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

942 Chapter 14 Partial Derivatives
x
z
x
FIGURE 5
FIGURE 5
7et140505
05/04/10
MasterID: 01599
y
r s r s
But, using the Chain Rule again (see Figure 5), we have
−
S −z −
−r −xD S − −z
−x −xD S −x
−r 1 − −z
−y −xD −y
−r − −2 z
−x s2rd 1
2
−
S −z −
−r −yD S − −z
−x −yD S −x
−r 1 − −z
−y −yD −y
−r −
−2 z
−y −x s2sd
−2 z
−x −y s2rd 1 −2 z
−y s2sd 2
Putting these expressions into Equation 5 and using the equality of the mixed
second-order derivatives, we obtain
− 2 z
−r 2
− 2 −z
−x 1 2r S2r −2 z
−x 2
− 2 −z
−x 1 4r 2 −2 z
−x 2
1 2s − 2 z
−y −xD 1 2sS2r
2D
− 2 z
−x −y 1 2s −2 z
−y
1 8rs − 2 z
−x −y 1 4s 2 −2 z
−y 2 ■
Implicit Differentiation
The Chain Rule can be used to give a more complete description of the process of implicit
differentiation that was introduced in Sections 3.5 and 14.3. We suppose that an equation
of the form Fsx, yd − 0 defines y implicitly as a differentiable function of x, that is,
y − f sxd, where Fsx, f sxdd − 0 for all x in the domain of f . If F is differentiable, we can
apply Case 1 of the Chain Rule to differentiate both sides of the equation Fsx, yd − 0
with respect to x. Since both x and y are functions of x, we obtain
−F
−x
dx
dx 1 −F
−y
dy
dx − 0
But dxydx − 1, so if −Fy−y ± 0 we solve for dyydx and obtain
6
dy
dx − 2
−F
−x
−F
−y
− 2
F x
F y
To derive this equation we assumed that Fsx, yd − 0 defines y implicitly as a function
of x. The Implicit Function Theorem, proved in advanced calculus, gives conditions
under which this assumption is valid: it states that if F is defined on a disk containing
sa, bd, where Fsa, bd − 0, F y sa, bd ± 0, and F x and F y are continuous on the disk, then
the equation Fsx, yd − 0 defines y as a function of x near the point sa, bd and the derivative
of this function is given by Equation 6.
EXAMPLE 8 Find y9 if x 3 1 y 3 − 6xy.
SOLUTION The given equation can be written as
The solution to Example 8 should be
compared to the one in Example 3.5.2.
so Equation 6 gives
Fsx, yd − x 3 1 y 3 2 6xy − 0
dy
dx − 2 F x 3x 2 2 6y
− 2
F y 3y 2 2 6x − 2 x 2 2 2y
y 2 2 2x
■
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