ECONOMIC APPLICATIONS OF IMPLICIT DIFFERENTIATION 1. Substitution of Inputs Let Q = F(L, K ) be the production function in terms of labor and capital. Consider the isoquant Q0 = F(L, K ) of equal production. (This is the level curve of the function.) Thinking of K as a function of L along the isoquant and using the chain rule, we get ∂ Q ∂ Q ∂ K 0 = + . ∂ L ∂ K ∂ L ∂ K The derivative measures the rate of substitution of capital for labor needed to keep the output constant. ∂ L For the Cobb-Douglas production function Q = cLa K b. If the quantity produced Q is fixed and K is considered a function of L, then differentiating ∂ 0 = cbLa K b ∂ L ∂ K = acLa−1 K b + bcLa K b−1 , so ∂ L ∂ K acLa−1 K b = − ∂ L bcLa K b−1 a K = − . b L Thus, the marginal rate of technical substitution of capital for labor is a constant multiple of the average of capital to labor. 2. Comparative-Static Analysis of Balancing Supply and Demand Assume the supply and demand functions for a single commodity are Q = S(P, T ) (supply) and Q = D(P, Y ) (demand), where Y is the income, T is the tax on the commodity, and P is the price. We do not assume a particular form of the supply and demand functions, but we do assume that the supply increases with price and decreases with tax while the demand decreases with price and increases with income: Assume that ∂ S ∂ S S = > 0, S = < 0, P ∂ P T ∂T ∂ D ∂ D D = < 0, D = > 0. P ∂ P Y ∂Y At an equilibrium state, supply and demand are balanced, so 0 = S(P, T ) − D(P, Y ). The partial derivative of the right hand side with respect to P is SP − DP > 0, so this equation determines the price P as a function of income Y and tax T while the system stays in an equilibrium state. Taking the 1 2 ECONOMIC APPLICATIONS OF IMPLICIT DIFFERENTIATION partial derivatives with respect to Y and T , we get ∂ P ∂ P 0 = S − D − D and P ∂Y P ∂Y Y ∂ P ∂ P 0 = S + S − D , P ∂T T P ∂T so ∂ P (S − D ) = D and P P ∂Y Y ∂ P (S − D ) = −S , P P ∂T T and ∂ P D = Y > 0 and ∂Y SP − DP ∂ P −S = T > 0. ∂T SP − DP Thus, the price increases with both an increase in income and tax. 3. General Method of Implicit Differentiation Following the book’s treatment of the general implicit function theorem, assume that there are m > 0 equations in n + m variables with n > 0, F1(x1,..., xn, y1,..., ym) = C1 F2(x1,..., xn, y1,..., ym) = C2 . Fm(x1,..., xn, y1,..., ym) = Cm such that at p = (a, b) ∂ F ∂ F 1 (p) ··· 1 (p) ∂y ∂y 1 m . 6= det . 0. ∂ F ∂ F m (p) ··· m (p) ∂y1 ∂ym By the implicit function theorem, there is a “implicitly defined function” y = h(x) such that C = F(x, h(x)) for all x near a. Thus, the (y1,..., ym) are the dependent variables and (x1,..., xn) are the independent variables. Let H(x) = (x, h(x)), so C = F(H(x)). Differentiating this equation with respect to x and using the chain rule, we get the matrix equation I 0 = DF(p)DH(a) = D F(p), D F(p) x y Dh(a) = Dx F(p) + Dy F(p)Dh(a) or ∂ F1 ∂ F1 ∂ F1 ∂ F1 ∂y1 ∂y1 (p) ··· (p) (p) ··· (p) (p) ··· (p) 0 ∂x ∂x ∂y ∂y ∂x ∂x 1 n 1 m 1 n . = . + . . (1) . . . . . 0 ∂ F ∂ F ∂ F ∂ F ∂y ∂y m (p) ··· m (p) m (p) ··· m (p) m (p) ··· m (p) ∂x1 ∂xn ∂y1 ∂ym ∂x1 ∂xn ECONOMIC APPLICATIONS OF IMPLICIT DIFFERENTIATION 3 Solving for the partial derivatives of the dependent variables and taking the inverse of the square matrix involving the partial derivatives of the Fi , we get −1 ∂y ∂y ∂ F1 ∂ F1 ∂ F ∂ F 1 (p) ··· 1 (p) (p) ··· (p) 1 (p) ··· 1 (p) ∂x ∂x ∂y ∂y ∂x ∂x 1 n 1 m 1 n . = − . . (2) . . . . ∂y ∂y ∂ F ∂ F ∂ F ∂ F m (p) ··· m (p) m (p) ··· m (p) m (p) ··· m (p) ∂x1 ∂xn ∂y1 ∂ym ∂x1 ∂xn In a given problem, we just write down the matrix equation (1) and solve them rather than use equation (2). The reason is that it is easier to remember the correct form for the matrix equation (1): The rows are determined by the coordinate functions (or equations) of the problem; the first matrix has the derivatives with respect to the independent variables; the second matrix has the derivatives with respect to the dependent variables and is multiplied by the unknown matrix of partial derivatives of the dependent variables with ∂ F ∂y respect to the independent variables. Notice in the matrix product the term i is multiplied by j as the ∂y j ∂x` chain rules says it must be. 4. Marginal Inputs of Prices A firm uses two inputs, q1 and q2 to produce a single output Q determined by a production function. Let p1 be the price of q1, and p2 be the price of q2; let 1 be the price of the output Q. (Either the price of Q is taken as the unit of money or we are looking at the ratios of the prices p j to the price of Q.) The profit is given by π = Q − p1q1 − p2q2. We will see in the section on maximization later in the quarter that the ∂π ∂π inputs that maximize profits satisfy = 0 and = 0. Calculating these partial derivatives, we get ∂q1 ∂q2 ∂π ∂ Q 0 = = − p1 and ∂q1 ∂q1 ∂π ∂ Q 0 = = − p2. ∂q2 ∂q2 1 1 = 3 2 For the Cobb-Douglas production function Q q1 q2 , these equations become 1 − 2 1 (3) 0 = q 3 q 2 − p 3 1 2 1 1 1 − 1 0 = q 3 q 2 − p . 2 1 2 2 We have two equations and four variables p1, p2, q1, and q2. If we specified values for the prices, then we would have two equations for the variables q1 and q2 which could determine their values. To see that these two equations in (3) determine the amounts of inputs q1 and q2 that optimize profit in terms of the prices p1 and p2, we check the determinant of the matrix of partial derivatives with respect to q1 and q2: 5 1 2 1 -2 - 3 2 1 - 3 - 2 q1 q2 q1 q2 9 6 - 4 2 1 1 - 4 det = q 3 q-1 − = q 3 q-1 6= 0. 1 - 2 - 1 -1 1 -3 1 2 1 2 q 3 q 2 q 3 q 2 36 36 36 6 1 2 4 1 2 4 ECONOMIC APPLICATIONS OF IMPLICIT DIFFERENTIATION Therefore, the partial derivatives of q1 and q2 with respect to p1 and p2 satisfy the following matrix equation: -2 - 5 1 1 - 2 - 1 ∂q ∂q q 3 q 2 q 3 q 2 1 1 9 1 2 6 1 2 ∂p ∂p 0 = -1 0 + 1 2 1 - 2 1 -1 1 -3 ∂q ∂q 0 0 -1 3 - 2 3 2 2 2 q1 q2 q1 q2 6 4 ∂p1 ∂p2 -1 ∂q ∂q -2 - 5 1 1 - 2 - 1 1 1 q 3 q 2 q 3 q 2 ∂p ∂p 9 1 2 6 1 2 1 2 = ∂q ∂q 1 - 2 1 -1 1 -3 2 2 3 - 2 3 2 q1 q2 q1 q2 ∂p1 ∂p2 6 4 -1 1 -3 -1 - 2 - 1 q 3 q 2 q 3 q 2 4 1 2 1 2 3 4 6 = 36 q q2 1 -1 - 2 - 1 -2 - 5 1 q 3 q 2 q 3 q 2 6 1 2 9 1 2 5 -1 2 1 3 2 3 2 -9 q1 q2 -6 q1 q2 = 2 1 -1 3 , 3 2 3 2 -6 q1 q2 -8 q1 q2 ∂q and all the partial derivatives i are negative. ∂p j 1 + 1 Because the sum of the exponents in the production function is less than one, 3 2 < 1, the production has decreasing return to scale. Although it is not obvious from the above treatment, this property caused the partial derivatives to be negative and has the effect that the amounts of the inputs decreases with an increase in either price. The details of this example are complicated. However, the outline of the method is not complicated. (a) First, some equations are derived (or given) relating several variables. In our example, (3) has two equations with four variables. (b) Second, thinking of these equations as defining some variables in terms of the others, we take partial derivatives. In our case, we take the partial derivatives with respect to p1 and p2. 5. Changing of Technology of Production A firm uses two inputs to produce an output. Assume the amounts of the inputs are x and y with p the price of x and q the price of y. The amount produced Q is determined by the Cobb-Douglas production function Q = xa yb, where the technology determines the exponents a and b. The firm can vary the exponent b while keeping a fixed by changing the technology. It wants to keep the amount produced fixed, Q0, and the cost of the inputs fixed px + qy = 125. What is the rate of change of the amounts at x = 5, y = 50, p = 5, q = 2, a = 1/3, and b = 2/3? a b Rather than use the equation Q0 = x y , we take its logarithm and obtain the two equations 125 = px + qy and ln(Q0) = a ln(x) + b ln(y).