Implicit Differentiation
The functions that we have met so far can be described by expressing one variable explicitly in terms of another
variable—for example,
y = or y = x sin x or, in general, y = f(x).
Some functions, however, are defined implicitly by a relation between x and y such as
x2 + y2 = 25 or
x3 + y3 = 6xy
Implicit Differentiation
In some cases it is possible to solve such an equation for y as an explicit function (or several functions) of x.
For instance, if we solve Equation 1 for y, we get
y = , so two of the functions determined by the
implicit Equation 1 are f(x) = and g(x) = .
Implicit Differentiation
The graphs of f and g are the upper and lower semicircles of the circle x2 + y2 = 25. (See Figure 1.)
Figure 1
Implicit Differentiation
It’s not easy to solve Equation 2 for y explicitly as a function of x by hand. (A computer algebra system has no trouble, but the expressions it obtains are very complicated.)
Nonetheless, (2) is the equation of a curve called the folium of Descartes shown in Figure 2 and it implicitly defines y as several functions of x.
Implicit Differentiation
The graphs of three such functions are shown in Figure 3.
When we say that f is a function defined implicitly by Equation 2, we mean that the equation
x3 + [f(x)]3 = 6xf(x) is true for all values of x in the domain of f.
Figure 3
Graphs of three functions defined by the folium of Descartes
Implicit Differentiation
Fortunately, we don’t need to solve an equation for y in terms of x in order to find the derivative of y. Instead we can use the method of implicit differentiation.
This consists of differentiating both sides of the equation with respect to x and then solving the resulting equation for y′.
In the examples and exercises of this section it is always assumed that the given equation determines y implicitly as a differentiable function of x so that the method of implicit differentiation can be applied.
Example 1
(a) If x2 + y2 = 25, find .
(b) Find an equation of the tangent to the circle x2 + y2 = 25 at the point (3, 4).
Solution 1:
(a) Differentiate both sides of the equation x2 + y2 = 25:
Example 1 – Solution
Remembering that y is a function of x and using the Chain Rule, we have
Thus
Now we solve this equation for dy/dx:
cont’d
Example 1 – Solution
(b) At the point (3, 4) we have x = 3 and y = 4, so
An equation of the tangent to the circle at (3, 4) is therefore y – 4 = (x – 3) or 3x + 4y = 25
Solution 2:
(b) Solving the equation x2 + y2 = 25 for y, we get y = . The point (3, 4) lies on the upper
semicircle y = and so we consider the function
f(x) = .
cont’d
Example 1 – Solution
Differentiating f using the Chain Rule, we have
So
and, as in Solution 1, an equation of the tangent is 3x + 4y = 25.
cont’d
