DERIVATIVE OF A FUNCTION Consider, again, the function (2.1) The slope of this function is defined as the change in the value of y divided by a change in the value of x, or the “rise” over the “run.” When defining the slope between two discrete points, the formula for the slope may be given as (2.6) Consider Figure 2.12, and use the foregoing definition to calculate the value of the slope of the cord AB. As point B is brought arbitrarily closer to point A, however, the value of the slope of AB approaches the value of the slope at the single point A, which would be equivalent to the slope of a tangent to the curve at that point. This procedure is greatly simplified, however, by first taking the derivative of the function and calculating its value, in this case, at x 1 . The first derivative of a function (dy/dx) is simply the slope of the func- tion when the interval along the horizontal axis (between x 1 and x 2 ) is made infinitesimally small.Technically,the derivative is the limit of the ratio Dy/Dx as Dx approaches zero, that is, (2.44) When the limit of a function as x Æ x 0 equals the value of the function at x 0 , the function is said to be continuous at x 0 , that is, lim xÆx 0 f(x) = f(x 0 ). dy dx y x x = Ê Ë ˆ ¯ Æ lim D D D 0 Slope = = - - = () - () - D D y x yy xx fx fx xx 21 21 21 21 yfx= () 62 introduction to mathematical economics y= f(x) y x 0 x 1 x 2 y 2 y 1 A B FIGURE 2.12 Discrete versus instanta- neous rates of change. Calculus offers a set of rules for using derivatives (slopes) for making optimizing decisions such as minimizing cost (TC) or maximizing total profit (p). RULES OF DIFFERENTIATION Having established that the derivative of a function is the limit of the ratio of the change in the dependent variable to the change in the inde- pendent variable, we will now enumerate some general rules of differenti- ation that will be of considerable value throughout the remainder of this course. It should be underscored that for a function to be differentiable at a point, it must be well defined; that is it must be continuous or “smooth.” It is not possible to find the derivative of a function that is discontinuous (i.e., has a “corner”) at that point. The interested student is referred to the selected adings at the end of this chapter for the proofs of these propositions. POWER-FUNCTION RULE A power function is of the form where a and b are real numbers. The rule for finding the derivative of a power function is (2.45) where f¢(x) is an alternative way to denote the first derivative. Example A special case of the power-function rule is the identity rule: Another special case of the power-function rule is the constant-function rule. Since x 0 = 1, then yfx ax a= () == 0 dy dx fx x x=¢ () = () == - 11 1 1 11 0 yfx x= () = dy dx fx x x= ¢ () = () = - 24 8 21 yx= 4 2 dy dx f x bax b =¢ () = -1 yfx ax b = () = rules of differentiation 63 Thus, Example SUMS AND DIFFERENCES RULE There are a number of economic and business relationships that are derived by combining one or more separate, but related, functions. A firm’s profit function, for example, is equal to the firm’s total revenue function minus the firm’s total cost function. If we define g and h to be functions of the variable x, then (2.46) Example Example a. Consider the general case of the linear function (2.5) b. y = f(x) = 5 - 4 c. From Problem 2.1 QP S =+10 2 QP D =-25 3 dy dx fx= ¢ () =-4 dy dx fx du dx dv dx bb= ¢ () =+=+=0 yfx gxhx abx= () = () + () =+ dy dx fx x= ¢ () =+22 yfx gxhx xx= () = () + () =+2 2 ugx xvhx x= () == () =2 2 ; dy dx fx du dx dv dx =¢ () =± yfx uvgxhx= () =±= () ± () ugxvhx= () = () ; dy dx fx= ¢ () =◊ () = - 01 5 0 01 y ==◊515 0 dy dx fx ax=¢ () = () = - 00 01 64 introduction to mathematical economics Example PRODUCT RULE Similarly, there are many relationships in business and economics that are defined as the product of two or more separate, but related, func- tions. The total revenue function of a monopolist, for example, is the product of price, which is a function of output, and output, which is a func- tion of itself. Again, if we define g and h to be functions of the variable x, then Further, let Although intuition would suggest that the derivative of a product is the product of the derivatives, this is not the case. The derivative of a product is defined as (2.47) Example Substituting into Equation (2.47) dy dx fx x x x x x= ¢ () =- () +- ()() =- +22324 1212 22 dv dx hx= ¢ () =-2 vx=- () 32 du dx gx x= ¢ () = 4 ugx x= () = 2 2 yx x=- () 232 2 dy dx fx u dv dx v du dx uh x vg x=¢ () = Ê Ë ˆ ¯ + Ê Ë ˆ ¯ =¢ () +¢ ( ) yfx uvgxhx= () == () ◊ () ugxvhx= () = () ; dy dx fx x x= ¢ () =-+012 18 1 0 2 yxxx=-++004 09 10 5 32 dQ dP S = 2 dQ dP D =-3 rules of differentiation 65 QUOTIENT RULE Even less intuitive than the product rule is the quotient rule. Again, defining g and h as functions of x, we write Further, let then (2.48) Example Substituting into Equation (2.48), we have Interestingly, in some instances it is convenient, and easier, to apply the product rule to such problems. This becomes apparent when we remember that Example yfx gx hx x x xx x x= () = () () == () = () [] - - 2 3 23 2 13 2 2 1 21 y u v uv== -1 dy dx fx xxx x xx x x x = ¢ () = - () () () = - = -22324 2 412 4 3 2 2 2 2 43 dv dx hx x= ¢ () = 4 vhx x= () = 2 2 du dx gx= ¢ () =-2 ugx x= () =-32 yfx x x= () =- () 32 2 2 dy dx fx v du dx u dv dx v h x dg x dx g x dh x dx hx hx g x gx h x hx =¢ () = () - () = () () [] - () () [] () = () ◊¢ () - () ◊¢ () () 2 2 2 yfx u v gx hx = () == () () ugxvhx= () = () ; 66 introduction to mathematical economics It is left to the student to demonstrate that the same result is derived by applying the quotient rule. CHAIN RULE Often in business and economics a variable that is a dependent variable in one function is an independent variable in another function. Output Q, for example, is the dependent variable in a perfectly competitive firm’s short-run production function where L represents the variable labor, and K 0 represents a constant amount of capital labor utilizes in the short run. On the other hand, output is the independent variable in the firm’s total revenue function where P is the (constant) selling price. In the example just given, we might be interested in determining how total revenue can be expected to change given a change in the firm’s labor usage. For this we require a technique for taking the derivative of one func- tion whose independent variable is the dependent variable of another func- tion. Here we might be interested in finding the derivative dTR/dL. To find this derivative value, we avail ourselves of the chain rule. Let y = f(u) and u = g(x). Substituting, we are able to write the composite function The chain rule asserts that (2.49) Applying the chain rule, we get Example yfu u= () =+ 3 10 yx= () +210 2 3 dTR dL dTR dQ dQ dL LL L= Ê Ë ˆ ¯ Ê Ë ˆ ¯ = () == 10 2 20 20 05 05 dy dx dy du du dx df u du dg x dx fu gx = Ê Ë ˆ ¯ Ê Ë ˆ ¯ = () È Î Í ˘ ˚ ˙ () È Î Í ˘ ˚ ˙ =¢ () ◊¢ () yfgx= () [] TR g Q PQ Q= () ==10 QfLK L= () =, . 0 05 4 dy dx fx x x x x= ¢ () =- () [] + ()() =- + = 2 13 13 4 23 43 23 22 1 rules of differentiation 67 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Now we consider the derivative of two important functions–the expo- nential function and the logarithmic function. The number e is the base of the natural exponential function, y = e x . The natural logarithmic function is y = log e x = ln x. The number e is itself generated as the limit to the series (2.50) To illustrate the practical importance of the number e, suppose, for example, that you were to invest $1 in a savings account that paid an inter- est rate of i percent. If interest was compounded continuously (see Chapter 12), the value of the deposit at the year end would be (2.51) Now, suppose that a deposit of D dollars was compounded continuously for n years. At the end of n years the deposit would be worth The derivative of the exponential function y = e x is (2.52) That is, the derivative of the exponential function is the exponential func- tion itself. The derivative of the natural logarithm of a variable with respect to that variable, on the other hand, is the reciprocal of that variable. That is, if then y x e = log dy dx de dx e x x = () = lim lim h h in h n t in D h D h De Æ• Æ• + Ê Ë ˆ ¯ È Î Í ˘ ˚ ˙ =+ Ê Ë ˆ ¯ È Î Í ˘ ˚ ˙ =1 1 1 1 lim h h i i h e Æ• + Ê Ë ˆ ¯ =1 e h h h =+ Ê Ë ˆ ¯ = Æ• lim . . . .1 1 2 71829 dy dx fx u x x x x x x= ¢ () = () = () ◊ () =◊=3 4 3 2 2 4 12 4 48 22 2 4 5 du dx gx x= ¢ () = 4 ugx x= () = 2 2 dy du fu u x x= ¢ () == () =332 12 22 2 4 68 introduction to mathematical economics (2.53) When more complicated functions are involved, we can apply the chain rule. Suppose, for example, that y = ln x 2 . Letting u = x 2 this becomes y = ln u. The derivative of y with respect to x then becomes It may also be demonstrated that the result for the derivative of an expo- nential function follows directly from a special relationship that exists between the exponential function and the logarithmic function. Given the function x = e y , then y = ln x. Moreover, if x = ln y, then y = e x . These func- tions are said to be reciprocal functions. When two functions are related in this way, the derivatives are also related; that is, dy/dx = 1/(dx/dy). Using this rule, we can prove the exponential function rule: Returning to the earlier discussion of continuous compounding, suppose that the value of an asset is given by where r is the rate of interest, t time, and D the initial value of the asset. The rate of change of the value of the asset over time is That is, the rate of change in the value of the asset is the rate of interest times the value of the asset at time t. INVERSE-FUNCTION RULE Earlier in this chapter we discussed the existence of inverse functions. It will be recalled that if the function y = f(x) is a one-to-one correspondence, then not only will a given value of x correspond to a unique value of y, but a given value of y will correspond to a unique value of x. In this case, the function f has the inverse function g(y) = f -1 (y) = x, which is also a one-to- one correspondence. Given an inverse function, its derivative is dDe dt D de dt e dD dt D de du du dt e dD dt De r e rDe rt rt rt u rt urt rt () = Ê Ë ˆ ¯ + Ê Ë ˆ ¯ = Ê Ë ˆ ¯ Ê Ë ˆ ¯ + Ê Ë ˆ ¯ =+ () =0 Dt De rt () = de dx dy dx d y dy y ye x x () == () === 11 1ln dy dx dy du du dx ux x x x = Ê Ë ˆ ¯ Ê Ë ˆ ¯ = ()() = Ê Ë ˆ ¯ () =12 1 2 2 2 dy dx dx dx x e = () = log 1 rules of differentiation 69 (2.54) Equation (2.54) asserts that the derivative of an inverse function is the reciprocal of the derivative of the original function. It will also be recalled that functions with a one-to-one correspondence are said to be monotonically increasing if x 2 > x 1 fi f(x 2 ) > f(x 1 ). Functions in which a one-to-one correspondence exist are said to be monotonically decreasing if x 2 > x 1 fi f(x 2 ) < f(x 1 ). In general, for an inverse function to exist, the original function must be monotonic. In other words, it is not possible to write x = g(y) = f -1 (y) until we have determined whether the function y = f(x) is monotonic. It is possible to determine whether a function is monotonic by examin- ing its first derivative. If the first derivative of the function is positive for all values of x, then the function y = f(x) is monotonically increasing. If the first derivative of the function is negative for all values of x, then the function y = f(x) is monotonically decreasing. Problem 2.5. Consider the function a. Is this function monotonic? b. If the function is monotonic, use the inverse-function rule to find dx/dy. Solution a. The derivative of this function is which is positive for all values of x. Thus, the function f(x) is a monoto- nically increasing function. b. Because f(x) is a monotonically increasing function, the inverse function g(y) = f -1 (y) exists. Thus, it is possible to use the inverse-function rule to determine the derivative of the inverse function, that is, It should be noted that the inverse-function rule may also be applied to nonmonotonic functions, provided the domain of the function is restricted. For example, y = f(x) = x 2 is nonmonotonic because its derivative does not have the same sign for all values of x. On the other hand, if the domain of this function is restricted to positive values for x, then dy/dx > 0. dx dy dy dx xx == ++ 11 47 46 dy dx fx x x=¢ () =+ +47 46 yfx x x x= () =+ +402 57 . ¢ () == = ¢ () gy dx dy dy dx f x 11 70 introduction to mathematical economics Problem 2.6. Consider the function a. Is this function monotonic? b. If the function is monotonic, use the inverse-function rule to find dx/dy. Solution a. The derivative of this function is This function is not monotonic, since the sign of dy/dx depends on whether x is positive or negative. On the other hand, the derivative is negative for all positive values for x. b. Because the derivative of f(x) is positive for all x > 0, then it is possible to use the inverse-function rule to determine the derivative of the inverse function, that is, for all x > 0. IMPLICIT DIFFERENTIATION The functions we have been discussing are referred to as explicit func- tions. Explicit functions are those in which the dependent variable is on the left-hand side of the equation and the independent variables are on the right-hand side. In many cases in business and economics, however, we may also be interested in what are called implicit functions. Implicit functions are those in which the dependent variable is also func- tionally related to one or more of the right-hand-side variables. Such func- tions often arise in economics as a result of some equilibrium condition that is imposed on a model. A common example of an implicit function in macroeconomic theory is in the definition of the equilibrium level of national income Y, which is given as the sum of consumption spending C, which is itself assumed to be a function of national income, net investment spending I, government expenditures G, and net exports X - M. This equi- librium condition is written (2.55) Clearly, any change in the value of Y must come about because of changes in any and all changes in the components of aggregate demand. The total derivative of this relationship may be written YCIG XM=+++ - () ¢ () == =¢ () = -+ () gy dx dy dy dx fy x 1 1 1 34 3 dy dx fx x x=¢ () =- - =- + () 34 34 33 yfx xx= () =- -3 4 implicit differentiation 71 [...]... for a maximum are given by 2 y = f11 < 0 ∂x 12 (2. 81a) 2 y = f 12 < 0 ∂x 22 (2. 81b) 2 2 2 2 Ê ∂ yˆÊ ∂ yˆ Ê ∂ y ˆ 2 = f11 f 22 - f 12 > 0 2 ¯Ë 2 Ë ∂x1 ∂x2 Ë ∂x1 ∂x2 ¯ (2. 81c) The second-order conditions for a minimum are given by: 2 y = f11 > 0 ∂x 12 (2. 82a) 2 y = f 22 > 0 ∂x 22 (2. 82b) 2 2 2 2 Ê ∂ yˆÊ ∂ yˆ Ê ∂ y ˆ 2 = f11 f 22 - f 12 > 0 2 ¯Ë 2 ¯ Ë ∂x1 ∂x2 Ë ∂x1 ∂x2 ¯ (2. 82c) Example Consider once again... = 20 Q - 3Q2 - 2Q2 = 20 Q - 5Q2 dp = 20 - 10Q = 0 (i.e., the first-order condition for a profit maximum) dQ Q = 2 units d2p = -10 < 0 (i.e., the second-order condition for a profit maximum) dQ 2 b p = 20 (2) - 5 (2) 2 = 40 - 20 = $20 c TR = PQ = 20 Q - 3Q2 = (20 - 3Q)Q P = 20 - 3Q = 20 - 3 (2) = 20 - 6 = $14 Problem 2. 8 Another monopolist has the following TR and TC functions: TR(Q) = 45Q - 0.5Q 2 TC (Q) = 2. .. 600(5) + 4,000 = 625 + 3,000 + 4,000 = $7, 624 c TC5 = Ú5 (50x + 600)dx 2 2 5 [ ] + 600(5) + 4, 000] - [25 (2) 2 = Ú 25 (5) + 600(5) + 4, 000 2 [ = 25 (5) 2 2 + 600 (2) + 4 , 000 ] = ( 625 + 3, 000 + 4, 000) - (100 + 1, 20 0 + 4, 000) = $7, 625 - $5, 300 + $2, 325 CHAPTER REVIEW Economic and business relationships may be represented in a variety of ways, including tables, charts, graphs, and algebraic expressions... statement? Explain CHAPTER EXERCISES 2. 1 Solve each of the following systems of equations and check your answers a 2x + y = 100 -4x + 2y = 40 b x - y = 20 1 (–)x - y = 0 3 c x2 - y = 20 x2 + y = 10 d 2x + y2 = 4 -x + y2 = 16 2. 2 Solve the following system of equations: x+ y+z=1 -x - Ê 1ˆ Ê 2 yz=4 Ë 2 Ë 3¯ 2x + 2 y - z = 5 95 chapter exercises 2. 3 Find the first derivatives and the indicated values of the... = f(x) = x4 - 2x3 + 3x2 - 5x + x-1 - x -2 + 24 Find f¢(-3), f¢(0), f¢(3) 2. 4 Find the second derivatives a y = f(x) = 8x + 3x2 - 13 b y = f(x) = ex + x2 - (2x - 4 )2 c y = f(x) = x/3 + 4 (x) - (x - 3 )2( x2 + 2) d y = f(x) = 2 loge(x2 + 4x) - [(x - 2) /(x + 3) ]2 2.5 The total cost function of a firm is given by: TC = 800 + 12Q + 0.018Q 2 where TC denotes total cost and Q denotes the quantity produced per... 57Q - 8Q 2 + Q 3 Find the p-maximizing output level Solution p = TR - TC = (45Q - 0.5Q 2 ) - (2 + 57Q - 8Q 2 + Q 3 ) = -2 - 12Q + 7.5Q 2 - Q 3 dp = - 12 + 15Q - 3Q2 = 0 (i.e., the first-order condition for a local dQ maximum) Utilizing the quadratic formula: 80 introduction to mathematical economics Q1 ,2 = = -15 ± 15 2 - 4(-3)(- 12) 2( -3) -15 ± 81 -15 ± 9 = -6 -6 Q1 = -15 - 9 -24 = =4 -6 -6 Q2 = -15 +... f¢(0), f¢ (2) , f¢( 12) b y = f(x) = (2 + x)(3 - x) Find f¢( -2) , f¢(3), f¢(6) c y = f(x) = (x - 2) /(x2 + 4) Find f¢(-4), f¢(0), f¢ (2) d y = f(x) = [(x - 1)/(x + 4)]3 Find f¢( -2) , f¢(0), f¢ (2) e y = f(x) = 2x2 - 4/x + (4x) - 3 (x) Find f¢(1), f¢(5), f¢(10) 3 f y = f(x) = 3 logex - (–)logex Find f¢(0), f¢(1), f¢(100) 4 x g y = f(x) = 6e Find f¢(-1), f¢(0), f¢(1), f¢ (2) h y = f(x) = x4 - 2x3 + 3x2 - 5x +... 2. 4 Maximum First-order condition Second-order condition Minimum dy =0 dx d2y 0 dx 2 profit maximization: the second-order condition 79 d2p = -6(1) + 18 = -6 + 18 = 12 > 0 dQ 2 which is a p minimum Problem 2. 7 A monopolist’s total revenue and total cost functions are TR(Q) = PQ = 20 Q - 3Q 2 TC (Q) = 2Q 2 a Determine the output level (Hint: p(Q) = TR(Q) - TC(Q)) that will maximize... relationship to marginal and average total cost FIGURE 2. 13 b= 0 Q1 Q 2 Q3 DTC TC 2 - TC1 = DQ Q2 - Q1 Q4 Q5 Q (2. 65) where the values where Q1 represents the initial value of output and Q2 represents the changed level of output Since the ray passes through the origin, then the initial values (Q1, TC1) are (0, 0) Setting TC2 = TC and Q2 = Q, Equation (2. 66) reduces to b = ATC = TC Q (2. 66) Of course, the... determine the values of P and A that maximize the firm’s total sales, Q, set the first partial derivatives in Equations (2. 69) and (2. 70) equal to zero 80 - 4P - A = 0 (2. 77) - P - 6 A + 100 = 0 (2. 78) Equations (2. 77) and (2. 78) are the first-order conditions for a maximum Solving these two linear equations simultaneously in two unknowns yields (in thousands of dollars) P = $16. 52 A = $13. 92 Substituting these . x= () = () () == () = () [] - - 2 3 23 2 13 2 2 1 21 y u v uv== -1 dy dx fx xxx x xx x x x = ¢ () = - () () () = - = -22 324 2 4 12 4 3 2 2 2 2 43 dv dx hx x= ¢ () = 4 vhx x= () = 2 2 du dx gx= ¢ () = -2 ugx x= () =- 32 yfx. as (2. 47) Example Substituting into Equation (2. 47) dy dx fx x x x x x= ¢ () =- () +- ()() =- +22 324 121 2 22 dv dx hx= ¢ () = -2 vx=- () 32 du dx gx x= ¢ () = 4 ugx x= () = 2 2 yx x=- () 23 2 2 dy dx fx. Æ• + Ê Ë ˆ ¯ È Î Í ˘ ˚ ˙ =+ Ê Ë ˆ ¯ È Î Í ˘ ˚ ˙ =1 1 1 1 lim h h i i h e Æ• + Ê Ë ˆ ¯ =1 e h h h =+ Ê Ë ˆ ¯ = Æ• lim . . . .1 1 2 71 829 dy dx fx u x x x x x x= ¢ () = () = () ◊ () =◊=3 4 3 2 2 4 12 4 48 22 2 4 5 du dx gx x= ¢ () = 4 ugx x= () = 2 2 dy du fu u x x= ¢ () == () =3 32 12 22 2 4 68 introduction