Implicit function

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Topics in calculus
Fundamental theorem Limits of functions Continuity Vector calculus</br>Tensor calculus Mean value theorem
Differentiation
Product rule Quotient rule Chain rule Implicit differentiation Taylor's theorem Related rates Table of derivatives
Integration
Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution

In mathematics, to give a function <math>f</math> implicitly is to give an equation <math>R(x,y) = S(x,y)</math> that at least in part has the same graph as <math>y = f(x)</math>. It can be useful to define a function <math>f</math> implicitly when there is no simple formula for <math>f(x)</math> so it is not convenient to give its graph in the form <math>y = f(x)</math>. If there is a way to rearrange the implicit equation, making the left hand side be <math>y</math> and the right hand side be a formula in <math>x</math> with no <math>y</math>, then the function can be explicitly defined.

1 Caveats
2 Implicit differentiation
3 Examples
4 Formula for two variables
5 Implicit function theorem
6 See also
7 External links

[edit] Caveats

Not every equation <math>R(x,y) = S(x, y)</math> has a graph that is the graph of a function. It might be necessary to use just part of the graph.

This may be true, as in the case of a graph that is a line; it may be true with some limitations, such as specifying that one cannot give a vertical line as a graph; it may be true with some limitations on the function domain, as when the relation is <math>x = C(y)</math> with <math>C</math> a cubic polynomial with a 'hump' in its graph; or it may be true only after also cutting <math>R</math> down to size, as in the case <math>x=y^2</math>. That is, an implicit function can sometimes be defined successfully only by modifying the relation by 'zooming in' to some part of the <math>x</math>-axis, and 'cutting back' unwanted function branches. A resulting formula may qualify as an ordinary explicit function.

[edit] Implicit differentiation

In calculus, implicit differentiation can be applied to implicit functions. This is by an application of the chain rule, to calculate derivatives <math>\begin{matrix}\frac{dy}{dx}\end{matrix}</math> without necessarily making <math>y</math> an explicit function of <math>x</math>. So, implicit differentiation is nothing more than a special case of the chain rule for derivatives.

[edit] Examples

Consider for example

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

Differentiation then gives <math>\frac{dy}{dx}=-1</math>. Equally, one can directly differentiate the implicit equation:

Solving for <math>\begin{matrix}\frac{dy}{dx}\end{matrix}</math>:

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

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

<math>f(x) = y = \pm\sqrt{\frac{8 - x^4}{2}}</math>,

and then differentiate this function. This creates two derivatives, one for <math>y > 0</math> and another for <math>y < 0</math>. Implicit differentiation avoids this.

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

thus,

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 <math>y^3 - y = x</math>. It is impossible to express <math>y</math> explicitly as a function of <math>x</math> (at least using elementary means, although the cubic formula will suffice for restricted values of <math>x</math> and <math>y</math>), which means that <math>\begin{matrix}\frac{dy}{dx}\end{matrix}</math> cannot be found by explicit differentiation. Using the implicit method, <math>\begin{matrix}\frac{dy}{dx}\end{matrix}</math> can be expressed;

factoring out <math>\frac{dy}{dx}</math> shows that

<math>\frac{dy}{dx}(3y^2 - 1) = 1</math> which yields the final answer

[edit] Formula for two variables

Suppose that y is an implicit function of x in the form <math>F(x, y) = 0</math>, and you want to find the implicit derivative <math>\begin{matrix}\frac{dy}{dx}\end{matrix}</math>. To avoid confusion in variables we temporarily introduce a dummy independent variable <math>x'</math>, and later let <math>x=x'</math>.

First, differentiating by <math>x'</math> using the chain rule:

<math>\frac{dF}{dx'} = \frac{\partial F}{\partial x}\frac{dx}{dx'} + \frac{\partial F}{\partial y}\frac{dy}{dx'}</math>

Now we make use of the fact that <math>x=x'</math>, and that F = 0:

<math>0 = \frac{\partial F}{\partial x}\cdot 1 + \frac{\partial F}{\partial y}\frac{dy}{dx}</math>

Now we solve for the general formula:

<math>\frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}</math>

You can verify that this works for all the examples above.

[edit] Implicit function theorem

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It has been suggested that this article or section be merged with Implicit_function_theorem. (Discuss)

It can be shown that if <math>R(x,y)</math> is given by a smooth submanifold <math>M</math> in <math>R^2</math>, and <math>(a,b)</math> is a point of this submanifold such that the tangent space there is not vertical, then <math>M</math> in some small enough neighbourhood of <math>(a,b)</math> is given by a parametrization <math>(x,f(x))</math> where <math>f</math> 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

the condition on <math>F</math> can be checked by means of partial derivatives.