A function given in the form of v = /(x), say, y = /(x) = 3xA (8.17)

is called an explicit junction, because the variable v is explicitly expressed as a function of x. If this function is written alternatively in the equivalent form y-3x4=0 («,17')

however, wre no longer have an explicit function. Rather, the function (8.17) is then only implicitly defined by the equation (8.17')- When we are (only) given an equation in the form of (8.17'). therefore, the function y = f{x) which it implies, and whose specific form may not even be known to us, is referred to as an implicit junction.

An equation in the form of (8.17') can be denoted in general by F(y, x) = 0, because its left side is a function of the two variables y and x. Note that we arc using the capital letter Fherc to distinguish it from the function /; the function F? representing the left-side expression in (8.17'), has two arguments, y and x, whereas the function /,1 representing the implicit function, has only one argument, x. There may, of course, be more than two arguments in the F function. For instance, wc may encounter an equation F(y. ,. -. . -*,„) = 0. Such an equation may also define an implicit function y = j\x\,..., xm).

The equivocal word may in the last sentence was used advisedly. For, whereas an explicit function, say, y = f(x), can always be transformed into an equation F(y, x) = 0 by simply transposing the fix) expression to the left side of the equals sign, the reverse transformation is not always possible. Indeed, in certain cases, a given equation in the form of F(y,x) = 0 may not implicitly define a function y = fix). For instance, the equation x2 -hy2 = 0 is satisfied only at the point of origin (0, 0), and hence yields no meaningful function to speak of. As another example, the equation

C ha pte r 8 Compara ri ve-Static A nalysis oj Gene) -at- Func tion Models 195

implies not a function, but a relation, because (8.18) plots as a circle, as shown in Fig. 8.6, so that no unique value ofj> corresponds to each value of x. Note, however, that if we restrict^ to nonnegative values, then we will have the upper half of the circic only, and that does constitute a function, namely, y = Similarly, the lower half of the circle, with y values nonpositive, constitutes another function, y — —s/9 — x2. In contrast, neither the left half nor the right half of the circic can qualify as a function.

In view of this uncertainty, it becomes of interest to ask whether there arc known general conditions under which we can be sure that a given equation in the form of

docs indeed define an implicit function y = /(xi,...,xM) (8.20)

locally, i.e., around some specific point in the domain. The answer to this lies in the so-called implicit-function theorem, which states that;

Given (8.19), if (tf) the function t has continuous partial derivatives Fv, F¡,____Fm, and if

(fc) at a point xio,. -. - xmo) satisfying the equation (8,19), Ft. is nonzero, then there exists an m-dimensional neighborhood of (xki, ..., .v„fo), N, in whichy is an implicitly defined function of the variables*!,..., in the fonn of (8,20), This implicit function satisfies

Vo = f(xin, ■ ■ -, ¿mil) ■ It also satisfies the equation (8.19) for every m-tuplc (x\_____ xm) in the neighborhood N—thereby giving (8.19) the status of an identity in that neighborhood. Moreover, the implicit function/is continuous and has continuous partial derivatives

Let us apply this theorem to the equation of the circle, (8.18), which contains only one x variable. First, we can duly verify that Fv = 2y and Fx == 2x are continuous, as required. Then we note that Fy is nonzero except when y — 0, that is, except at the leftmost point (-3, 0) and the rightmost point (3, 0) on the circle. Thus, around any point on the circle except (-3, 0) and (3, 0), we can construct a neighborhood in which the equation (8.18) defines an implicit function y = /'(*). This is easily verifiable in Fig, 8,6, where it is indeed possible to draw, say, a rectangle around any point on the circic- except (-3, 0) and (3, 0)—such that the portion of the circlc enclosed therein will constitute the graph of a function, with a unique^ value for each value of x in that rectangle.

Several things should be noted about the implicit-function theorem. First, the conditions cited in the theorem are in the nature of sufficient (but not ncccssary) conditions. This means that if we happen to find Fv = 0 at a point satisfying (8.19), we cannot use the theorem to deny the existence of an implicit function around that point. For such a function may in fact exist (see Exercise 8.5-7)/ Second, even if an implicit function/is assured to exist, the theorem gives no cluc as to the specific form the function/takes. Nor, for that matter, docs it tell us the exact size of the neighborhood jV in which the implicit function is defined. However, despite these limitations, this theorem is one of great importance. For whenever the conditions of the theorem are satisfied, it now becomes meaningful to talk about and make use of a function such as (8.20), even if our model may contain an equation (8,19) which is difficult or impossible to solve explicitly for y in terms of the x variables. Moreover, since the theorem also guarantees the existence of the partial derivatives /i, -.., fm, it is now also meaningful to talk about these derivatives of the implicit function.

If the equation F{y\ x\9 ...,xm) = 0 can be solved for >\ wc can explicitly write out the function y = f(x\____fxm), and find its derivatives by the methods learned before. For instance, (8.18) can be solved to yield two .separate functions v+ = + V9 x2 [upper half of circle]

v = -\/9 x2 [lower half of circle] and their derivatives can be found as follows:

ax ax

d-f = AH9 _ x2} 1/2] = _ 1(9 _ ¿rV2{-2x) dx dx x —x

But what if the given equation, F(y\ ,..., ~y„t) = 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 fory 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

T On thé other hand, if Fy = 0 in an entire neighborhood, then it can be concluded that no implicit function is defined in that neighborhood. By the same token if fy = 0 identically, then no implicit function exists anywhere.

Chapter 8 Comparative-Static Analysis of General-Function Models 197

equal, their respective total differentials must be equal;* (2) differentiation of an expression that involves y, n,...,^ will yield an expression involving the differentials dy, dx\, *.., dxnt; and (3) the differentia! of y> dy, can be substituted out, so the fact that we cannot solve for/ does not matter.

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 — i/0, or

Since the implicit function/ = f(xu ■ ■ -, xm) has the total differential dy = J\ dxi + f2dx2^----+ fn dx„

we can substitute this dy expression into (8,22) to get (after collecting terms)

(Fy /, + Fx) dx} + (Fy h + F2) dx2 + >- + (FM fm + Fm) dxm = 0 (8,22')

The fact that all the dxcan vary independently from one another means that, for the equation (8.22') to hold, each parenthesized expression must individually vanish; i.e., we must have

Dividing through by Fvt and solving for fif we obtain the so-called implicit-function rule for finding the partial derivative / of the implicit function y = f(x\, ««■, xm):

dXi Fy

In the simple case where the given equation is F(y, x) = 0, the rule gives f = ~T <8'23'>

dx Fy r Take, for example, the identity x2 - y2 = (x + y)(x - y)

This is an identity because the two sides are equal for any values of x and y that one may assign. Taking the total differential of each side, we have d(left side) = 2x dx - 2y dy rf(right side) = (x - y) d(x -1- 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 - Z for instance, is valid only for y = ±1. The total differentials of the two sides are d(left side) = 2x dx - 2y dy ¿(right side) = 2x dx + 2y dy which are not equal. Note, in particular, that they are not equal even at y = ±1.

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 ty always appears in the denominator of the ratio. This being the case, it is not admissible to have Fv = 0. Since the implicit-function theorem specifies that Fv / 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

Example 2

Find dy/dx for the implicit function defined by (8.17'). Since f(y, x) takes the form of y - 3x4; we have, by (8.23'), dy Fx -12x3 , dx Fy 1

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

Find dy/dx for the implicit functions defined by the equation of the circle (8,18). This time we have f{y, x) = x2 + y2 - 9; thus Fy = 2y and Fx = 2x. By (8.23'), 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 (8.18') and their derivatives in (8.21). If we substitute y4" for y in the implicit-function-rule result dy/dx — -x/y, we will indeed obtain the derivative dyh ¡dx as shown in (8.21); similarly, the substitution of y~ for y will yield the other derivative in (8.21). Thus our earlier assertion is duly verified.

Find dy/dx for any implicit function(s) that may be defined by the equation F(yf x, w) = ylx2 + w* + yxw- 3-0. This equation is not easily solved for y> But since Fy, FXt 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/tlx. By (8.23), moreover, we can immediately write

At the point (1, 1, 1), this derivative has the value -

Assume that the equation f(Q, K,L) = 0 implicitly defines a production function Q = f{K, i). Let us find a way of expressing the marginal physical products MPP* and MPP^ in relation to the function F. Since the marginal products are simply the partial derivatives DQ/dK and ilQfiïL, we can apply the implicit-function rule and write i>Q

dl Fn

* The restriction y / 0 is of course perfectly consistent with our earlier discussion of the equation (8.18) that follows the statement of the implicit-function theorem.

Chapter 8 Comparai ive-Sla ! ic A naiysis of General-Function Models 199

Aside from these, we can obtain yet another partial derivative, dK _ Ft from the equation F(Q, K, i) = 0. What is the economic meaning of dK/SL? The partial sign implies that the other variable, Q, is being held constant; it follows that the changes in Kand L described by this derivative are in the nature of "compensatory" charges 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 drawn with the ft variable on the vertical axis and the I variable on the horizontal axis. As a matter of fact, the derivative UK/¡it is the measure of the slope of such an isoquant, which is negative in the normal case. The absolute value of l)K j\)Lf on the other hand, is the measure of the morginaf rote of technical substitution between the two inputs.

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