Xy

But what if the given equation, F(y, x,,..., xm) = 0, cannot be solved for y explicitly? In this case, if under the terms of the implicit-function theorem an implicit function is known to exist, we can still obtain the desired derivatives without having to solve for y first. To do this, we make use of the so-called "implicit-function rule"—a rule that can give us the derivatives of every implicit function defined by the given equation. The development of this rule depends on the following basic facts: (1) if two expressions are identically equal, their respective total differentials must be equal;* (2) differentiation of an expression that involves y,x,,..., xm will yield an expression involving the differentials dy, dxv..., dxm; and (3) if we divide dy by dx^, and let all the other differentials (dx2,..., dxm) be zero, the quotient can be interpreted as the partial derivative

* Take, for example, the identity

This is an identity because the two sides are equal for any values of x and y that one may assign, faking the total differential of each side, we have

¿(right side) = (x - y) d(x + y) + (x + y) d(x - y)

The two results are indeed equal. If two expressions are not identically equal, but are equal only for certain specific values of the variables, however, their total differentials will not be equal. The equation x2 - y2 = x2 + y2 - 2

for instance, is valid only for y = + 1. The total differentials of the two sides are

¿(right side) = 2x ¿x + 2y dy which are not equal. Note, in particular, that they are not even equal at^ — +1.

dy/dx x; similar derivatives can be obtained if we divide dy by dx2, etc. Applying these facts to the equation F(y, x,,..., xm) = 0—which, we recall, has the status of an identity in the neighborhood N in which the implicit function is defined—we can write dF = d0, or

Suppose that only y and xx are allowed to vary (only dy and dxt are not set equal to zero). Then the above equation reduces to Fydy + F, dxx = 0. Upon dividing through by dxx, and solving for dy/dx,, we then get dy_ dxx dx F

other variables constant 1 y

By similar means, we can derive all the other partial derivatives of the implicit function /. These may conveniently be summarized in a general rule—the implicit-function rule—as follows: Given F(y, xx,..., xm) = 0, if an implicit function y = f(xx,..., xm) exists, then the partial derivatives of / are

In the simple case where the given equation is F( y, x) = 0, the rule gives

What this rule states is that, even if the specific form of the implicit function is not known to us, we can nevertheless find its derivative(s) by taking the negative of the ratio of a pair of partial derivatives of the F function which appears in the given equation that defines the implicit function. Observe that Fy always appears in the denominator of the ratio. This being the case, it is not admissible to have F = 0. Since the implicit-function theorem specifies that F + 0 at the point around which the implicit function is defined, the problem of a zero denominator is automatically taken care of in the relevant neighborhood of that point.

Example 1 Find dy/dx for the implicit function defined by (8.14'). Since F(y, x) takes the form of y - 3x4, we have, by (8.19'), dy ^ -U*3 2 , dx ( Fv 1

In this particular case, we can easily solve the given equation for y, to get y = 3x4. Thus the correctness of the above derivative is easily verified.

Example 2 Find dy/dx for the implicit functions defined by the equation of the circle (8.15). This time we have F(y, x) = x2 + y2 - 9; thus Fy = 2y and Fx = 2x.

By (8.19'), the desired derivative is

Earlier, it was asserted that the implicit-function rule gives us the derivative of every implicit function defined by a given equation. Let us verify this with the two functions in (8.15') and their derivatives in (8.18). If we substitute y+ for y in the implicit-function-rule result dy/dx = —x/y, we will indeed obtain the derivative dy+/dx as shown in (8.18); similarly, the substitution of y~ for y will yield the other derivative in (8.18). Thus our earlier assertion is duly verified.

Example 3 Find dy/dx for any implicit function(s) that may be defined by the equation F(y, x, w) = y3x2 + w3 + yxw —3 = 0. This equation is not easily solved for y. But since Fv, Fx, and Fw are all obviously continuous, and since Fy = 3y2x2 + xw is indeed nonzero at a point such as (1,1,1) which satisfies the given equation, an implicit function y = f(x,w) assuredly exists around that point at least. It is thus meaningful to talk about the derivative dy/dx. By (8.19), moreover, we can immediately write dy = Fx = 2y3x + yw dx Fy 3y2x2 + xw

At the point (1,1,1), this derivative has the value — J.

Example 4 Assume that the equation F(Q, K, L) = 0 implicitly defines a production function Q = f(K, L). Let us find a way of expressing the marginal physical products MPPA and MPP, in relation to the function F. Since the marginalproducts are simply the partial derivatives dQ/dK and dQ/dL, we can apply the implicit-function rule and write

Aside from these, we can obtain yet another partial derivative, dL fk from the equation F(Q, K, L) = 0. What is the economic meaning of 3K/3L1 The partial sign implies that the other variable, Q, is being held constant; it follows that the changes in K and L described by this derivative are in the nature of "compensatory" changes designed to keep the output Q constant at a specified level. These are therefore the type of changes pertaining to movements along a production isoquant. As a matter of fact, the derivative dK/dL is the measure of the slope of such an isoquant, which is negative in the normal case. The absolute

* The restriction y j= 0 is of course perfectly consistent with our earlier discussion of the equation <8.15) that follows the statement of the implicit-function theorem.

value of dK/dL, on the other hand, is the measure of the marginal rate of technical substitution between the two inputs.

Extension to the Simultaneous-Equation Case

The implicit-function theorem also comes in a more general and powerful version that deals with the conditions under which a set of simultaneous equations

F\yx,...,yn\xx,...,xm) = 0 (8.20) F2(yi>--->y*>Xi---*Xm) = 0

F"(yx,...,y„; xx,..., xj = 0 will assuredly define a set of implicit functions*

The generalized version of the theorem states that:

Given the equation system (8.20), if (a) the functions F\..., F" all have continuous partial derivatives with respect to all the y and x variables, and if (b) at a point (yxo,..., yn0; xx0,...,xm0) satisfying (8.20), the following Jacobian determinant is nonzero:

d{F\...,F")

dFl

dFx

dFx

dyx

dy2

dyn

dF2

dF2

dF2

dyx

dy2

Sy„

3F"

9F"

dF"

dyx

dy2

d}'n

then there exists an /»-dimensional neighborhood of (xxo,..., xm0), N, in which the variablesyx,..., yn are functions of the variables x],..., xm in the form of (8.21). These implicit functions satisfy

.Vio = / (-^10'• • ■ > xmo) yn0 = /"(*,(>,■ • • ' Xmo)

* To view it another way, what these conditions serve to do is to assure us that the n equations in (8.20) can in principle be solved for the n variables—yx,..., y„—even if we may not be able to obtain the solution (8.21) in an explicit form.

They also satisfy (8.20) for every w-tuple (xx,..., xm) in the neighborhood N —thereby giving (8.20) the status of a set of identities as far as this neighborhood is concerned. Moreover, the implicit functions /',...,/" are continuous and have continuous partial derivatives with respect to all the x variables.

As in the single-equation case, it is possible to find the partial derivatives of the implicit functions directly from the n equations in (8.20), without having to solve them for the y variables. Taking advantage of the fact that, in the neighborhood N, (8.20) has the status of identities, we can take the total differential of each of these, and write dFj = 0 (j = 1,2,...,«). The result is a set of equations involving the differentials dyx,..., dyn and dxx,..., dxm. Specifically, after transposing the dxi terms to the right of the equals signs, we have dF' J dFx J dF1 ,

—<fy i + "5—dy2 + ---+——dy dyi 3y2 3yn dF{ J 3F1 , ^

-5—dyx + —dy2 + ■ • • + -r—dy„ dy. dy2 dyn dF2 J 8F2 , dF" , dF" J dF" ,

-5—dy, + -5—dy2 + • • • + -5—dyn dy 1 dy2 dy„

Since all the partial derivatives appearing in (8.22) will take specific (constant) values when evaluated at the point (yxo,..., y„0; x10,..., xm0)—the point around which the implicit functions (8.21) are defined—we have here a system of n linear equations, in which the differentials dyj (considered to be endogenous) are expressed in terms of the differentials dxt (considered to be exogenous). Now, suppose that we let all the differentials dxt be zero except dxt (that is, only .t, is allowed to vary); then all the terms involving dx2,..., dxm will drop out of the system. Suppose, further, that we divide each remaining term by dxx\ then there will emerge the expressions dyx/dxx,..., dyn/dxx. These, however, should be interpreted as partial derivatives of (8.21) because all the x variables have been held constant except xx. Thus, by taking the steps just described, we are led to the desired partial derivatives of the implicit functions. Note that, in fact, we can obtain in one fell swoop a total of n of these (here, they are dyx/dxx,..., dyn/dxx).

What results from the above-cited steps is the following linear system:

<hi dyI \ dx, J ' dy2 \ dx, 3F2 I dy, \ 8F2 [ dy7

miiy*

dF2 dx.

dF"

dx dF"

dF"

hr, dx, dF" dx, dy, \ dx, J dy2 \ dx, j dyn v , where, for visual clarity, we have placed parentheses around those derivatives for which we are seeking a solution, to distinguish them from the other derivatives that are now considered to be constants. In matrix notation, this system can be written as

dF1

dFx

dFx

dy,

dy2

dyn

8F2

dF2

dF2

dy,

dy2

dyn

dF"

dF"

dF"

dy,

dy2

0 0

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