h + 2 f(2+h)–f(2) Slopes and Derivatives
  • 53. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. Slopes and Derivatives
  • 54. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2, 2) 2 y = x2–2x+2 Slopes and Derivatives
  • 55. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. The cord is fixed at the base–point (2, 2) and the other end depends on the value h. (2, 2) 2 y = x2–2x+2 Slopes and Derivatives
  • 56. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) The cord is fixed at the base–point (2, 2) and the other end depends on the value h. y = x2–2x+2 Slopes and Derivatives
  • 57. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2+h, f(2+h) (2, 2) 2 2 + h h f(2+h)–f(2) The cord is fixed at the base–point (2, 2) and the other end depends on the value h. y = x2–2x+2 Slopes and Derivatives
  • 58. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2+h, f(2+h) (2, 2) 2 2 + h f(2+h)–f(2) The cord is fixed at the base–point (2, 2) and the other end depends on the value h. As h varies, we obtained different cords with different slopes. h y = x2–2x+2 Slopes and Derivatives
  • 59. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2+h, f(2+h) (2, 2) 2 2 + h f(2+h)–f(2) The cord is fixed at the base–point (2, 2) and the other end depends on the value h. As h varies, we obtained different cords with different slopes. As h gets larger, the cords deviates away from the tangent line at (2, 2) and as h gets smaller the corresponding cords swing and settle toward the tangent line. h y = x2–2x+2 Slopes and Derivatives
  • 60. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2+h, f(2+h) (2, 2) 2 2 + h f(2+h)–f(2) The cord is fixed at the base–point (2, 2) and the other end depends on the value h. As h varies, we obtained different cords with different slopes. As h gets larger, the cords deviates away from the tangent line at (2, 2) and as h gets smaller the corresponding cords swing and settle toward the tangent line. These cords have slopes 2 + h. slope = 2 + h h y = x2–2x+2 Slopes and Derivatives
  • 61. c. Now we deduce the slope at the point (2, 2) in the following geometric argument. (2+h, f(2+h) (2, 2) 2 2 + h slope = 2 + h f(2+h)–f(2) The cord is fixed at the base–point (2, 2) and the other end depends on the value h. As h varies, we obtained different cords with different slopes. As h gets larger, the cords deviates away from the tangent line at (2, 2) and as h gets smaller the corresponding cords swing and settle toward the tangent line. These cords have slopes 2 + h. Hence as h “shrinks” to 0, the slope or the derivative at (2, 2) must be 2. h y = x2–2x+2 Slopes and Derivatives
  • 62. d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. Slopes and Derivatives
  • 63. Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. Slopes and Derivatives
  • 64. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. y = x2–2x+2 Slopes and Derivatives
  • 65. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). The difference quotient is h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. f(x+h) – f(x) h y = x2–2x+2 Slopes and Derivatives
  • 66. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). The difference quotient is h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. f(x+h) – f(x) h = (x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2] h y = x2–2x+2 Slopes and Derivatives
  • 67. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). The difference quotient is h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. f(x+h) – f(x) h = (x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2] h 2xh – 2h + h2 h = y = x2–2x+2 Slopes and Derivatives
  • 68. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). The difference quotient is h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. f(x+h) – f(x) h = (x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2] h 2xh – 2h + h2 h = 2x – 2 + h = y = x2–2x+2 slope = 2x–2+h Slopes and Derivatives
  • 69. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). The difference quotient is h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. f(x+h) – f(x) h = (x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2] h 2xh – 2h + h2 h = 2x – 2 + h = As h shrinks to 0, the slopes of the cords approach the value 2x – 2 y = x2–2x+2 slope = 2x–2+h Slopes and Derivatives
  • 70. (x+h, f(x+h) (x, f(x)) x x + h f(x+h)–f(x) Connect the cord at the base–point (x, f(x)) to the other end point at (x+h, f(x+h)). The difference quotient is h d. Simplify the difference quotient of f(x) then find the slope at the point (x, f(x)) – as a formula in x. f(x+h) – f(x) h = (x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2] h 2xh – 2h + h2 h = 2x – 2 + h. = As h shrinks to 0, the slopes of the cords approach the value 2x – 2 which must be the slope at (x, f(x)). y = x2–2x+2 slope = 2x–2+h Slopes and Derivatives
  • 71. Let’s summarize the result. Slopes and Derivatives
  • 72. Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. Slopes and Derivatives
  • 73. Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. y = x2–2x+2 Slopes and Derivatives
  • 74. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. y = x2–2x+2 Slopes and Derivatives
  • 75. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. y = x2–2x+2 Slopes and Derivatives
  • 76. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. y = x2–2x+2 slope at x = 2x – 2 Slopes and Derivatives
  • 77. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) y = x2–2x+2 slope at x = 2x – 2 Slopes and Derivatives
  • 78. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) – it’s read as “f prime of x”. y = x2–2x+2 slope at x = 2x – 2 Slopes and Derivatives
  • 79. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) – it’s read as “f prime of x”. So the derivative of f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2. y = x2–2x+2 slope at x = 2x – 2 Slopes and Derivatives
  • 80. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) – it’s read as “f prime of x”. So the derivative of f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2. y = x2–2x+2 The name derivative came from the fact that f’(x) = 2x – 2 is derived from f(x) = x2 – 2x + 2. slope at x = 2x – 2 Slopes and Derivatives
  • 81. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) – it’s read as “f prime of x”. So the derivative of f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2. y = x2–2x+2 The name derivative came from the fact that f’(x) = 2x – 2 is derived from f(x) = x2 – 2x + 2. The derivative f ’(x) = 2x – 2 tells us the slopes of f(x) = x2 – 2x + 2. slope at x = 2x – 2 Slopes and Derivatives
  • 82. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) – it’s read as “f prime of x”. So the derivative of f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2. y = x2–2x+2 The name derivative came from the fact that f’(x) = 2x – 2 is derived from f(x) = x2 – 2x + 2. The derivative f ’(x) = 2x – 2 tells us the slopes of f(x) = x2 – 2x + 2. For example the slope at x = 2 is f ’(2) = 2, at x = 3 is f ’(3) = 4, etc… slope at x = 2x – 2 Slopes and Derivatives
  • 83. (x, f(x)) x Let’s summarize the result. If f(x) = x2 – 2x + 2, then the slope at the point (x, f(x)) is 2x – 2. This slope–formula is the derivative of f(x) and it is written as f ’(x) – it’s read as “f prime of x”. So the derivative of f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2. y = x2–2x+2 The name derivative came from the fact that f’(x) = 2x – 2 is derived from f(x) = x2 – 2x + 2. The derivative f ’(x) = 2x – 2 tells us the slopes of f(x) = x2 – 2x + 2. For example the slope at x = 2 is f ’(2) = 2, at x = 3 is f ’(3) = 4, etc… The derivative f ’(x) is the extension of the concept of slopes of lines to curves. slope at x = 2x – 2 Slopes and Derivatives
  • 84. The first topic of the calculus course is derivatives. Slopes and Derivatives
  • 85. The first topic of the calculus course is derivatives. We will examine derivative algebraically by * developing the language of derivative which makes the concept more rigorous Slopes and Derivatives
  • 86. The first topic of the calculus course is derivatives. We will examine derivative algebraically by * developing the language of derivative which makes the concept more rigorous * developing the computation techniques for finding derivatives of all elementary functions Slopes and Derivatives
  • 87. The first topic of the calculus course is derivatives. We will examine derivative algebraically by * developing the language of derivative which makes the concept more rigorous * developing the computation techniques for finding derivatives of all elementary functions Slopes and Derivatives We will examine derivative geometrically
  • 88. The first topic of the calculus course is derivatives. We will examine derivative algebraically by * developing the language of derivative which makes the concept more rigorous * developing the computation techniques for finding derivatives of all elementary functions Slopes and Derivatives We will examine derivative geometrically by * developing the relations between derivatives f ’(x) and the shape of the graph of y = f(x)
  • 89. The first topic of the calculus course is derivatives. We will examine derivative algebraically by * developing the language of derivative which makes the concept more rigorous * developing the computation techniques for finding derivatives of all elementary functions Slopes and Derivatives We will examine derivative geometrically by * developing the relations between derivatives f ’(x) and the shape of the graph of y = f(x) * developing the computation procedures using f ’(x) to locate special positions on y = f(x)
  • 90. The first topic of the calculus course is derivatives. We will examine derivative algebraically by * developing the language of derivative which makes the concept more rigorous * developing the computation techniques for finding derivatives of all elementary functions Slopes and Derivatives We will examine derivative geometrically by * developing the relations between derivatives f ’(x) and the shape of the graph of y = f(x) * developing the computation procedures using f ’(x) to locate special positions on y = f(x) We will also investigate the applications of the derivatives which include problems of optimization, rates of change and numerical methods.