# Implicit Equation

In mathematics , an implicit equation is a relation of the form r ( x 1 , …, x n ) = 0 , where r is a function of many variables (often a polynomial ). For example, the implicit equation of the unit circle is 2 + 2 – 1 = 0 . An implicit function is a function that is defined by an implicit equation, one related to a variable, treated as the value of the function, the other treated as an argument .  : For example, the equation 2 + 2 – 1 = 0

defines the unit circle of y as an implicit function of x if -1 x 1 , and a relative bounds of y for negative values.

Provides the implicit function theorem whereby certain types of relations define relations defined as an implicit function, i.e. the zero set of pointer function of some consecutive differential multivariate function.

## Example

### Reverse 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 that gives a solution to the equation .

y=g(x)

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

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

G is an implicit definition defined as the inverse of -1 . For some functions g , −1 ( y ) can be written explicitly as a closed-form expression –

for example, if g ( x ) = 2 x − 1 , then g −1 ( y ) ) =1/2( Y + 1)

However, this is often not possible, or simply by introducing a new notation (as in the product log example below).

Intuitively, an inverse function is obtained from g by changing 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 .

### Algebraic function

An algebraic function is a function that satisfies a polynomial equation whose coefficients are the polynomials themselves. For example, an algebraic function in a 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 i ( x ) are polynomial functions of x . The solution of this algebraic function can be written on the right hand side of the equation y = f ( x ) . Written as such, 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 on the left side of the unit circle equation:

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

Solving for y gives an obvious 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 quadratic , cubic and quartile equations in y , the same is not true for quintic and higher degree equations in general, 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 .

## Warnings

Not every equation R ( x , y ) = 0 implies a graph of a single-valued function, the circle equation being a prime example. Another example is an implicit function given by x – C ( y ) = 0 where C is

a cubic polynomial with a “hump” in its graph. Thus, for an implicit function to be a real (single-valued) function it may be necessary to use only 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 off” some unwanted function branches. then yAn equation expressing as an implicit function of the other variables can be written.

There may be other distortions in the defined equation R ( x , y ) = 0 . For example, the equation x = 0 does not mean that a function f ( x ) gives a solution to y ; This is a vertical line. To avoid this kind of problem, various constraints are often imposed on the acceptable types of equations or domains . The implicit function theorem provides a similar approach in dealing with these types of distortions.

## Inherent discrimination

In calculus , a method called implicit differentiation uses chain rules to differentiate implicitly defined functions.

To differentiate an implicit function y ( x ) defined by the equation R ( x , y ) = 0 , it is usually not possible to explicitly solve for y and then differentiate. Instead, one can fully differentiate R ( x , y ) = 0 with respect to x and y and then solve the resulting linear equation DY / dx

Obviously to get the derivative in terms of x and y . Even when it is possible to explicitly solve the original equation, the formula resulting from the total variance is generally very simple and easy to use.

### Example

#### EXAMPLE 1

Consider

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

This equation is easy to solve for y , giving

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

where is the explicit form of the right-side function y ( x ) . differentiation then gives DY/dx= -1 ।

Alternatively, one can completely disassemble 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}}}

olve for DY/dx gives

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

#### 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 clearly differentiate this with respect to x , one must first obtain

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

And then isolate this function. This makes two derivatives: one for y 0 and the other for y < 0 .

It’s easy enough to explicitly separate 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 explicitly solve for y , and implicit differentiation is the only viable method of differentiation. An example equation is

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

It is impossible to express y algebraically explicitly as a

function of x , and therefore one cannot find dY/dx by clear differentiation. Using the implicit method, dY/dx can be obtained by differentiating the equation to get

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

Where from dx/dx= 1 . factoring out dY/dx shows that

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

which gives result

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

for which it is defined

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

### General Formula for Derivation of Implicit Function

If R ( x , y ) = 0 , the derivative of the implicit function y ( x ) is given by

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

where Rx and Ry represent the partial derivatives of R with respect to x and y .

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

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

Therefore

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

which, when resolved dY/dx, gives the above expression.

## Implicit work theorem

• The unit circle can be defined implicitly as the set of points (x, y) satisfying x2 + y2 = 1. Around point A, y can be expressed as an implicit function y(x). (Unlike in many cases, here this function can be made explicit as g1(x) = √1 − x2.) No such function exists around point B, where the tangent space is vertical.

Let R(x, y) be a differentiable function of two variables, and (a, b) be a pair of real numbers such that R(ab) = 0. If ∂R/∂y ≠ 0, then R(xy) = 0 defines an implicit function that is differentiable in some small enough neighbourhood of (a, b); in other words, there is a differentiable function f that is defined and differentiable in some neighbourhood of a, such that R(x, f(x)) = 0 for x in this neighbourhood. The condition ∂R/∂y ≠ 0 means that (ab) is

a regular point of the implicit curve of implicit equation R(x, y) = 0 where the tangent is not vertical.

In a less technical language, implicit functions exist and can be differentiated, if the curve has a non-vertical tangent

## In algebraic geometry

Consider the relation of the form R ( 1 , …, n ) = 0 , where R is a multivariate polynomial. The set of values ​​of the variables satisfying this relation is called an implicit curve if n = 2 and an implicit surface if n = 3 . Implicit equations are the basis of algebraic geometry, whose basic subjects of study are the simultaneous solutions of several implicit equations whose left-hand sides are polynomials. These sets of simultaneous solutions are called affine algebraic sets.

## In differential equations

Solutions of differential equations are usually expressed by an implicit function. 

## Applications in Economics

### Marginal rate of substitution

In economics, when the level set r ( x , y ) = 0 is an indifference curve for quantities X and Y of the two goods consumed, the absolute value of the implicit derivative dY/dxThe marginal rate of substitution of the two goods is interpreted as: how much y must be gained in order to be indifferent to the loss of one unit of x .

### Marginal Rate of Technical Substitution

Similarly, sometimes the level set r ( l , k ) is an isoquant showing different combinations of quantities used l of labor and k of physical capital each of which will result in the production of the same given quantity of output of some good. Absolute value of the derivative in this caseDK/DailyThe marginal rate of technical substitution between two factors of production is defined as: how much more capital the firm must use to produce the same amount of labor with one less unit of labor.

### Customization

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

Furthermore, the effects of the parameters of the problem on x * —the partial derivative of the implicit function—can be expressed as the total derivatives of the system of first-order conditions found using the total discriminant.