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h'(x) = f'(g(x))g'(x)    

Alternately, if we let y = g(x), z = f (y), then we may write the formula in the following way (using the alternate notation for derivatives):

=    

This is easy to remember, because it looks like the dy are quantities that cancel. While convenient, one must be careful to realize that dy is just a notational device; it does not represent a number and cannot be haphazardly manipulated as such.

Implicit Differentiation

Sometimes we encounter an equation relating two variables that does not come from a function. One familiar example is the equation for a unit circle, x2 + y2 = 1. While this equation is not a function in itself, its graph of its solutions is made up of the graph of two functions defined on the interval [- 1, 1]: f (x) = and g(x) = - . These functions are said to be implicit functions for the equation.

In the case of the unit circle, we were able to write down the implicit functions explicitly, but this is not always possible. As an example, consider the equation x2y2 = x + y, the graph of whose solutions resembles an "infinite boomerang," displayed below.

Figure %: Plot of x2y2 = x + y

It is not possible to find a simple formula for x or y, so we cannot write down the implicit functions. But we still may want to know the slope of the graph at a particular point, that is, the derivative of an implicit function at that point. Implicit differentiation allows us to do this.

The idea is to differentiate both sides of the equation with respect to x (using the chain rule where necessary). The two sides must remain equal under this differentiation. Then we solve for y'(x) in terms of x and y. The fact that we need to know both the x- and y-coordinates of a point in order to compute the derivative should come as no surprise, since two different points on the graph may very well have the same x- coordinate. The full set of solutions to an equation is not, in general, the graph of a function.

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