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Partial 1 Derivatives Thursday Oct 19, 2000 While derivatives are of interest to economists, as our above examples show, they are often not applicable because most of the mathematical functions used to describe economic concepts are multivariate functions. E.g., we assumed above that cost was soley a function of output, so speci...ed the cost function as c = c(x). But, more generally, minimum cost is not just a function of how many units of output the ...rm chooses to produce but also a function of input prices. Consider the cost function for a ...rm that uses two inputs, l and k (labor and capital) which it purchases in a competitive market at the prices w and r, where r is the rental price of capital. The ...rm's cost function indenti...es the minimum cost of producing x units out output at prices w and r. That is c = c(x; w; r) In the simple form c (x), we were implicitly holding w and r constant at some speci...c amounts. Cost more generally is a function of the output level and the input prices. Note that for a ...rm that buys inputs in a competitive market w and r are exogenous. Can we take the derivative of c = c(x; w; r) with respect to x, w or r? Strictly speaking NO. However, one can take the partial derivative of the cost function with respect to any of the three variables. When one partially di erentiates a function of several variables with respect to one of those variables, just treat the other variables as constants and use the rules for di erentiating functions of one variable. We denote the partial derivative of c (x; w; r) with respect to x as @c(x; r) w; @x Note the di erent notation for a partial derivative. Further note that the notation c0 (x; w; r) is ambiguous so is replaced with cx (x; w; r) for @c(x;w;r) , cw (x; w; r) for @c(x;w;r) and cr (x; w; r) for @c(x;w;r) @x @w @r 1 Putting all this a little more formally and generally. Consider the function y = f (x1 ; x2 ; :::xN ) = f(x) where a bolded x, x denotes a vector of x's. Consider the simpliest case y = f (x1 ; x2 ) = f (x) What is the partial derivative of f (x1 ; x2 ) with respect to x1 evaluated at (x0 ; x0 )? de...nition 1 2 @f (x0 ; x0 ) f (x0 + t; x0 ) f(x0 ; x0 ) 1 2 1 2 1 2 = fx1 (x0 ; x0 ) = lim 1 2 t!0 @x1 t What is the partial derivative of f (x1 ; x2 ) with respect to x2 evaluated at (x0 ; x0 )? de...nition 1 2 @f (x0 ; x0 ) f (x0 ; x0 + t) f(x0 ; x0 ) 1 2 1 2 1 2 = fx2 (x0 ; x0 ) = lim 1 2 t!0 @x2 t Show it graphically Do a simple example. Assume y = f(x1 ; x2 ) = ax1 x2 In which case a(x0 + t)x0 ) ax0 x0 @f (x0 ; x0 ) 1 2 1 2 1 2 = fx1 (x0 ; x0 ) = lim 1 2 t!0 @x1 t ax0 x0 + atx0 ax0 x0 atx0 2 1 2 2 = lim = lim ax0 = ax0 = lim 1 2 2 2 t!0 t!0 t t!0 t As with derivatives, we could use our basic de...nition of a partial derivative to ...nd partial derivatives. But, in general, if we are taking the derivative of a multivariate function with respect to one variable, we can treat the others as constants and use are derived rules for di erentiating a function of a single variable. E.g. ...nd the derivatives of z = f (x; y) = x + 2x1=2 y 1=2 + y; x 0, y 0 fx = 1 + (1=2)(2)x 1=2 y 1=2 + 0 = 1 + x 1=2 y1=2 fy = 1 + x1=2 y 1=2 2

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