Implicit function

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Topics in calculus

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

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[edit] Caveats

Not every equation R(x,y) = S(x,y) 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 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.

[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 \begin{matrix}\frac{dy}{dx}\end{matrix} without necessarily making y an explicit function of x. Therefore, implicit differentiation is nothing more than a special case of the chain rule for derivatives.

[edit] Examples

Consider for example

y + x = -4 \,

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

f(x) = y = -x - 4 \,

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

\frac{dy}{dx} + \frac{dx}{dx} = \frac{d}{dx}(-4)
\frac{dy}{dx} + 1 = 0

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

\frac{dy}{dx} = -1.

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(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;

4x^3 + 4y\frac{dy}{dx} = 0

thus,

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

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 y3 − y = x. It is impossible to express y explicitly as a function of x (at least using elementary means, although the cubic formula will suffice for restricted values of x and y), which means that \begin{matrix}\frac{dy}{dx}\end{matrix} cannot be found by explicit differentiation. Using the implicit method, \begin{matrix}\frac{dy}{dx}\end{matrix} can be expressed;

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

factoring out \frac{dy}{dx} shows that

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

[edit] Formula for two variables

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

First, differentiating by x' using the chain rule:

\frac{dF}{dx'} = \frac{\partial F}{\partial x}\frac{dx}{dx'} + \frac{\partial F}{\partial y}\frac{dy}{dx'}

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

0 = \frac{\partial F}{\partial x}\cdot 1 + \frac{\partial F}{\partial y}\frac{dy}{dx}

Now we solve for the general formula:

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

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

[edit] Implicit function theorem

In multivariable calculus, a branch of mathematics, the implicit function theorem is a tool which allows relations to be converted to functions. It does this by representing the relation as the graph of a function. There may not be a single function whose graph is the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function.

[edit] Example

Consider the unit circle. If we define the function f as f(x,y) = x2 + y2 − 1, then the relation f(x,y) = 0 cuts out the unit circle. Explicitly, the unit circle is the set {(x,y) | f(x,y) = 0}. There is no way to represent the unit circle as the graph of a function y = g(x) because for each choice of x \in (-1,1), there are two choices of y, namely \sqrt{1-x^2} and -\sqrt{1-x^2}.

However, it is possible to represent part of the circle as a function. If we let g_1(x) = \sqrt{1-x^2} for − 1 < x < 1, then the graph of y = g1(x) gives the upper half of the circle. Similarly, if g_2(x) = -\sqrt{1-x^2}, then the graph of y = g2(x) gives the lower half of the circle.

It is not possible to find a function which will cut out a neighbourhood of (1,0) or ( − 1,0). Any neighbourhood of (1,0) or ( − 1,0) contains both the upper and lower halves of the circle. Because functions must be single-valued, there is no way of writing both the upper and lower halves using one function y = g(x). Consequently there is no function whose graph looks like a neighbourhood of (1,0) or ( − 1,0). In this case the conclusion of the implicit function theorem fails.

The purpose of the implicit function theorem is to tell us the existence of functions like g1(x) and g2(x) in situations where we cannot write down explicit formulas. It guarantees that g1(x) and g2(x) are differentiable, and it even works in situations where we do not have a formula for f(x,y).<>

[edit] Statement of the theorem

Let f : Rn+m → Rm be a continuously differentiable function. We think of Rn+m as the cartesian product Rn × Rm, and we write a point of this product as (x1, ..., xn, y1, ..., ym). f is the given relation. Our goal is to construct a function g : Rn → Rm whose graph is the set of all (x1, ..., xn, y1, ..., ym) such that f(x1, ..., xn, y1, ..., ym) = 0.

As noted above, this may not always be possible, so we will fix a point (a1, ..., an, b1, ..., bm) which satisfies f(a1, ..., an, b1, ..., bm) = 0, and we will ask for a g that works near the point (a1, ..., an, b1, ..., bm). In other words, we want an open set U of Rn, an open set V of Rm, and a function g : U → V such that the graph of g equals the relation f = 0 on U × V. In symbols,

\{ (x_1, \ldots, x_n, g(x_1, \ldots, x_n)) \} = \{ (x_1, \ldots, x_n, y_1, \ldots, y_m) | f(x_1, \ldots, x_n, y_1, \ldots, y_m) = 0 \} \cap (U \times V)

To state the implicit function theorem, we need the Jacobian, also called the differential or total derivative, of f. This is the matrix of partial derivatives of f. Abbreviating (a_1, \ldots, a_n, b_1, \ldots, b_m) to (a,b), the Jacobian matrix is

\begin{matrix} (Df)(a,b) & = & \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(a,b) & \cdots & \frac{\partial f_1}{\partial x_n}(a,b) & \frac{\partial f_1}{\partial y_1}(a,b) & \cdots & \frac{\partial f_1}{\partial y_m}(a,b)\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ \frac{\partial f_m}{\partial x_1}(a,b) & \cdots & \frac{\partial f_m}{\partial x_n}(a,b) & \frac{\partial f_m}{\partial y_1}(a,b) & \cdots & \frac{\partial f_m}{\partial y_m}(a,b)\\ \end{bmatrix}\\ & = & \begin{bmatrix} X & | & Y \end{bmatrix}\\ \end{matrix}

where X is the matrix of partial derivatives in the x's and Y is the matrix of partial derivatives in the y's. The implicit function theorem says that if Y is an invertible matrix, then there are U, V, and g as desired. Writing all the hypotheses together gives the following statement.

Let f : \bold R^{n+m} \rightarrow \bold R^m be a continuously differentiable function, and let \bold R^{n+m} have coordinates (x_1, \ldots, x_n, y_1, \ldots, y_m). Fix a point (a_1, \ldots, a_n, b_1, \ldots, b_m) = (a,b). If the matrix \begin{bmatrix} (\partial f_i / \partial y_j)(a,b) \end{bmatrix} is invertible, then there exists an open set U containing (a_1, \ldots, a_n), an open set V containing (b_1, \ldots, b_m), and a differentiable function g : U \rightarrow V such that \{ (x_1, \ldots, x_n, g(x_1, \ldots, x_n)) \} = \{ (x_1, \ldots, x_n, y_1, \ldots, y_m) | f(x_1, \ldots, x_n, y_1, \ldots, y_m) = 0 \} \cap (U \times V).

[edit] References

[edit] External links

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