怎么求偏导数的，很急，第七题的第三小题和第五小题

如题所述

第1个回答推荐于2016-06-22

(3)
z=lntan(x/y)
∂z/∂x = [1/tan(x/y)] ∂/∂x ( tan(x/y)
= [1/tan(x/y)] (sec(x/y))^2 . ∂/∂x (x/y)
= [1/tan(x/y)] (sec(x/y))^2 . (1/y)

∂z/∂y = [1/tan(x/y)] (sec(x/y))^2 . ∂/∂y (x/y)
= [1/tan(x/y)] (sec(x/y))^2 . (-x/y^2)
(5)
z= sin(xy) + [cos(xy)]^2
∂z/∂x =cos(xy) . ∂/∂x (xy) + 2[cos(xy)] ∂/∂x(cos(xy))
=cos(xy) . (y) + 2[cos(xy)] (-sin(xy)) ∂/∂x (xy)
=cos(xy) . (y) - 2[cos(xy)] (sin(xy)) (y)

∂z/∂y =cos(xy) . ∂/∂y (xy) -2[cos(xy)] (sin(xy)) ∂/∂y (xy)
=cos(xy) . (x) -2[cos(xy)] (sin(xy)) (x)本回答被提问者和网友采纳

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第七题偏导数怎么求,过程答：答案如下

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