Implicit function

Topics in Calculus

Fundamental theorem | Function | Limits of functions | Continuity | Calculus with polynomials

Differentiation

Product rule | Quotient rule | Chain rule | Implicit differentiation | Taylor's theorem

Integration
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Vector Calculus
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In mathematics, to give an implicit function f is to give the graph of a function, as a relation. That is, one produces somehow a binary relation R(x,y) in the hope that it suffices to define y as a conventional function of x. Of course when there is a direct way to re-arrange the formula, making y the subject, this is straightforward.

Contents

Caveats

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 x = C(y) with C a cubic polynomial with a 'hump' in its graph; or it may be true only after also cutting R down to size, as in the case x = y2. That is, an implicit function can sometimes be defined successfully only by modifying the relation by 'zooming in' to some part of the x-axis, and 'cutting back' unwanted function branches. A resulting formula may qualify as an ordinary explicit function.

Implicit differentiation

In calculus, implicit differentiation can be applied to implicit functions. This is by an application of the chain rule, to calculate derivatives dy/dx without necessarily making y an explicit function of x.

Examples

Consider for example

y + x = -4.

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

f(x) = y = −x - 4;

such differentiation would result in a value of −1. Equally, one can use implicit differentiation;

dy/dx + dx/dx = 0 = dy/dx + 1; dy/dx = -1.

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

x4 + 2y2 = 8.

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

f(x) = y = \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. Implicit differentiation avoids this.

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

4x3 + 4y(dy/dx) = 0;

thus,

dy/dx = −4x3 / 4y = −x3 / y.

Implicit function theorem

It can be shown that if R(x,y) is given by a smooth submanifold M in R2, and (a,b) is a point of this submanifold such that the tangent space there is not vertical, then M in some small enough neighbourhood of (a,b) is given by a parametrization (x,f(x)) 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(x,y) = 0

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

For the important generalisation to functions of several variables, see implicit function theorem.

Read about complex implicit functions at www.riemannsurfaces.info

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