Implicit function

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In mathematics, an implicit equation is a relation of the form ${\displaystyle R(x_{1},\dots ,x_{n})=0,}$ where R is a function of several variables (often a polynomial). For example, the implicit equation of the unit circle is ${\displaystyle x^{2}+y^{2}-1=0.}$

An implicit function is a function that is defined by an implicit equation, that relates one of the variables, considered as the value of the function, with the others considered as the arguments.^{[1]: 204–206} For example, the equation ${\displaystyle x^{2}+y^{2}-1=0}$ of the unit circle defines y as an implicit function of x if −1 ≤ x ≤ 1, and y is restricted to nonnegative values.

The

.

Examples

Inverse functions

A common type of implicit function is an

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

example below).

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

Example: The

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

Algebraic functions

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

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

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.

Caveats

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

having a "hump" in its graph. Thus, for an implicit function to be a true (single-valued) 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. Then an equation expressing y as an implicit function of the other variables can be written.

The defining equation R(x, y) = 0 can also have other pathologies. For example, the equation x = 0 does not imply a function f(x) giving solutions for y 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

provides a uniform way of handling these sorts of pathologies.

Implicit differentiation

In calculus, a method called implicit differentiation makes use of the chain rule to differentiate implicitly defined functions.

To differentiate an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y and then differentiate. Instead, one can

R(x, y) = 0 with respect to x and y and then solve the resulting linear equation for dy/dx to explicitly get the derivative in terms of x and y. Even when it is possible to explicitly solve the original equation, the formula resulting from total differentiation is, in general, much simpler and easier to use.

Examples

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

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

General formula for derivative of implicit function

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.

Implicit function theorem

Let R(x, y) be a differentiable function of two variables, and (a, b) be a pair of real numbers such that R(a, b) = 0. If ∂R/∂y ≠ 0, then R(x, y) = 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 (a, b) 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.^{[2]: §11.5}

In algebraic geometry

Consider a

.

In differential equations

The solutions of differential equations generally appear expressed by an implicit function.^[3]

Applications in economics

Marginal rate of substitution

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.

Marginal rate of technical substitution

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.

Optimization

Often in

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.

See also

References

Further reading

• ISBN 0-521-28952-1
  .
• ISBN 0-07-054235-X
  .
• Simon, Carl P.;
  ISBN 0-393-95733-0
  .

External links

• Archived at Ghostarchive and the Wayback Machine: "Implicit Differentiation, What's Going on Here?". 3Blue1Brown. Essence of Calculus. May 3, 2017 – via YouTube.