© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 115 Chapter 1 The Derivative.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 2 of 115  The Slope of a Straight Line  The Slope of a Curve at a Point  The Derivative  Limits and the Derivative  Some Rules for Differentiation  More About Derivatives  The Derivative as a Rate of Change Chapter Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 3 of 115 DefinitionExample Equations of Nonvertical Lines: A nonvertical line L has an equation of the form The number m is called the slope of L and the point (0, b) is called the y-intercept. The equation above is called the slope-intercept equation of L. For this line, m = 3 and b = -4. Nonvertical Lines

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 4 of 115 Lines – Positive SlopeEXAMPLE The following are graphs of equations of lines that have positive slopes.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 5 of 115 Lines – Negative SlopeEXAMPLE The following are graphs of equations of lines that have negative slopes.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 6 of 115 Interpretation of a GraphEXAMPLE SOLUTION 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. First, let’s graph the line to help us understand the exercise. (80, 460)

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 7 of 115 Interpretation of a Graph 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. CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 10 of 115 Finding Slope and y -intercept of a LineEXAMPLE SOLUTION Find the slope and y-intercept of the line 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.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 11 of 115 Sketching Graphs of LinesEXAMPLE SOLUTION Sketch the graph of the line passing through (-1, 1) with slope ½. 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)

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 12 of 115 Sketching Graphs of Lines Now we connect the two points that have already been determined, since two points determine a straight line. CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 13 of 115 Making Equations of LinesEXAMPLE SOLUTION Find an equation of the line that passes through the points (-1/2, 0) and (1, 2). 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).

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 14 of 115 Making Equations of LinesCONTINUED This is the equation from Property 3. (x 1, y 1 ) = (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.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 15 of 115 Making Equations of LinesEXAMPLE SOLUTION Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x. 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. This is Property 5. (slope of a line)(slope of a new line) = -1 The slope of one line is 2 and the slope of the desired line is denoted by m. 2m = -1 Divide.m = -0.5 Now we can find the equation of the desired line using Property 3.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 16 of 115 Making Equations of LinesCONTINUED This is the equation from Property 3. (x 1, y 1 ) = (2, 0) and m = -0.5. Distribute.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 17 of 115 Slope as a Rate of ChangeEXAMPLE SOLUTION Compute the rate of change of the function over the given intervals. 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.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 18 of 115 Tangent Lines DefinitionExample Tangent Line to a Circle at a Point P: The straight line that touches the circle at just the one point P

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 19 of 115 Slope of a Curve & Tangent Lines DefinitionExample The Slope of a Curve at a Point P: The slope of the tangent line to the curve at P (Enlargements)

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 20 of 115 Slope of a GraphEXAMPLE SOLUTION Estimate the slope of the curve at the designated point P. 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

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 22 of 115 Equation & Slope of a Tangent LineEXAMPLE Find the slope of the tangent line to the graph of y = x 2 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 = x 2 at the point (x, y) is 2x. The x-coordinate of (-0.4, 0.16) is -0.4, so the slope of y = x 2 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:

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 23 of 115 The Derivative DefinitionExample Derivative: The slope formula for a function y = f (x), denoted: Given the function f (x) = x 3, the derivative is

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 24 of 115 Differentiation DefinitionExample Differentiation: The process of computing a derivative. No example will be given at this time since we do not yet know how to compute derivatives. But don’t worry, you’ll soon be able to do basic differentiation in your sleep.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 26 of 115 Differentiation ExamplesEXAMPLE SOLUTION Find the derivative of 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) = x r. Use the Power Rule where r = -1/7 and then simplify.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 27 of 115 Differentiation ExamplesEXAMPLE SOLUTION Find the slope of the curve y = x 5 at x = -2. This is the given function. We must first find the derivative of 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).

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 29 of 115 Equation of the Tangent LineEXAMPLE SOLUTION Find the equation of the tangent line to the graph of f (x) = 3x at x = 4. This is the given function. We must first find the derivative of 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.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 30 of 115 Leibniz Notation for Derivatives Ultimately, this notation is a better and more effective notation for working with derivatives.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 32 of 115 Calculating Derivatives Via the Difference QuotientEXAMPLE SOLUTION Apply the three-step method to compute the derivative of the following function: This is the difference quotient. STEP 1: We calculate the difference quotient and simplify as much as possible. Evaluate f (x + h) and f (x). Simplify.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 33 of 115 Calculating Derivatives Via the Difference Quotient STEP 2: As h approaches zero, the expression -2x – h approaches -2x, and hence the difference quotient approaches -2x. Factor. CONTINUED Cancel and simplify. STEP 3: Since the difference quotient approaches the derivative of f (x) = -x 2 + 2 as h approaches zero, we conclude that

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 35 of 115 Limit Calculation of the DerivativeEXAMPLE SOLUTION Using limits, apply the three-step method to compute the derivative of the following function: This is the difference quotient. Evaluate f (x + h) and f (x). Simplify. STEP 1:

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 36 of 115 Limit Calculation of the Derivative Factor. CONTINUED 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

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 39 of 115 Using Limits to Calculate a DerivativeEXAMPLE SOLUTION Use limits to compute the derivative for the function We must calculate

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 40 of 115 Using Limits to Calculate a DerivativeCONTINUED Now that replacing h with 0 will not cause the denominator to be equal to 0, we use Limit Theorem VIII.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 41 of 115 More Work With Derivatives and LimitsEXAMPLE SOLUTION Match the limit with a derivative. Then find the limit by computing the derivative. The idea here is to identify the given limit as a derivative given by for a specific choice of f and x. Toward this end, let us rewrite the limit as follows. Now go back to. Take f (x) = 1/x and evaluate according to the limit definition of the derivative:

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 42 of 115 More Work With Derivatives and Limits On the right side we have the desired limit; while on the left side can be computed using the power rule (where r = -1): Hence, CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 43 of 115 Limits as x Increases Without BoundEXAMPLE SOLUTION Calculate the following limit. Both 10x + 100 and x 2 – 30 increase without bound as x does. To determine the limit of their quotient, we employ an algebraic trick. Divide both numerator and denominator by x 2 (since the highest power of x in either the numerator or the denominator is 2) to obtain As x increases without bound, 10/x approaches 0, 100/x 2 approaches 0, and 30/x 2 approaches 0. Therefore, as x increases without bound, 10/x + 100/x 2 approaches 0 + 0 = 0 and 1 - 30/x 2 approaches 1 – 0 = 1. Therefore,

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 46 of 115 DifferentiationSOLUTION Differentiate This is the given function. EXAMPLE We begin to differentiate. Rewrite the rational expression with a negative exponent. Use the General Power Rule taking x 3 + x + 1 to be g(x). Use the Sum Rule.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 48 of 115 DifferentiationSOLUTION Differentiate This is the given function. EXAMPLE We begin to differentiate. Rewrite the rational expression with a negative exponent. Use the General Power Rule taking to be g(x). Use the Constant Multiple Rule.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 49 of 115 Differentiation Use the Sum Rule and rewrite as x 1/2. CONTINUED Differentiate. Simplify.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 51 of 115 The Derivative as a Rate of ChangeSOLUTION Let S(x) represent the total sales (in thousands of dollars) for month x in the year 2005 at a certain department store. Represent each statement below by an equation involving S or. EXAMPLE (a) The sales at the end of January reached $120,560 and were rising at the rate of $1500 per month. (b) At the end of March, the sales for this month dropped to $80,000 and were falling by about $200 a day (Use 1 month = 30 days). (a) Since the sales at the end of January (the first month, so x = 1) reached $120,560 and S(x) represents the amount of sales for a given month, we have: S(1) = 120,560. Further, since the rate of change of sales (rate of change means we will use the derivative of S(x)) for the month of January is a positive $1500 per month, we have: (b) At the end of March (the third month, so x = 3), the sales dropped to $80,000. Therefore, sales for the month of March was $80,000. That is: S(3) = 80,000. Additionally, since sales were dropping by $200 per day during March, this means that the rate of change of the function S(x) was (30 days) x (-200 dollars) = -6000 dollars per month. Therefore, we have:

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 52 of 115 Differentiating Various Independent VariablesSOLUTION Find the first derivative. EXAMPLE We first note that the independent variable is t and the dependent variable is T. This is significant inasmuch as they are considered to be two totally different variables, just as x and y are different from each other. We now proceed to differentiate the function. This is the given function. We begin to differentiate. Use the Sum Rule. Use the General Power Rule.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 53 of 115 Differentiating Various Independent Variables Finish differentiating. CONTINUED Simplify.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 54 of 115 Second DerivativesSOLUTION Find the first and second derivatives. EXAMPLE This is the given function. This is the first derivative. This is the second derivative.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 55 of 115 Second Derivatives Evaluated at a PointSOLUTION Compute the following. EXAMPLE Compute the first derivative. Compute the second derivative. Evaluate the second derivative at x = 2.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 57 of 115 Average Rate of ChangeSOLUTION Suppose that f (x) = -6/x. What is the average rate of change of f (x) over the interval 1 to 1.2? EXAMPLE The average rate of change over the interval is

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 59 of 115 Instantaneous Rate of ChangeSOLUTION Suppose that f (x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1? EXAMPLE The rate of change of f (x) at x = 1 is equal to. We have That is, the rate of change is 6 units per unit change in x.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 60 of 115 Average Velocity DefinitionExample Average Velocity: Given a position function s(t), the average velocity from time t = a to t = a + h is Suppose a car is 3 miles from its starting point after 5 minutes and 7 miles from its starting point after an additional 6 minutes (after a total of 11 minutes). The average velocity of the car between the two given locations is miles per minute where a = 5 and h = 6.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 61 of 115 Position, Velocity & Acceleration s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 62 of 115 Position, Velocity & AccelerationSOLUTION A toy rocket fired straight up into the air has height s(t) = 160t – 16t 2 feet after t seconds. EXAMPLE (a) What is the rocket’s initial velocity (when t = 0)? (b) What is the velocity after 2 seconds? (c) What is the acceleration when t = 3? (d) At what time will the rocket hit the ground? (e) At what velocity will the rocket be traveling just as it smashes into the ground? (a) To determine what the rocket’s initial velocity is, we must first have a velocity function and then evaluate t = 0 in that function. This is the given position function. Differentiate to get the velocity function.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 63 of 115 Position, Velocity & Acceleration Therefore, the initial velocity of the rocket is 160 feet per second. Now replace t with 0 and evaluate. CONTINUED (b) To determine the velocity after 2 seconds we evaluate v(2). This is the velocity function. Replace t with 2 and evaluate. Therefore, the velocity of the rocket after 2 seconds is 96 feet per second. (c) To determine the acceleration when t = 3, we must first find the acceleration function. This is the velocity function. Differentiate to get the acceleration function. Since the acceleration function is a constant function, the acceleration of the rocket is a constant -32 ft/s 2. Therefore, the acceleration when t = 3 is -32ft/s 2.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 64 of 115 Position, Velocity & AccelerationCONTINUED Therefore, the rocket will be 0 feet above the ground at times t = 0 and t = 10. t = 0 corresponds to when the rocket first began its flight, so that would not be the solution. So, the rocket hit the ground after 10 seconds, when t – 10 = 0. (d) To determine at what time the rocket will hit the ground we must determine what we know about the position, velocity, or acceleration of the rocket when the rocket hits the ground. We know that at the time of impact, the position of the rocket will be 0 feet above the ground. Therefore, we will use the position function and replace s(t) with 0. This is the given position function. Replace s(t) with 0. Factor 16 out of both terms on the right and divide both sides by 16. Factor. Solve for t.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 65 of 115 Position, Velocity & AccelerationCONTINUED Therefore, when the rocket hits the ground, it will be have a velocity of -160 ft/s. That is, it will be traveling 160 ft/s in the downward direction. (e) To determine at what velocity the rocket will be traveling just as it smashes into the ground, we must use the velocity function. The question is, what do we use for t? From part (d), we know that the rocket will hit the ground at t = 10 seconds. Therefore, we will find v(10). This is the velocity function. Replace t with 10 and evaluate.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 115 Chapter 1 The Derivative.

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© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 115 Chapter 1 The Derivative.

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Presentation on theme: "© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 115 Chapter 1 The Derivative."— Presentation transcript:

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