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

940 Chapter 14 Partial Derivatives
EXAMPLE 3 If z − e x sin y, where x − st 2 and y − s 2 t, find −zy−s and −zy−t.
SOLUTION Applying Case 2 of the Chain Rule, we get
−z
−s − −z
−x
−x
−s 1 −z
−y
− t 2 e st 2 sin ss 2 td 1 2ste st 2 cosss 2 td
−z
−t − −z
−x
−x
−t 1 −z
−y
−y
−s − se x sin ydst 2 d 1 se x cos yds2std
−y
−t − se x sin yds2std 1 se x cos ydss 2 d
− 2ste st 2 sin ss 2 td 1 s 2 e st 2 cosss 2 td ■
x
s
z
x
x
x
t
FIGURE 22
z
y
s
z
y
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y
s t s t
y
t
Case 2 of the Chain Rule contains three types of variables: s and t are independent
variables, x and y are called intermediate variables, and z is the dependent variable.
Notice that Theorem 3 has one term for each intermediate variable and each of these
terms resembles the one-dimensional Chain Rule in Equation 1.
To remember the Chain Rule, it’s helpful to draw the tree diagram in Figure 2. We
draw branches from the dependent variable z to the intermediate variables x and y to
indicate that z is a function of x and y. Then we draw branches from x and y to the independent
variables s and t. On each branch we write the corresponding partial derivative.
To find −zy−s, we find the product of the partial derivatives along each path from z to s
and then add these products:
−z
−s − −z
−x
−x
−s 1 −z
−y
Similarly, we find −zy−t by using the paths from z to t.
Now we consider the general situation in which a dependent variable u is a function
of n intermediate variables x 1 , . . . , x n , each of which is, in turn, a function of m independent
variables t 1 , . . . , t m . Notice that there are n terms, one for each intermediate variable.
The proof is similar to that of Case 1.
−y
−s
4 The Chain Rule (General Version) Suppose that u is a differentiable function
of the n variables x 1 , x 2 , . . . , x n and each x j is a differentiable function of
the m variables t 1 , t 2 , . . . , t m . Then u is a function of t 1 , t 2 , . . . , t m and
−u
− −u −x 1
−t i −x 1 −t i
for each i − 1, 2, . . . , m.
1 −u
−x 2
−x 2
−t i
1 ∙ ∙ ∙ 1 −u
−x n
−x n
−t i
x
FIGURE 33
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y
w
u v u v u v u v
z
t
EXAMPLE 4 Write out the Chain Rule for the case where w − f sx, y, z, td and
x − xsu, vd, y − ysu, vd, z − zsu, vd, and t − tsu, vd.
SOLUTION We apply Theorem 4 with n − 4 and m − 2. Figure 3 shows the tree diagram.
Although we haven’t written the derivatives on the branches, it’s understood that
if a branch leads from y to u, then the partial derivative for that branch is −yy−u. With
the aid of the tree diagram, we can now write the required expressions:
−w
−u − −w
−x
−w
−v − −w
−x
−x
−u 1 −w
−y
−x
−v 1 −w
−y
−y
−u 1 −w
−z
−y
−v 1 −w
−z
−z
−u 1 −w
−t
−z
−v 1 −w
−t
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−t
−u
−t
−v
■