Mathematical methods for economic theory

Martin J. Osborne

2.3 Derivatives of functions defined implicitly

One parameter

The equilibrium value of a variable x in some economic models is the solution of an equation of the form
f(xp) = 0,
where f is a function and p is a parameter. In such a case, we would sometimes like to know how the equilibrium value of x depends on the parameter. For example, does it increase or decrease when the value of the parameter increases?

Typically, we make assumptions about the form of the function f—for example, we might assume that it is increasing in x and decreasing in p—but do not assume that it takes a specific form. Thus typically we cannot solve explicitly for x as a function of p (i.e. we cannot write x = g(p) for some function g).

We say that the equation

f(xp) = 0 for all p
defines x implicitly as a function of p. We may emphasize this fact by writing f(x(p), p) = 0 for all p.

Before trying to determine how a solution for x depends on p, we should ask whether, for each value of p, the equation has a solution. Certainly not all such equations have solutions. The equation x2 + 1 = 0, for example, has no (real) solution. Even a single linear equation may have no solution in the relevant range. If, for example, the value of x is restricted to be nonnegative number (perhaps it is the quantity of a good), then for p > 0 the equation x + p = 0 has no solution.

If a single equation in a single variable has a solution, we may be able to use the Intermediate Value Theorem to show that it does. Assume that the function f is continuous, the possible values of x lie between x1 and x2, and for some value of p we have f(x1p) < 0 and f(x2p) > 0, or alternatively f(x1p) > 0 and f(x2p) < 0. Then the Intermediate Value Theorem tells us that there exists a value of x between x1 and x2 for which f(xp) = 0. (Note that even if these conditions are not satisfied, the equation may have a solution.)

If we cannot appeal to the Intermediate Value Theorem (because, for example, f is not continuous, or does not satisfy the appropriate conditions), we may be able to argue that a solution exists by appealing to the particular features of our equation.

Putting aside the question of whether the equation has a solution, consider the question of how a solution, if one exists, depends on the parameter p. Even if we cannot explicitly solve for x, we can find how a solution, if it exists, depends on p—we can find the derivative of x with respect to p, if it exists—by differentiating the equation that defines it. The principle to use is simple: if you want to find a derivative, differentiate!

Differentiating both sides of the equation f(x(p), p) = 0, using the chain rule, we get

f'1(x(p), p)x'(p) + f'2(x(p), p) = 0,
so that
x'(p) = −f'2(x(p), p)/f'1(x(p), p)
if f'1(x(p), p) ≠ 0. Notice that even though you cannot isolate x in the original equation, after differentiating the equation you can isolate the derivative of x in terms of the partial derivatives of f, which is what you want.

This calculation tells you, for example, that if f is an increasing function of both its arguments (f'1(xp) > 0 and f'2(xp) > 0 for all (xp)), then x is a decreasing function of p.

Application: slopes of level curves

The equation f(xy) = c of the level curve of the function f for the value c defines y implicitly as a function of x: we can write
f(xg(x)) = c for all x.
If the function g is differentiable, what is its derivative g'(x), the slope of the level curve at x? If f is differentiable and we differentiate both sides of the identity f(xg(x)) = c with respect to x we obtain
f'1(xg(x)) + f'2(xg(x))g'(x) = 0,
so that if f'2(xg(x)) ≠ 0 we can isolate g'(x):
g'(x) = −
f'1(xg(x))
f'2(xg(x))
or, in different notation,
dy
dx
 = −
f'1(xy)
f'2(xy)
.
Here is a precise result.
Proposition (Implicit function theorem)  
Let f be a continuously differentiable function of two variables defined on an open set S. If f'2(x0y0) ≠ 0 then there exists a continuously differentiable function g of a single variable defined on an open interval I containing x0 such that f(xg(x)) = f(x0y0) for all x ∈ I, and g'(x0), the slope of the level curve of f for the value f(x0y0) at the point (x0y0), is
f'1(x0y0)
f'2(x0y0)
.
Source  
For a proof, see Rudin (1964), Theorem 9.18 (p. 196), Rudin (1976), Theorem 9.28 (p. 224), or Apostol (1974), Theorem 13.7 (p. 374).

We deduce that the equation of the tangent to the level curve at (x0y0) is

y − y0 = −
f'1(x0y0)
f'2(x0y0)
·(x − x0).
(Remember that the equation of a line through (x0y0) with slope m is given by y − y0 = m(x − x0).) Thus the equation of the tangent may alternatively be written as
f'1(x0y0)(x − x0) + f'2(x0y0)(y − y0) = 0,
or
(f'1(x0y0), f'2(x0y0))
left parenthesis x − x0 right parenthesis
y − y0
 = 0.

The vector (f'1(x0y0), f'2(x0y0)) is called the gradient vector and is denoted ∇f(x0y0).

Let (xy) ≠ (x0y0) be a point on the tangent at (x0y0). Then the vector

left parenthesis x − x0 right parenthesis
y − y0
is parallel to the tangent. The previous displayed equation, in which the product of this vector with the gradient vector is 0, shows that the two vectors are orthogonal (the angle between them is 90°). Thus the gradient vector is orthogonal to the tangent, as illustrated in the following figure.

x y (x0, y0) (f1'(x0, y0), f2'(x0, y0)) f(x, y) = c y = g(x)

One can compute the second derivative of the level curve as well as the first derivative, by differentiating once again.

Many parameters

Suppose that the equilibrium value of the variable x is the solution of an equation of the form
f(xp) = 0,
where p is a vector of parameters—p = (p1, ..., pn), say. By differentiating the equation with respect to pi, holding all the other parameters fixed, we may determine how x varies with pi.

Recording the dependence of x on p explicitly in the notation, we have

f(x(p), p) = 0 for all p.
Differentiating this identity with respect to pi we have
f'x(x(p), p)x'i(p) + f'pi(x(p), p) = 0
so that if f'x(x(p), p) ≠ 0 we have
x'i(p) = −
f'pi(x(p), p)
f'x(x(p), p)
.

Example
Consider the competitive firm studied previously that uses a single input to produce a single output with the differentiable production function f, facing the price w for the input and the price p for output. Denote by z(wp) its profit-maximizing input for any pair (wp). We know that z(wp) satisfies the first-order condition
pf'(z(wp)) − w = 0 for all (wp).
How does z depend on w and p?

Differentiating with respect to w the equation that z(wp) satisfies we get

pf"(z(wp))z'w(wp) − 1 = 0.
Thus if f"(z(wp)) ≠ 0 then
z'w(wp) = 
1
pf"(z(wp))
.
We know that f"(z(w,p)) ≤ 0 given that z(wp) is a maximizer, so that if f"(z(wp)) ≠ 0 we conclude that z'w(wp) < 0, which makes sense: as the input price increases, the firm's optimal output decreases.

A similar calculation yields

z'p(wp) = −
f'(z(wp))
pf"(z(wp))
,
which for the same reason is positive.