Definitions

# Implicit function

In mathematics, an implicit function is a generalization for the concept of a function in which the dependent variable has not been given "explicitly" in terms of the independent variable. To give a function f explicitly is to provide a prescription for determining the value of the function y in terms of the input value x:
y = f(x).
By contrast, the function is implicit if the value of y is obtained from x by solving an equation of the form:
R(x,y) = 0.
Formally, a function f:XY is said to be an implicit function if it satisfies the equation
R(x,f(x)) = 0
for all xX.

### Introduction

Implicit functions can often be useful in situations where it is inconvenient to solve explicitly an equation of the form R(x,y) = 0 for y in terms of x. Even if it is possible to rearrange this equation to obtain y as an explicit function f(x), it may not be desirable to do so since the expression of f may be much more complicated than the expression of R. In other situations, the equation R(x,y) = 0 may fail to define a function at all, and rather defines a kind of multiple-valued function. Nevertheless, in many situations, it is still possible to work with implicit functions. Some techniques from calculus, such as differentiation, can be performed with relative ease using implicit differentiation.

The implicit function theorem provides a link between implicit and explicit functions. It states that if the equation R(x, y) = 0 satisfies some mild conditions on its partial derivatives, then one can in principle solve this equation for y, at least over some small interval. Geometrically, the graph defined by R(x,y) = 0 will overlap locally with the graph of a function y = f(x).

Various numerical methods exist for solving the equation R(x,y)=0 to find an approximation to the implicit function y. Many of these methods are iterative in that they produce successively better approximations, so that a prescribed accuracy can be achieved. Many of these iterative methods are based on some form of Newton's method.

## Examples

### Inverse functions

Implicit functions commonly arise as one way of describing the notion of an inverse function. If f is a function, then the inverse function of f is a solution of the equation
$x=f\left(y\right) implies y = f^\left\{-1\right\}\left(x\right)$
for y in terms of x. Intuitively, an inverse function is obtained from f by interchanging the roles of the dependent and independent variables. Stated another way, the inverse function is the solution y of the equation
$R\left(x,y\right) = x-f\left(y\right) = 0.$

Examples.

1. The natural logarithm y = ln(x) is the solution of the equation x - ey = 0.
2. The product log is an implicit function given by x - y ey = 0.

### Algebraic functions

An algebraic function is a solution y for an equation R(x,y) = 0 where R is a polynomial of two variables. Algebraic functions play an important role in mathematical analysis and algebraic geometry. A simple example of an algebraic function is given by the unit circle:
$x^2+y^2-1=0.$
Solving for y gives
$y=pmsqrt\left\{1-x^2\right\}.$
Note that there are two "branches" to the implicit function: one where the sign is positive and the other where it is negative. Both branches are thought of as belonging to the implicit function. In this way, implicit functions can be multiple-valued.

## Caveats

Not every equation $R\left(x,y\right) = 0$ has a graph that is the graph of a function, the circle equation being one prominent example. Another example is an implicit function given by x - C(y) = 0 where C is a cubic polynomial having a "hump" in its graph. Thus, for an implicit function to be a true function it might be necessary to use just part of the graph. An implicit function can sometimes be successfully defined as a true function only after "zooming in" on some part of the x-axis and "cutting away" some unwanted function branches. A resulting formula may only then qualify as a legitimate explicit function.

The defining equation R = 0 can also have other pathologies. For example, the implicit equation x = 0 does not define a function at all; it is a vertical line. In order to avoid a problem like this, various constraints are frequently imposed on the allowable sorts of equations or on the domain. The implicit function theorem provides a uniform way of handling these sorts of pathologies.

## Implicit differentiation

In calculus, a method called implicit differentiation can be applied to implicitly defined functions. This method is an application of the chain rule allowing one to calculate the derivative of a function given implicitly.

As explained in the introduction, $y$ can be given as a function of $x$ implicitly rather than explicitly. When we have an equation $R\left(x,y\right)=0$, we may be able to solve it for $y$ and then differentiate. However, sometimes it is simpler to differentiate $R\left(x,y\right)$ with respect to $x$ and then solve for $dy/dx$.

### Examples

1. Consider for example

$y + x = -4 ,$

This function normally can be manipulated by using algebra to change this equation to an explicit function:

$f\left(x\right) = y = -x - 4 ,$

Differentiation then gives $frac\left\{dy\right\}\left\{dx\right\}=-1$. Alternatively, one can differentiate the equation:

$frac\left\{dy\right\}\left\{dx\right\} + frac\left\{dx\right\}\left\{dx\right\} = frac\left\{d\right\}\left\{dx\right\}\left(-4\right)$

$frac\left\{dy\right\}\left\{dx\right\} + 1 = 0$

Solving for $begin\left\{matrix\right\}frac\left\{dy\right\}\left\{dx\right\}end\left\{matrix\right\}$:

$frac\left\{dy\right\}\left\{dx\right\} = -1.$

2. An example of an implicit function, for which implicit differentiation might be easier than attempting to use explicit differentiation, is

$x^4 + 2y^2 = 8 ,$

In order to differentiate this explicitly, one would have to obtain (via algebra)

$f\left(x\right) = y = pmsqrt\left\{frac\left\{8 - x^4\right\}\left\{2\right\}\right\}$,

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

One might find it substantially easier to implicitly differentiate the implicit function;

$4x^3 + 4yfrac\left\{dy\right\}\left\{dx\right\} = 0$

thus,

$frac\left\{dy\right\}\left\{dx\right\} = frac\left\{-4x^3\right\}\left\{4y\right\} = frac\left\{-x^3\right\}\left\{y\right\}$

3. Sometimes standard explicit differentiation cannot be used and, in order to obtain the derivative, another method such as implicit differentiation must be employed. An example of such a case is the implicit function $y^3 - y = x$. It is impossible to express $y$ explicitly as a function of $x$ and $begin\left\{matrix\right\}frac\left\{dy\right\}\left\{dx\right\}end\left\{matrix\right\}$ therefore cannot be found by explicit differentiation. Using the implicit method, $begin\left\{matrix\right\}frac\left\{dy\right\}\left\{dx\right\}end\left\{matrix\right\}$ can be expressed:

$3y^2frac\left\{dy\right\}\left\{dx\right\} - frac\left\{dy\right\}\left\{dx\right\} = 1$

factoring out $frac\left\{dy\right\}\left\{dx\right\}$ shows that

$frac\left\{dy\right\}\left\{dx\right\}\left(3y^2 - 1\right) = 1$

$frac\left\{dy\right\}\left\{dx\right\}=frac\left\{1\right\}\left\{3y^\left\{2\right\}-1\right\}$ where $y ne pmfrac\left\{1\right\}\left\{sqrt\left\{3\right\}\right\}$

### Formula for two variables

"The Implicit Function Theorem states that if $F$ is defined on an open disk containing $\left(a,b\right)$, where $F\left(a,b\right)=0$, $F_y \left(a,b\right) not = 0$, and $F_x$ and $F_y$ are continuous on the disk, then the equation $F\left(x,y\right) = 0$ defines $y$ as a function of $x$ near the point $\left(a,b\right)$ and the derivative of this function is given by..."

$frac\left\{dy\right\}\left\{dx\right\} = -frac\left\{partial F / partial x\right\}\left\{partial F / partial y\right\} = -frac \left\{F_x\right\}\left\{F_y\right\}$.
$F_\left\{var\right\}$ indicates the derivative of $F$ with respect to $\left\{var\right\}$

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

$frac\left\{partial F\right\}\left\{partial x\right\} frac\left\{dx\right\}\left\{dx\right\} + frac\left\{partial F\right\}\left\{partial y\right\} frac\left\{dy\right\}\left\{dx\right\} = 0 = frac\left\{partial F\right\}\left\{partial x\right\} + frac\left\{partial F\right\}\left\{partial y\right\} frac\left\{dy\right\}\left\{dx\right\}$.

## Implicit function theorem

It can be shown that if $R\left(x,y\right)$ is given by a smooth submanifold $M$ in $R^2$, and $\left(a,b\right)$ is a point of this submanifold such that the tangent space there is not vertical (that is $frac\left\{partial R\right\}\left\{partial y\right\}ne0$), then $M$ in some small enough neighbourhood of $\left(a,b\right)$ is given by a parametrization $\left(x,f\left(x\right)\right)$ where $f$ is a smooth function. In less technical language, implicit functions exist and can be differentiated, unless the tangent to the supposed graph would be vertical. In the standard case where we are given an equation

$F\left(x,y\right) = 0$

the condition on $F$ can be checked by means of partial derivatives.