8.3
Implicit functions
So far we have been working with functions where the dependent variable can be explicitly written in terms of the independent variables, i.e., in the form y=f(x1, . . . , xn). Compare this form to the case g(x1, . . . , xn, y) = 0: if this equation assigns a unique value toy for each (x1, . . . , xn), we say that this equation definesy as an implicit function of (x1, . . . , xn).
For illustration, note that in the previous sections, we had discussed when an equation like y =f(x) defines the inverse of f, i.e., x as a function of y. If an inverse of f exists, we could say that the equation y = f(x) defined
x as an implicit function of y. But in many occasions, we may not be able to find an ”explicit” inverse function. We would still like to find the rate of change in the dependent variable with respect to the independent variables.
Example 8.3.1. • The equation xey = 1 defines y as an implicit func-tion of x. But we can solve for y as an explicit function of x as
y= logx.
• The equationx2+ 3xy+ 4y2 = 0 definesy as an implicit function ofx. We can still solve fory as a correspondence using the methods to solve a quadratic equation: y = 3±p7
8 x.
• There is no general method to solve an equation like 4exy 4x2 5y= 0. But note that ifx= 0, y = 4
5, if x= 1, y= 0 etc., thus defining y as a function ofx.
Remark. In Microeconomics, we often find functions that are implicitly defined. For instance, for di↵erentiable utility functions, Walrasian demand functions are defined as implicit functions of prices and income by the first order conditions of the utility maximization problem. We would like to find the rate of change of these demand functions with respect to prices and income, even if we are unable to solve the demand functions explicitly.
It is important to note that an expression like g(x, y) = c does not nec-essarily imply that y would be a function of x. If g(x, y) is C1, we can pick a solution of g(x, y) = c, say, (x⇤, y⇤). We will be able to define y as an implicit function ofx, sayy(x) at the neighbourhood of (x⇤, y⇤), if the graph
of g(x, y) = cdoes not change its slope at (x⇤, y⇤) in a way that for a value
Suppose giveng(x, y) = c, whereg isC1, and a particular solution of this equation, (x⇤, y⇤), we can define y as a continuously di↵erentiable function
ofx, y(x) near (x⇤, y⇤). Then y0(x) can be expressed in terms of the partials
of g:
@g(x⇤, y(x⇤))
@x dx+
@g(x⇤, y(x⇤))
@y dy(x
⇤) = 0
)y0(x⇤) =
@g(x⇤,y(x⇤))
@x
@g(x⇤,y(x⇤))
@y
Therefore, if y(x) solves g(x, y) = c and is C1, a necessary condition is
@g(x⇤,y(x⇤))
@y 6= 0 .
The implicit function theorem given below states that this necessary con-dition is also sufficient.
Theorem 8.7 (Implicit function theorem for one exogenous variable). Let
g(x, y) be a C1 function on an open ball around (x⇤, y⇤) in R2 such that
g(x⇤, y⇤) =c. If @g(x⇤,y⇤)
@y 6= 0 then there exists aC
1 functiony=y(x) defined
on an open intervalE around x⇤ such that
• g(x, y(x)) =c, for all x2E;
• y(x⇤) =y⇤;
• y0(x⇤) =
@g(x⇤,y⇤) @x
@g(x⇤,y⇤) @y
.
Note that for the equation g(x, y) =x2+y2 = 4, g(x, y) =x2 +y2 isC1 at (0,2) and (2,0), both of which satisfy the equation. Since @g(x,y)@y = 2y,
@g(0,2)
@y = 4 there exists a C
1 function y(x) in the neighbourhood of (0,2)
attaining slope @g@(0x,2)
@g(0,2) @y
= 0 at (0,2). But since @g(2,0)@y = 0, we cannot define
such a function at (2,0).
8.3.1
Implicit functions of several variables
Theorem 8.7 extends in a straightforward manner to implicit functions of several variables.
Theorem 8.8 (Implicit function theorem for multiple exogenous variables).
Letg(x1, . . . , xn, y) be a C1 function on an open ball around (x⇤1, . . . , x⇤n, y⇤) inRn+1 such that g(x⇤
1, . . . , x⇤n, y⇤) =c. If
@g(x⇤1,...,x⇤n,y⇤)
@y 6= 0, then there exists aC1 functiony=y(x
• g(x1, . . . , xn, y(x1, . . . , xn)) =c, for all (x1, . . . , xn)2B;
• y(x⇤1, . . . , x⇤n) =y⇤;
• @y(x⇤1,...,x⇤n) @xi =
@g(x⇤1,...,x⇤n,y⇤) @xi @g(x⇤1,...,x⇤n,y⇤)
@y
.
8.3.2
Systems of Implicit Functions
A system of m equations in m+n unknowns given by
g1(x1, . . . , xm+n) = 0 ... ... ...
gm(x1, . . . , xm+n) = 0
is called a system of implicit functions if there exists a partition of the vari-ables into independent and dependent varivari-ables, such that if the values of the independent variables are substituted into the above equations, the resulting system can be uniquely solved for the corresponding values of the dependent variables.
For instance, consider the following system ofmlinear equations inm+n
unknowns:
a11x1+· · ·+a1,m+nxm+n = 0
... ... ...
am1x1+· · ·+am,m+nxm+n = 0
Suppose x1, . . . , xm are treated as endogenous variables andxm+1, . . . , xm+n are treated as exogenous variables. Then we can define each endogenous variable as a function for given values of the endogenous variables if the matrix of coefficients of the endogenous variables is invertible, i.e.,
a11 · · · a1m ... ... ...
am1 · · · amm
6
= 0
If the above condition is fulfilled, then we can obtain the endogenous variables at any given value of exogenous variables xm+1, . . . , xm+n by
2 6 4
x1 ...
xm
3 7 5=
2 6 4
a11 · · · a1m ... ... ...
am1 · · · amm
3 7 5
12
6 4
a1,m+1xm+1 · · · a1,m+nxm+n
... ... ...
am,m+1xm+1 · · · am,m+nxm+n
The following Theorem generalizes the insight for systems of implicit linear functions to systems of implicit nonlinear functions:
Theorem 8.9 (Generalized Implicit Function Theorem). LetF1, . . . , Fm be
C1 functions from Rm+n toR. Consider the system of equations
F1(y
1, . . . , ym, x1, . . . , xn) = c1 ... ... ... ...
Fm(y1, . . . , ym, x1, . . . , xn) = cm
as possibly defining y1, . . . , ym as implicit functions of x1, . . . , xn. Suppose that (y⇤
1, . . . , y⇤m, x⇤1, . . . , x⇤n) is a solution of the above system of equations. If the determinant of the matrix
2 6 4
@F1
@y1 · · ·
@F1
@ym
... . .. ...
@Fm @y1 · · ·
@Fm @ym
3 7 5
evaluated at (y⇤
1, . . . , ym⇤, x⇤1, . . . , x⇤n) is nonzero, then there existC1 functions
y1 =f1(x1, . . . , xn) ... ... ...
ym =fm(x1, . . . , xn)
defined on an open ballB around (x⇤
1, . . . , x⇤n) such that
F1(f1(x), . . . , fm(x), x1, . . . , xn) =c1 ... ... ... ...
Fm(f1(x), . . . , fm(x), x1, . . . , xn) =cm
for all x= (x1, . . . , xn)2B and
y⇤1 =f1(x⇤1, . . . , x⇤n) ... ... ...
ym⇤ =fm(x⇤1, . . . , x⇤n).
Further, the partial derivatives @yk @xh(y
⇤,x⇤) = @fk @xh(x
substitutingdxh = 1 anddxj = 0 for j 6=h in
@F1 @y1
dy1+· · · @F1 @ym
dym+ @F1
@x1
dx1+· · · @F1 @xn
dxn = 0
... ... ... ...
@Fm @y1
dy1+· · · @Fm
@ym
dym+ @Fm
@x1
dx1+· · · @Fm
@xn
dxn = 0