Implicit function

A common type of implicit function is an inverse function. Not all functions have a unique inverse function. If g is a function of x that has a unique inverse, then the inverse function of g, called g⁻¹, is the unique function giving a solution of the equation

${\displaystyle y=g(x)}$

for x in terms of y. This solution can then be written as

${\displaystyle x=g^{-1}(y)\,.}$

Defining g⁻¹ as the inverse of g is an implicit definition. For some functions g, g⁻¹(y) can be written out explicitly as a closed-form expression — for instance, if g(x) = 2x − 1, then g⁻¹(y) = 1/2(y + 1). However, this is often not possible, or only by introducing a new notation (as in the product log example below).

Intuitively, an inverse function is obtained from g by interchanging the roles of the dependent and independent variables.

Example. The product log is an implicit function giving the solution for x of the equation y − xe^x = 0.

An algebraic function is a function that satisfies a polynomial equation whose coefficients are themselves polynomials. For example, an algebraic function in one variable x gives a solution for y of an equation

${\displaystyle a_{n}(x)y^{n}+a_{n-1}(x)y^{n-1}+\cdots +a_{0}(x)=0\,,}$

where the coefficients a_i(x) are polynomial functions of x. This algebraic function can be written as the right side of the solution equation y = f(x). Written like this, f is a multi-valued implicit function.

Algebraic functions play an important role in mathematical analysis and algebraic geometry. A simple example of an algebraic function is given by the left side of the unit circle equation:

${\displaystyle x^{2}+y^{2}-1=0\,.}$

Solving for y gives an explicit solution:

${\displaystyle y=\pm {\sqrt {1-x^{2}}}\,.}$

But even without specifying this explicit solution, it is possible to refer to the implicit solution of the unit circle equation as y = f(x), where f is the multi-valued implicit function.

While explicit solutions can be found for equations that are quadratic, cubic, and quartic in y, the same is not in general true for quintic and higher degree equations, such as

${\displaystyle y^{5}+2y^{4}-7y^{3}+3y^{2}-6y-x=0\,.}$

Nevertheless, one can still refer to the implicit solution y = f(x) involving the multi-valued implicit function f.

Example 1. Consider

${\displaystyle y+x+5=0\,.}$

This equation is easy to solve for y, giving

${\displaystyle y=-x-5\,,}$

where the right side is the explicit form of the function y(x). Differentiation then gives dy/dx = −1.

Alternatively, one can totally differentiate the original equation:

${\displaystyle {\begin{aligned}{\frac {dy}{dx}}+{\frac {dx}{dx}}+{\frac {d}{dx}}(5)&=0\,;\\[6px]{\frac {dy}{dx}}+1+0&=0\,.\end{aligned}}}$

Solving for dy/dx gives

${\displaystyle {\frac {dy}{dx}}=-1\,,}$

the same answer as obtained previously.

Example 2. An example of an implicit function for which implicit differentiation is easier than using explicit differentiation is the function y(x) defined by the equation

${\displaystyle x^{4}+2y^{2}=8\,.}$

To differentiate this explicitly with respect to x, one has first to get

${\displaystyle y(x)=\pm {\sqrt {\frac {8-x^{4}}{2}}}\,,}$

and then differentiate this function. This creates two derivatives: one for y ≥ 0 and another for y < 0.

It is substantially easier to implicitly differentiate the original equation:

${\displaystyle 4x^{3}+4y{\frac {dy}{dx}}=0\,,}$

giving

${\displaystyle {\frac {dy}{dx}}={\frac {-4x^{3}}{4y}}=-{\frac {x^{3}}{y}}\,.}$

Example 3. Often, it is difficult or impossible to solve explicitly for y, and implicit differentiation is the only feasible method of differentiation. An example is the equation

${\displaystyle y^{5}-y=x\,.}$

It is impossible to algebraically express y explicitly as a function of x, and therefore one cannot find dy/dx by explicit differentiation. Using the implicit method, dy/dx can be obtained by differentiating the equation to obtain

${\displaystyle 5y^{4}{\frac {dy}{dx}}-{\frac {dy}{dx}}={\frac {dx}{dx}}\,,}$

where dx/dx = 1. Factoring out dy/dx shows that

${\displaystyle \left(5y^{4}-1\right){\frac {dy}{dx}}=1\,,}$

which yields the result

${\displaystyle {\frac {dy}{dx}}={\frac {1}{5y^{4}-1}}\,,}$

which is defined for

${\displaystyle y\neq \pm {\frac {1}{\sqrt[{4}]{5}}}\quad {\text{and}}\quad y\neq \pm {\frac {i}{\sqrt[{4}]{5}}}\,.}$

If R(x, y) = 0, the derivative of the implicit function y(x) is given by[2]^:§11.5

${\displaystyle {\frac {dy}{dx}}=-{\frac {\,{\frac {\partial R}{\partial x}}\,}{\frac {\partial R}{\partial y}}}=-{\frac {R_{x}}{R_{y}}}\,,}$

where R_x and R_y indicate the partial derivatives of R with respect to x and y.

The above formula comes from using the generalized chain rule to obtain the total derivative — with respect to x — of both sides of R(x, y) = 0:

${\displaystyle {\frac {\partial R}{\partial x}}{\frac {dx}{dx}}+{\frac {\partial R}{\partial y}}{\frac {dy}{dx}}=0\,,}$

hence

${\displaystyle {\frac {\partial R}{\partial x}}+{\frac {\partial R}{\partial y}}{\frac {dy}{dx}}=0\,,}$

which, when solved for dy/dx, gives the expression above.

In economics, when the level set R(x, y) = 0 is an indifference curve for the quantities x and y consumed of two goods, the absolute value of the implicit derivative dy/dx is interpreted as the marginal rate of substitution of the two goods: how much more of y one must receive in order to be indifferent to a loss of one unit of x.

Similarly, sometimes the level set R(L, K) is an isoquant showing various combinations of utilized quantities L of labor and K of physical capital each of which would result in the production of the same given quantity of output of some good. In this case the absolute value of the implicit derivative dK/dL is interpreted as the marginal rate of technical substitution between the two factors of production: how much more capital the firm must use to produce the same amount of output with one less unit of labor.

Often in economic theory, some function such as a utility function or a profit function is to be maximized with respect to a choice vector x even though the objective function has not been restricted to any specific functional form. The implicit function theorem guarantees that the first-order conditions of the optimization define an implicit function for each element of the optimal vector x* of the choice vector x. When profit is being maximized, typically the resulting implicit functions are the labor demand function and the supply functions of various goods. When utility is being maximized, typically the resulting implicit functions are the labor supply function and the demand functions for various goods.

Moreover, the influence of the problem's parameters on x* — the partial derivatives of the implicit function — can be expressed as total derivatives of the system of first-order conditions found using total differentiation.