James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 3.5 Implicit Differentiation 209
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. The graphs of three such
functions are shown in Figure 3. When we say that f is a function defined implicitly by
Equa tion 2, we mean that the equation
x 3 1 f f sxdg 3 − 6x f sxd
is true for all values of x in the domain of f .
y
˛+Á=6xy
y
y
y
0
x
0
x
0 x
0
x
FIGURE 2 The folium of Descartes
FIGURE 3 Graphs of three functions defined by the folium of Descartes
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 y9. 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 x 2 1 y 2 − 25, find dy
dx .
(b) Find an equation of the tangent to the circle x 2 1 y 2 − 25 at the point s3, 4d.
SOLUTION 1
(a) Differentiate both sides of the equation x 2 1 y 2 − 25:
d
dx sx 2 1 y 2 d − d
dx s25d
d
dx sx 2 d 1 d
dx sy2 d − 0
Remembering that y is a function of x and using the Chain Rule, we have
d
dx sy2 d − d dy sy2 d dy dy
− 2y
dx dx
Thus
2x 1 2y dy
dx − 0
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