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

SectioN 14.5 The Chain Rule 941
u
EXAMPLE 5 If u − x 4 y 1 y 2 z 3 , where x − rse t , y − rs 2 e 2t , and z − r 2 s sin t, find the
value of −uy−s when r − 2, s − 1, t − 0.
SOLUTION With the help of the tree diagram in Figure 4, we have
x
r s t r s t r s
FIGURE 44
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MasterID: 01598
y
z
t
−u
−s − −u
−x
−x
−s 1 −u
−y
−y
−s 1 −u
−z
−z
−s
− s4x 3 ydsre t d 1 sx 4 1 2yz 3 ds2rse 2t d 1 s3y 2 z 2 dsr 2 sin td
When r − 2, s − 1, and t − 0, we have x − 2, y − 2, and z − 0, so
−u
−s
− s64ds2d 1 s16ds4d 1 s0ds0d − 192
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EXAMPLE 6 If tss, td − f ss 2 2 t 2 , t 2 2 s 2 d and f is differentiable, show that t
satisfies the equation
t −t
−s 1 s −t
−t − 0
SOLUTION Let x − s 2 2 t 2 and y − t 2 2 s 2 . Then tss, td − f sx, yd and the Chain Rule
gives
−t
−s − −f
−x
−x
−s 1 −f
−y
−y
−s − −f −f
s2sd 1
−x −y s22sd
Therefore
−t
−t − −f
−x
−x
−t 1 −f
−y
−y
−t − −f −f
s22td 1
−x −y s2td
t −t
−s 1 s −t
−t − S2st −f −f
2 2st 1S22st
−x −yD −f −f
1 2st − 0
−x −yD
■
EXAMPLE 7 If z − f sx, yd has continuous second-order partial derivatives and
x − r 2 1 s 2 and y − 2rs, find (a) −zy−r and (b) − 2 zy−r 2 .
SOLUTIoN
(a) The Chain Rule gives
−z
−r − −z
−x
−x
−r 1 −z
−y
−y
−r − −z −z
s2rd 1
−x −y s2sd
(b) Applying the Product Rule to the expression in part (a), we get
5
− 2 z
−r 2
− −rS2r − −z −z
1 2s
−x −yD
− 2 S −z
−x 1 2r − −z 1 2s
−r −xD S − −z
−r −yD
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