1. 1 2. Economic Applications of Single- Variable Calculus Derivative Origins Single Variable Derivatives Economic uses of Derivatives

2. 2 2. Economic Applications of Single- Variable Calculus 2.1 Derivatives of Single- Variable Functions 2.2 Applications using Derivatives

3. 3 2. Economic Applications of Single-Variable Calculus In economics, derivatives are used in various ways:  Marginal amounts (slope)  Maximization  Minimization  Sketching Graphs  Estimation

4. 4 2.1 Derivatives of Single-Variable Functions Slope: -consider the following graph -the curved movement between A and B is approximated by the red straight line

5. 5 2.1 – Quadratic example Slope Approximation: Slope = a decrease of 65 units over 4 time periods, or an average decrease of 16.25

6. 6 2.1 – Which is the slope at point B? Slope = rise/run =Δq/ Δp = (q 1 -q 0 )/(p 1 -p 0 ) Slope AB =(q 1 -q 0 )/(p 1 -p 0 ) =(35-100)/(3-1) =-65/2 =-32.5 Slope BC =(q 1 -q 0 )/(p 1 -p 0 ) =(20-35)/(5-3) =-15/2 =-7.5

7. 7 2.1 Derivatives of Single-Variable Functions -the slopes of these secants (AB and BC) reveal the rate of change of q in response to a change in p -these slopes change as you move along the curve -in order to find the slope AT B, one must use an INSTANTANEOUS SLOPE -slope of a tangent line

8. 8 2.1 – Tangents The green tangent line represents the instantaneous slope

9. 9 2.1 Instantaneous Slope To calculate an instantaneous slope (using calculus), you need: 1)A function 2)A continuous function 3)A smooth continuous function

10. 10 2.1.1 – A Function Definitions: -A function is any rule that assigns a maximum and minimum of one value to a range of another value -ie y=f(x) assigns one value (y) to each x -note that the same y can apply to many x’s, but each x has only one y -ie: y=x 1/2 is not a function x = argument of the function (domain of function) f(x) or y = range of function

11. 11 Function: Each X Has 1 Y y=0+2sin(2pi*x/14)+2cos(2pi*x/14) -4 -3 -2 0 1 2 3 4 135791113151719 x y x

12. 12 Not a Function: Here each x Corresponds to 2 y values Often called the straight line test

13. 13 2.1.1 – Continuous -if a function f(x) draws close to one finite number L for all values of x as x draws closer to but does not equal a, we say: lim f(x) = L x-> a A function is continuous iff (if and only iff) i) f(x) exists at x=a ii) Lim f(x) exists x->a iii) Lim f(x) = f(a) x->a

14. 14 2.1.1 – Limits and Continuity In other words: i)The point must exist ii)Points before and after must exist iii)These points must all be joined Or simply: The graph can be drawn without lifting one’s pencil.

15. 15 2.1.2 Smooth -in order for a derivative to exist, a function must be continuous and “smooth” (have only one tangent)

16. 16 2.1.2 Derivatives -if a derivative exists, it can be expressed in many different forms: a)dy/dx b)df(x)/dx c)f ’(x) d)F x (x) e)y’

17. 17 2.1.2 Derivatives and Limits -a derivative (instantaneous slope) is derived using limits: This method is known as differentiation by first principles, and determines the slope between A and B as AB collapses to a point (A)

18. 18 2.1.2 Rules of Derivatives -although first principles always work, the following rules are more economical: 1) Constant Rule If f(x)=k (k is a constant), f ‘(x) = 0 2) General Rule If f(x) = ax+b (a and b are constants) f ‘ (x) = a

19. 19 2.1.2 Examples of Derivatives 1) Constant Rule If f(x)=27 f ‘(x) = 0 2) General Rule If f(x) = 3x+12 f ‘(x) = 3

20. 20 2.1.2 Rules of Derivatives 3) Power Rule If f(x) = kx n, f ‘(x) = nkx n-1 4) Addition Rule If f(x) = g(x) + h(x), f ‘(x) = g’(x) + h’(x)

21. 21 2.1.2 Examples of Derivatives 3) Power Rule If f(x) = -9x 7, f ‘(x) = 7(-9)x 7-1 =-63x 6 4) Addition Rule If f(x) = 32x -9x 2 f ‘(x) = 32-18x

22. 22 2.1.2 Rules of Derivatives 5) Product Rule If f(x) =g(x)h(x), f ‘(x) = g’(x)h(x) + h’(x)g(x) -order doesn’t matter 6) Quotient Rule If f(x) =g(x)/h(x), f ‘(x) = {g’(x)h(x)-h’(x)g(x)}/{h(x) 2 } -order matters -derived from product rule (implicit derivative)

23. 23 2.1.2 Rules of Derivatives 5) Product Rule If f(x) =(12x+6)x 3 f ‘(x) = 12x 3 + (12x+6)3x 2 = 48x 3 + 18x 2 6) Quotient Rule If f(x) =(12x+1)/x 2 f ‘(x) = {12x 2 – (12x+1)2x}/x 4 = [-12x 2 -2x]/x 4 = [-12x-2]/x 3

24. 24 2.1.2 Rules of Derivatives 7) Power Function Rule If f(x) = [g(x)] n, f ‘(x) = n[g(x)] n-1 g’(x) -work from the outside in -special case of the chain rule 8) Chain Rule If f(x) = f(g(x)), let y=f(u) and u=g(x), then dy/dx = dy/du X du/dx

25. 25 2.1.2 Rules of Derivatives 7) Power Function Rule If f(x) = [3x+12] 4, f ‘(x) = 4[3x+12] 3 3 = 12 [3x+12] 3 8) Chain Rule If f(x) = (6x 2 +2x) 3, let y=u 3 and u=6x 2 +2x, dy/dx = dy/du X du/dx = 3u 2 (12x+2) = 3(6x 2 +2x) 2 (12x+2)

26. 26 2.1.2 More Exciting Derivatives 1) Inverses If f(x) = 1/x= x -1, f ‘(x) = -x -2 =-1/x 2 1b) Inverses and the Chain Rule If f(x) = 1/g(x)= g(x) -1, f ‘(x) = -g(x) -2 g’(x)=-1/g(x) 2 g’(x)

27. 27 2.1.2 – More Exciting Derivatives 2) Natural Logs If y=ln(x), y’ = 1/x -chain rule may apply If y=ln(x 2 ) y’ = (1/x 2 )2x = 2/x

28. 28 2.1.2 – More Exciting Derivatives 3) Trig. Functions If y = sin (x), y’ = cos(x) If y = cos(x) y’ = -sin(x) -We see this relationship graphically:

29. 29 2.1.2 – More Derivatives Reminder: derivatives reflect slope:

30. 30 2.1.2 – More Derivatives 3b) Trig. Functions – Chain Rule If y = sin 2 (3x+2), y’= 2sin(3x+2)cos(3x+2)3 Exercises: y=ln(2sin(x) -2cos 2 (x-1/x)) y=sin 3 (3x+2)ln(4x-7/x 3 ) 5 y=ln([3x+4]sin(x)) / cos(12xln(x))

31. 31 2.1.2 – More Derivatives 4) Exponents If y = b x y’ = b x ln(b) Therefore If y = e x y’ = e x

32. 32 2.1.2 – More Derivatives 4b) Exponents and chain rule If y = b kx y’ = b kx ln(b)k Or more generally: If y = b g(x) y’ = b g(x) ln(b) X dg(x)/dx

33. 33 2.1.2 – More Derivatives 4b) Exponents and chain rule If y = 5 2x y’ = 5 2x ln(5)2 Or more complicated: If y = 5 sin(x) y’ = 5 sin(x) ln(5) * cos(x)

34. 34 2.1.2.1 – Higher Order Derivatives -First order derivates (y’), show us the slope of a graph -Second order derivatives measure the instantaneous change in y’, or the slope of the slope -or the change in the slope: -(Higher-order derivates are also possible)

35. 35 Here the slope increases as t increases, transitioning from a negative slope to a positive slope. A second derivative would be positive, and confirm a minimum point on the graph. 2.1.2.1 Second Derivatives

36. 36 Here, the slope moves from positive to negative, decreasing over time. A second derivative would be negative and indicate a maximum point on the graph. 2.1.2.1 Second Derivatives

37. 37 2.1.2.1 – Second Order Derivatives To take a second order derivative: 1)Apply derivative rules to a function 2)Simplify if possible 3)Apply derivative rules to the answer to (1) Second order differentiation can be shown a variety of ways: a)d 2 y/dx 2 b) d 2 f(x)/dx 2 c)f ’’(x)d) f xx (x) e) y’’

38. 38 2.1.2.1 – Second Derivative Examples y=12x 3 +2x+11 y’=36x 2 +2 y’’=72x y=sin(x 2 ) y’=cos(x 2 )2x y’’=-sin(x 2 )2x(2x)+cos(x 2 )2 y’’’=-cos(x 2 )2x(4x 2 )-sin(x 2 )8x-sin(x 2 )2x(2) =-cos(x 2 )8x 3 -sin(x 2 )12x

39. 39 2.1.2.2 – Implicit Differentiation So far we’ve examined cases where our function is expressed: y=f(x) ie: y=7x+9x 2 -14 Yet often equations are expressed: 14=7x+9x 2 -y Which requires implicit differentiation. -In this case, y can be isolated. Often, this is not the case

40. 40 2.1.2.2 – Implicit Differentiation Rules 1)Take the derivative of EACH term on both sides. 2)Differentiate y as you would x, except that every time you differentiate y, you obtain dy/dx (or y’) Ie: 14=7x+9x 2 -y d(14)/dx=d(7x)/dx+d(9x 2 )/dx-dy/dx 0 = 7 + 18x – y’ y’=7+18x

41. 41 2.1.2.2 – Implicit Differentiation Examples Sometimes isolating y’ requires algebra: xy=15+x y+xy’=0+1 xy’=1-y y’=(1-y)/x (this can be simplified to remove y) = [1-(15+x)/x]/x = (x-15-x)/x 2 =(-15)/x 2

42. 42 2.1.2.2 – Implicit Differentiation Examples x 2 -2xy+y 2 =1 d(x 2 )/dx+d(2xy)/dx+d(y 2 )/dx=d1/dx 2x-2y-2xy’+2yy’=0 y’(2y-2x)=2y-2x y’=(2y-2x) / (2y-2x) y’=1

43. 43 2.1.2.2 – Implicit Differentiation Examples Using the implicit form has advantages: 3x+7y 8 =18 3+56y 7 y’=0 56y 7 y’=-3 y’=-3/56y 7 vrs. y=[(18-3x)/7) 1/8 y’=1/8 * [(18-3x)/7) -7/8 * 1/7 * (-3) Which simplifies to the above.

44. 44 2.2.1 Derivative Applications - Graphs Derivatives can be used to sketch functions: First Derivative: -First derivative indicates slope -if y’>0, function slopes upwards -if y’<0, function slopes downwards -if y’=0, function is horizontal -slope may change over time -doesn’t give shape of graph

45. 45 2.2.1 Positive Slope Graphs Linear, Quadratic, and Lin-Log Graphs

46. 46 2.2.1 Derivative Applications - Graphs Next, shape/concavity must be determined Second Derivative: -Second derivative indicates concavity -if y’’>0, slope is increasing (convex) -if y’’<0, slope is decreasing (concave, like a hill or a cave) -if y’’=0, slope is constant (or an inflection point occurs, see later)

47. 47 2.2.1 Sample Graphs x’’= 2, slope is increasing; graph is convex

48. 48 2.2.1 – Sample Graphs x’’=-2, slope is decreasing; graph is concave

49. 49 2.2.1 Derivative Applications - Graphs Maxima/minima can aid in drawing graphs Maximum Point: If1) f(a)’=0, and 2) f(a)’’<0, -graph has a maximum point (peak) at x=a Minimum Point: If1) f(a)’=0, and 2) f(a)’’>0, -graph has a minimum point (valley) at x=a

50. 50 2.2.1 Sample Graphs x’’= 2, slope is increasing; graph is convex x’=-10+2t=0 t=5

51. 51 2.2.1 – Sample Graphs x’’=-2, slope is decreasing; graph is concave x’=10-2t=0 t=5

52. 52 2.2.1 Derivative Applications - Graphs Inflection Points: If1) f(a)’’=0, and 2) the graph is not a straight line -then an inflection point occurs -(where the graph switches between convex and concave)

53. 53 2.2.1 Derivatives and Graphing Cyclical case:

54. 54 2.2.1 Derivative Applications - Graphs 7 Graphing Steps: i)Evaluate f(x) at extreme points (x=0, ∞, - ∞, or a variety of values) ii)Determine where f(x)=0 iii)Calculate slope: f ’(x) - and determine where it is positive and negative iv)Identify possible maximum and minimum co-ordinates where f ‘(x)=0. (Don’t just find the x values)

55. 55 2.2.1 Derivative Applications - Graphs 7 Graphing Steps: v) Calculate the second derivative – f ‘’(x) and use it to determine max/min in iv vi) Using the second derivative, determine the curvature (concave or convex) at other points vii) Check for inflection points where f ‘’(x)=0

56. 56 2.2.1 Graphing Example 1 y=(x-5) 2 -3 i)f(0)=22, f(∞)= ∞, f(-∞)=∞ ii)y=0 when (x-5) 2 =3 (x-5) = ± 3 1/2 x = ± 3 1/2 +5 x = 6.7, 3.3 (x-intercepts) iii) y’=2(x-5) y’>0 when x>5 y’<0 when x<5

57. 57 2.2.1 Graphing Example 1 y=(x-5) 2 -3 iv) y’=0 when x=5 f(5)=(5-5) 2 -3=-3 (5,-3) is a potential max/min v) y’’=2, (5,-3) is a minimum vi) Function is always positive, it is always convex vii) y’’ never equals zero

58. 58 2.2.1 Graphing Example 1 (0,22) (3.3,0)(6.7,0) (5,-3)

59. 59 2.2.1 Graphing Example 2 y=(x+1)(x-3)=x 2 -2x-3 i)f(0)=-3, f(∞)= ∞, f(-∞)=∞ ii)y=0 when (x+1)(x-3) =0 x = 3,-1 (x-intercepts) iii) y’=2x-2 y’>0 when x>1 y’<0 when x<1

60. 60 2.2.1 Graphing Example 2 y=(x+1)(x-3)=x 2 -2x-3 iv) y’=0 when x=1 f(1)=1 2 -2(1)-3=-4 (1,-4) is a potential max/min v) y’’=2, x=1 is a minimum vi) Function is always positive, it is always convex vii) y’’ never equals zero

61. 61 2.2.1 Graphing Example 2

62. 62 2.2.2 Optimization Some claim economists have 3 jobs: 1)Analyze what has happened (past) 2)Describe the current economy (present) 3)Advise on future decisions (future) For #3, an economist must first calculate the best possible result.

63. 63 2.2.2 Optimization Optimization falls into two categories: 1)Maximization 1)Maximization (ie: production, profits, utility, happiness, grades, health, employment, etc.) 2)Minimization 2)Minimization (ie: costs, pollution, disutility, unemployment, sickness, homework, etc.)

64. 64 2.2.2 Optimizing in 3 Steps There are three steps for optimization: 1)FIRST ORDER CONDITION (FOC) Find where f’(x)=0. These are potential maxima/minima. 2)SECOND ORDER CONDITION (SOC) Evaluate f’’(x) at your potential maxima/minima. This determines if (1)’s solutions are maxima/minima/inflection points 3)Co-Ordinates Obtain the co-ordinates of your maxima/minima

65. 65 2.2.2 Optimizing Example 1 Cooking is tricky – too long spent cooking, and it burns, too little time spent cooking – and some of it is raw and inedible. Production of oatmeal is expressed as: Let x=15+10t-t 2 x = bowls of oatmeal t = 5 minute intervals of time Maximize Oatmeal Production

66. 66 2.2.2 Optimous Oatmeal Let x=15+10t-t 2 FOC: x’ = 10-2t = 0 10= 2t 5= t SOC: x’’ = -2 x’’ < 0, concave, maximum

67. 67 2.2.2 Optimous Oatmeal Let x=15+10t-t 2 Co-ordinates: x(5)= 15+10(5)-5 2 x(5)= 15+50-25 = 40 Gourmet oatmeal production is maximized at 40 bowls when 25 minutes (5X5) are spent cooking (All else held equal).

68. 68 2.2.2 – Oatmeal for everyone Production is maximized at (5,40).

69. 69 2.2.2 Marriage and Motorcycles Steve wants to buy a new motorbike. Being a married man however, he knows that his utility is directly tied to his wife’s opinion of the idea. Furthermore, he knows it’s best to bring it up to Denise (his wife) when she’s at her weakest. Denise’s daily opposition to a motorcycle is expressed as x=-cos(tπ/12) Where t = hour of the day (0-24) When should Steve ask Denise?

70. 70 2.2.2 M & M Let x=-cos(tπ/12) FOC: x’ = sin(tπ/12)π/12 = 0 0= sin(tπ/12) This occurs when t π/12 = 0, π, 2π t= 0, 12, 24 2 possible mimima (0=24 on the clock)

71. 71 2.2.2 Early Bird Gets the Motorcycle Let x=-cos(tπ/12) SOC: x’’= cos(tπ/12)(π/12) 2 x’’> 0 when 0≤t<6, 18<t≤24 -convex, min (0, 24 are acceptable) x’’<0 when 6<t<18 -concave, max (12 is out) 0 or 24 6 12 18 + -

72. 72 2.2.2 Early Bird Gets the Motorcycle Let x=-cos(tπ/12) Results: x(0)=-1 x(24)=-1 Denise has her least resistance (of -1) at midnight. Steve’s best move is to bring up the motorcycle when Denise is tired or asleep.

73. 73 2.2.2 Early Bird Gets the Motorcycle Let x=-cos(tπ/12)

74. 74 2.2.2 Necessary and Sufficient The FOC provides a NECESSARY condition for a maximum or minimum. The FOC is not a SUFFICIENT condition for a maximum or minimum. The FOC and SOC together are NECESSARY AND SUFFICIENT conditions for a max. or min.

75. 75 2.2.2 Marginal Concepts Marginal Profit = Change in profit from the sale/production of one extra unit. MP=dπ/dq If Marginal Profit >0, quantity should increase; as the next unit will increase profit. If Marginal Profit <0, quantity should decrease, as the last unit decreased profit

76. 76 2.2.2 Marginal Concepts Quantity is therefore optimized when Marginal Profit=0; Ie: when dπ/dq=0. SOC still confirms that this is a maximum (ie: that the previous unit increased profit and the next unit will decrease profit.)

77. 77 2.2.2 Marginal Concepts Alternately, remember that Profit = Total Revenue – Total Costs Or π = TR-TC therefore Mπ=MR-MC (dπ/dq=dTR/dq-dTC/dq) So Mπ = 0 is equivalent to saying that MR-MC =0 MR=MC

78. 78 2.2.2 Marrigal Example Your significant other shows you a new, horrible outfit they just bought and asks how they look. You think of 6 possible lies: 1)You look amazing 2)You have such great taste 3)That colour really brings out your eyes 4)Neon is in this year 5)I should get a matching outfit 6)You should wear that to my office party

79. 79 2.2.2 Marrigal Example The benefit lying is TR=2L 3 +60L The cost of lying is TC=21L 2 +100 Where L = number of lies How many lies should you tell?

80. 80 2.2.2 Marrigal Example Profit=TR-TC Profit=2L 3 +60L-[21L 2 +100] Profit=2L 3 -21L 2 +60L-100 FOC: Mπ=6L 2 -42L+60 Mπ=6(L 2 -7L+10)

81. 81 2.2.2 Marrigal Example Mπ=6(L 2 -7L+10) Solving using the quadratic formula: L=-b±(b 2 -4ac) 1/2 / 2a L=-7±(49-40) 1/2 /(2) L=(-7±3)/2 = 2, 5

82. 82 2.2.2 Marrigal Example Profit=2L 3 -21L 2 +60L-100 Mπ=6L 2 -42L+60 SOC: π’’=12L-42 π’’(2)=12(2)-42=-18, concave MAX π’’(5)=12(5)-42=18, convex MIN You should tell 2 lies

83. 83 2.2.2 Marrigal Example Profit=2L 3 -21L 2 +60L-100 Profit(2)=2(2) 3 -21(2) 2 +60(2)-100 Profit(2)=16-84+120-100 Profit(2)=-48 You are minimizing the damage at -48 by telling two lies.

84. 84 2.2.2 Constrained Optimization Thus far we have considered UNCONSTRAINED optimization. (No budget constraint, time constraint, savings constraint, etc.) We can also deal with CONSTRAINED optimization. Ie: Maximize utility with respect to a set income. Maximize income with respect to a 24 hour day. Maximize lie effectiveness while keeping a straight face. To use this, we use a Lagrangean. (chapter 4)

85. 85 2.2.3 Elasticities We have already seen how the derivative, or the slope, can change as x and y change -even if a slope is constant, changes can have different impacts at different points -For example, given a linear demand for Xbox 720’s, a $100 price increases affects profits differently at different starting prices:

86. 86 2.2.3 Xbox 720 Example Price increase from $0 to $100 New Income

87. 87 2.2.3 Xbox 720- Example Price increase from $500 to $600 Old Income New Income

88. 88 2.2.3 Elasticities $0 to $100 Old Revenue: $0 New Revenue:4.5 million sold X $100 each $450 million (INCREASE) $500 to $600 Old Revenue:2.5 million sold X $500 each $1.25 Billion New Revenue:1.5 million sold X 600 each $0.9 Billion (DECREASE)

89. 89 2.2.3 Elasticities -to avoid this problem, economists often utilize ELASTICITIES -elasticities deal with PERCENTAGES and are therefore more useful across a variety of points on a curve ELASTICITY = a PROPORTIONAL change in y from a PROPORTIONAL change in x Example: elasticity of demand: E = Δy/y / Δx/x = (Δy/Δx) (x/y) = (dy/dx) (x/y)

90. 90 2.2.3 Elasticity Example 1 Let y=12x+7 Find elasticities at x= 5, and 10 1)dy/dx = 12 2)f(5) = 12(5) + 7 = 67 3)f(10) = 12(10) +7 = 127 Next we apply the formula:

91. 91 2.2.3 Elasticity Example 1 Let y=12x+7 Find elasticities at (x,y)= (5,67) and (10,127) E = dy/dx * x/y 1)E (5)= 12 * 5/67 = 0.90 2)E (10)= 12 * 10/127 = 0.94

92. 92 2.2.3 Elasticity Interpretation What does an elasticity of X mean? => for a 1% increase in x (or the independent variable), there will be a X% increase in y (or the dependent variable) In our example, a 1% increase in x caused a: 1)0.90% increase in y 2)0.94% increase in y Question: Is an increase in y good or bad?

93. 93 2.2.3 Inelastic vrs. Elastic: The Boxer’s or Briefs debate An elasticity of less than 1 (in absolute terms) is inelastic. That is, y responds less than x in percentage terms. An elasticity of greater than 1 (in absolute terms) is elastic. That is, y responds more than x in percentage terms. -This has policy implications…

94. 94 2.2.3 Elastic Xboxes Let q=5,000,000-5,000p Or q=5,000-5p Where q is Xbox’s demanded in 1000’s p is price of an Xbox Find elasticities at p=0, 100, and 500 1)dq/dp = -5 2)Q(0) =5,000 3)Q(100)=4,500 4)Q(500)=2,500

95. 95 2.2.3 Elastic Xboxes Find elasticities at p,q =(0,5000), (100,4500) and (500,2500) E = dq/dp * p/q 1)E = -5* 0/5000 = 0 2)E = -5 * 100/4500 = -0.11 (inelastic) 3)E = -5 * 500/2500 = -1 (unit elastic)

96. 96 2.2.3 Elastic Xboxes From these values, we know that demand for Xboxes 720’s is INELASTIC below $500 and ELASTIC above $500 How does this impact revenue? Total Revenue = p*q(p) dTR/dp = q(p)+p*dq/dp = q( 1+p/q*dq/dp) = q (1+ E )

97. 97 2.2.3 Making Microsoft Money If E = -1, dTR/dp = 0; a small change in price won’t affect revenue If | E | 0, small increases in prices increase revenue If | E | > 1 (elastic), dTR/dp<0, small increases in prices decrease revenue Therefore, price increases are revenue enhancing up to a price of $500.

98. 98 2.2.3 Assuming costs are constant: If demand is inelastic Raise Price If demand is elastic Decrease Price If demand is unit elastic Price is perfect (usually)

99. 99 2.2.3 More Elasticity Exercises Let q = 100-2p 1)Find Elasticities at p=5, 20 and 40 2)Formulate Economic Advice at these points Let q = 200+2p-4p 2 1)Find Elasticities at p=0, 5 and 10 2)Formulate Economic Advice at these points

100. 100 2.2.3 Elastic Logs ANOTHER reason to use logs in economic formulae is to more easily calculate elasticities: E = dy/dx * x/y = (1/y) dy/dx (x) = (dlny/dy) dy/dx (dx/dlnx) = (dlny/dlnx) dx/dx (dy/dy) = (dlny/dlnx)

101. 101 2.2.3 Examples are a log’s best friend Let ln(q) = -1ln(p) E = dln(q)/dln(p) = -1 Hence demand is unit elastic and change in price would not affect total revenue.

102. 102 2.2.3 Log Elasticity Exercises Let ln(q) = 100+ln(30/p) 1)Find Elasticities at p=5, 10 and 20 2)Formulate Economic Advice at these points Let ln(q) = 1/2 ln(p 2 ) 1)Find Elasticities at p=0.5, 1, and 2

103. 103 2.2.4 – Linear Approximation Derivatives can be used to approximate a function in a linear fashion: The point f(x) can be estimated as f(a) plus a distance along the red secant:

104. 104 2.2.4 – Linear Approximation Example Note that true values are f(3.1)=19.22, f(4)=32

105. 105 2.2.4 – Quadratic Approximation Note that the approximated values differ from the true values as we move away from our starting point. A more accurate estimation would be a quadratic TAYLOR SERIES EXPANSION:

106. 106 2.2.4 – Quadratic Approximation The TAYLOR SERIES isn’t designed for simple equations, it’s designed for more complex ones:

107. 107 2.2.4 – Quadratic Approximation 2 Approximate instead f(x)=x 0.5 at a=4 Since 5 0.5 =2.236, this is a good approximation