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Chapter 1 The Derivative

Chapter 1 The Derivative. Chapter Outline. The Slope of a Straight Line The Slope of a Curve at a Point The Derivative Limits and the Derivative Differentiability and Continuity Some Rules for Differentiation More About Derivatives The Derivative as a Rate of Change. § 1.1.

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Chapter 1 The Derivative

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  1. Chapter 1The Derivative

  2. Chapter Outline • The Slope of a Straight Line • The Slope of a Curve at a Point • The Derivative • Limits and the Derivative • Differentiability and Continuity • Some Rules for Differentiation • More About Derivatives • The Derivative as a Rate of Change

  3. §1.1 The Slope of a Straight Line

  4. Section Outline • Nonvertical Lines • Positive and Negative Slopes of Lines • Interpretation of a Graph • Properties of the Slope of a Nonvertical Line • Finding the Slope and y-Intercept of a Line • Sketching the Graph of a Line • Making Equations of Lines • Slope as a Rate of Change

  5. Nonvertical Lines

  6. Lines – Positive Slope EXAMPLE The following are graphs of equations of lines that have positive slopes.

  7. Lines – Negative Slope EXAMPLE The following are graphs of equations of lines that have negative slopes.

  8. Interpretation of a Graph EXAMPLE A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then her pay is y = 5x + 60 dollars. Give an interpretation of the slope and the y-intercept of this straight line. SOLUTION First, let’s graph the line to help us understand the exercise. (80, 460)

  9. Interpretation of a Graph CONTINUED The slope is 5, or 5/1. Since the numerator of this fraction represents the amount of change in her pay relative to the amount of change in her sales, the denominator, for every 1 sale that she makes, her pay increases by 5 dollars. The y-intercept is 60 and occurs on the graph at the point (0, 60). This point suggests that when she has executed 0 sales, her pay is 60 dollars. This $60 could be referred to as her base pay.

  10. Properties of the Slope of a Nonvertical Line

  11. Properties of the Slope of a Line

  12. Finding Slope and y-intercept of a Line EXAMPLE Find the slope and y-intercept of the line SOLUTION First, we write the equation in slope-intercept form. This is the given equation. Divide both terms of the numerator of the right side by 3. Rewrite Since the number being multiplied by x is 1/3, 1/3 is the slope of the line. Since the other 1/3 is the number being added to the term containing x, 1/3, or (0, 1/3), is the y-intercept. Incidentally, it was a complete coincidence that the slope and y-intercept were the same number. This does not normally occur.

  13. Sketching Graphs of Lines EXAMPLE Sketch the graph of the line passing through (-1, 1) with slope ½. SOLUTION We use Slope Property 1. We begin at the given point (-1, 1) and from there, move up one unit and to the right two units to find another point on the line. (-1, 1) (-1, 1)

  14. Sketching Graphs of Lines CONTINUED Now we connect the two points that have already been determined, since two points determine a straight line.

  15. Making Equations of Lines EXAMPLE Find an equation of the line that passes through the points (-1/2, 0) and (1, 2). SOLUTION To find an equation of the line that passes through those two points, we need a point (we already have two) and a slope. We do not yet have the slope so we must find it. Using the two points we will determine the slope by using Slope Property 2. We now use Slope Property 3 to find an equation of the line. To use this property we need the slope and a point. We can use either of the two points that were initially provided. We’ll use the second (the first would work just as well).

  16. Making Equations of Lines CONTINUED This is the equation from Property 3. (x1, y1) = (1, 2) and m = 4/3. Distribute. Add 2 to both sides of the equation. NOTE: Technically, we could have stopped when the equation looked like since it is an equation that represents the line and is equivalent to our final equation.

  17. Making Equations of Lines EXAMPLE Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x. SOLUTION To find an equation of the line, we need a point (we already have one) and a slope. We do not yet have the slope so we must find it. We know that the line we desire is perpendicular to the line y = 2x. Using Slope Property 5, we know that the product of the slope of the line desired and the slope of the line y = 2x is -1. We recognize that the line y = 2x is in slope-intercept form and therefore the slope of the line is 2. We can now find the slope of the line that we desire. Let the slope of the new line be m. (slope of a line)(slope of a new line) = -1 This is Property 5. 2m = -1 The slope of one line is 2 and the slope of the desired line is denoted by m. m = -0.5 Divide. Now we can find the equation of the desired line using Property 3.

  18. Making Equations of Lines CONTINUED This is the equation from Property 3. (x1, y1) = (2, 0) and m = -0.5. Distribute.

  19. Slope as a Rate of Change EXAMPLE Compute the rate of change of the function over the given intervals. SOLUTION We first get y by itself in the given equation. This is the given equation. Subtract 2x from both sides. Since this is clearly a linear function (since it’s now in slope-intercept form) it has constant slope, namely -2. Therefore, by definition, it also has a constant rate of change, -2. Therefore, no matter what interval is considered for this function, the rate of change will be -2. Therefore the answer, for both intervals, is -2.

  20. §1.2 The Slope of a Curve at a Point

  21. Section Outline • Tangent Lines • Slopes of Curves • Slope of a Curve as a Rate of Change • Interpreting the Slope of a Graph • Finding the Equation and Slope of the Tangent Line of a Curve

  22. Tangent Lines

  23. Slope of a Curve & Tangent Lines

  24. Slope of a Graph EXAMPLE Estimate the slope of the curve at the designated point P. SOLUTION The slope of a graph at a point is by definition the slope of the tangent line at that point. The figure above shows that the tangent line at P rises one unit for each unit change in x. Thus the slope of the tangent line at P is

  25. Slope of a Curve: Rate of Change

  26. Interpreting Slope of a Graph EXAMPLE • Refer to the figure below to decide whether the following statements about the debt per capita are correct or not. Justify your answers . • The debt per capita rose at a faster rate in 1980 than in 2000. • The debt per capita was almost constant up until the mid-1970s and then rose at an almost constant rate from the mid-1970s to the mid-1980s.

  27. Interpreting Slope of a Graph CONTINUED SOLUTION (a) The slope of the graph in 1980 is marked in red and the slope of the graph in 2000 is marked in blue, using tangent lines. It appears that the slope of the red line is the steeper of the two. Therefore, it is true that the debt per capita rose at a faster rate in 1980.

  28. Interpreting Slope of a Graph CONTINUED (b) Since the graph is a straight, nearly horizontal line from 1950 until the mid-1970s, marked in red, it is therefore true that the debt per capita was almost constant until the mid-1970s. Further, since the graph is a nearly straight line from the mid-1970s to the mid-1980s, marked in blue, it is therefore true that the debt per capita rose at an almost constant rate during those years.

  29. Equation & Slope of a Tangent Line EXAMPLE Find the slope of the tangent line to the graph of y = x2 at the point (-0.4, 0.16) and then write the corresponding equation of the tangent line. SOLUTION The slope of the graph of y = x2 at the point (x, y) is 2x.The x-coordinate of (-0.4, 0.16) is -0.4, so the slope of y = x2 at this point is 2(-0.4) = -0.8. We shall write the equation of the tangent line in point-slope form. The point is (-0.4, 0.16) and the slope (which we just found) is -0.8. Hence the equation is:

  30. §1.3 The Derivative

  31. Section Outline • The Derivative • Differentiation • Slope and the Derivative • Equation of the Tangent Line to the Graph of y =f(x) at (a,f(a)) • Leibniz Notation for Derivatives • Calculating Derivatives Via the Difference Quotient • Differentiable • Limit Definition of the Derivative • Limit Calculation of the Derivative

  32. The Derivative

  33. Differentiation

  34. Differentiation Examples These examples can be summarized by the following rule.

  35. Differentiation Examples EXAMPLE Find the derivative of SOLUTION This is the given equation. Rewrite the denominator as an exponent. Rewrite with a negative exponent. What we’ve done so far has been done for the sole purpose of rewriting the function in the form of f(x) = xr. Use the Power Rule where r = -1/7 and then simplify.

  36. Differentiation Examples EXAMPLE Find the slope of the curve y = x5 at x = -2. SOLUTION We must first find the derivative of the given function. This is the given function. Use the Power Rule. Since the derivative function yields information about the slope of the original function, we can now use to determine the slope of the original function at x = -2. Replace x with -2. Evaluate. Therefore, the slope of the original function at x = -2 is 80 (or 80/1).

  37. Equation of the Tangent Line to the Graph of y = f(x) at the point (a, f(a))

  38. Equation of the Tangent Line EXAMPLE Find the equation of the tangent line to the graph of f(x) = 3x at x = 4. SOLUTION We must first find the derivative of the given function. This is the given function. Differentiate. Notice that in this case the derivative function is a constant function, 3. Therefore, at x = 4, or any other value, the value of the derivative will be 3. So now we use the Equation of the Tangent Line that we just saw. This is the Equation of the Tangent Line. f(4) = 12 and Simplify.

  39. Leibniz Notation for Derivatives Ultimately, this notation is a better and more effective notation for working with derivatives.

  40. Calculating Derivatives Via the Difference Quotient The Difference Quotient is

  41. Calculating Derivatives Via the Difference Quotient EXAMPLE Apply the three-step method to compute the derivative of the following function: SOLUTION STEP 1: We calculate the difference quotient and simplify as much as possible. This is the difference quotient. Evaluate f(x + h) and f(x). Simplify. Simplify. Simplify.

  42. Calculating Derivatives Via the Difference Quotient CONTINUED Factor. Cancel and simplify. STEP 2: As h approaches zero, the expression -2x – h approaches -2x, and hence the difference quotient approaches -2x. STEP 3: Since the difference quotient approaches the derivative of f(x) = -x2 + 2 as h approaches zero, we conclude that

  43. Differentiable

  44. Limit Definition of the Derivative

  45. Limit Calculation of the Derivative EXAMPLE Using limits, apply the three-step method to compute the derivative of the following function: SOLUTION This is the difference quotient. STEP 1: Evaluate f(x + h) and f(x). Simplify. Simplify. Simplify.

  46. Limit Calculation of the Derivative CONTINUED Factor. Cancel and simplify. STEP 2: As h approaches 0, the expression -2x – h + 0.5 approaches -2x + 0.5. STEP 3: Since the difference quotient approaches , we conclude that

  47. §1.4 Limits and the Derivative

  48. Section Outline • Definition of the Limit • Finding Limits • Limit Theorems • Using Limits to Calculate a Derivative • Limits as x Increases Without Bound

  49. Definition of the Limit

  50. Finding Limits EXAMPLE Determine whether the limit exists. If it does, compute it. SOLUTION Let us make a table of values of x approaching 4 and the corresponding values of x3 – 7. As x approaches 4, it appears that x3 – 7 approaches 57. In terms of our notation,

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