Mathematics
Undergraduate
なぜこの問題を偏微分すると、ZxもZyも同じ答えになるのですか?
1
(2) z=
√x + y
(2)(1/2)=-
1
であるから,
2x =
2t√t
1
2(x+y)√x+y'
1
zy=-2(x+y)√x + y
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