Derivatives of implicit functions

Definition

A function defined by an equation of the form f(x, y) = 0 [in general, f(x1, x2, ... , xn) = 0 ]. If y is thought of as the dependent variable, f(x, y) = 0 is said to define y as an implicit function of x.

Implicit functions

Let y be related to x by the equation

f(x, y) = 0
../../images/implicit_function.gif

and suppose the locus is that shown in figure.

We cannot say that y is a function of x since at a particular value of x there is more than one value of y (because, in the figure, a line perpendicular to the x axis intersects the locus at more than one point) and a function is, by definition, single-valued. Although equation above does not define y as a function of x, we can say that on certain judiciously chosen segments of the locus y can be considered to be a single-valued function of x [expressible as y = f(x)]. For example, the segment P1P2 could be separated out as defining a function y = f(x). As a consequence, it is customary to say that equation defines y implicitly as a function of x; and we refer to y as an implicit function of x.

Implicit function theorem

The implicit function theorem provides a condition under which a relation defines an implicit function. It states that if the left hand side of the equation R(x, y) = 0 is differentiable and satisfies some mild condition on its partial derivatives at some point (a, b) such that R(a, b) = 0, then it defines a function y = f(x) over some interval containing a. Geometrically, the graph defined by R(x,y) = 0 will overlap locally with the graph of some equation y = f(x).

Derivaties of implicit functions

Consider the locus of f(x, y) = 0 shown in the figure. Let us ask the following question: “At a particular point on the locus what is the value of the quantity dy/dx?” This question can be answered at all points on the locus except points P1, P2, P3 and P4 (at these points the quantity dy/dx does not exist – it becomes infinite) and the answer is:

\dfrac{dy}{dx} = - \dfrac{\dfrac{\partial f(x,y)}{\partial x}}{\dfrac{\partial f(x,y)}{\partial y}}

If we have an equation of the type f(x, y) = 0, and certain conditions are met, we can view one of the variables as a function of the other in the vicinity of a particular point (x0, y0) that satisfies the equation. The conditions that must be met are stated in the implicit function theorem.

Differentiation of implicit functions

If we have an equation such as f(x, y, ... , u) = 0 which defines a variable as a function of others implicitly, there are two techniques for computing derivatives.

Direct differentiation

Given a particular variable to be considered as the dependent variable, if it is possible to solve the equation for the dependent variable in terms of the independent variables, we can compute the derivative directly by formula.

Example. Compute \frac{dy}{dx} for the equation y - 3x^2 + 5x + 1 = 0 .

Solution. Solve the equation for y to get

y = 3x^2 - 5x - 1

and compute the derivative directly as \frac{dy}{dx} = 6x - 5.

Implicit differentiation

Decide which variable is to be considered the dependent variable and which the independent. Say y is to be considered the dependent variable in f(x, y) = 0. Regarding y as the dependent variable, differentiate the equation as it stands with respect to the independent variable x and then solve the resulting relation for \frac{dy}{dx}. This method is known as implicit differentiation.

Example. Compute \frac{dy}{dx} for the equation x^5 + x^2y^3 - y^6 + 7 = 0 .

Solution. Differentiating implicitly, we get

\dfrac{dx^5}{dx}+\big(y^3\dfrac{dx^2}{dx}+x^2\dfrac{dy^3}{dx}\big)-\dfrac{dy^6}{dx}+\dfrac{d(7)}{dx}=0
5x^4 + 2xy^3 + 3x^2y^2\dfrac{dy}{dx} - 6y^5\dfrac{dy}{dx} = 0

Solving this for \frac{dy}{dx} gives

\dfrac{dy}{dx} = \dfrac{5x^4+2xy^3}{6y^5-3x^2y^2}

Because in most cases it is difficult or impossible to solve for the dependent variable, we usually use the method of implicit differentiation.