=2e^(2x)siny/cosy
δz/δy = （(δu/δy)v - uδv/δy)/v^2 =[e^xcosy e^(-x)cosy +e^xsiny e^(-x)siny]/e^(-2x)(cosy)^2
=e^(2x)(secy)^2追问
麻烦可以手写一下吗？谢谢
追答不可以，都这样了还不自己看
本回答被提问者和网友采纳z=u/v, u=e^x.siny , v=e^(-x).cosy, ∂z/∂x, ∂z/∂y
solution:
u=e^x.siny
∂u/∂x = e^x.siny (1)
∂u/∂y = e^x.cosy (2)
v=e^(-x).cosy
∂v/∂x = -e^(-x).cosy (3)
∂v/∂y = -e^(-x). siny (4)
z=u/v
∂z/∂x
=(1/v).∂u/∂x - (u/v^2).∂v/∂x
=[1/(e^(-x).cosy) ] .[e^x.siny] - [ e^x.siny/( e^(-2x).(cosy)^2 ) ]. (-e^(-x).cosy )
=e^(2x).tany + e^(2x).tany
=2e^(2x).tany
∂z/∂y
=(1/v).∂u/∂y - (u/v^2).∂v/∂y
=[1/(e^(-x).cosy) ] .[ e^x.cosy ] - [e^x.siny/(e^(-2x).(cosy)^2) ] .[-e^(-x). siny ]
=e^(2x) + e^(2x). (tany)^2
=e^(2x) . (secy)^2
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