y=ln tan x/2 -cosx lntanx
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发布时间：2024-10-18 21:39
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时间：4分钟前
dz =[1/tan(x/y)]*sec^2(x/y) *[dx/y + (-x/y^2)dy]
dz/dx = 2 /[y *sin(2x/y)]
dz/dy= - 2x/[y^2*sin(2x/y)]
dz/dx =d (ln tan x/y）/dx
=(1/tan( x/y)) *d (tan x/y)/dx
=(1/tan( x/y)) *sec^2( x/y)*d(x/y)
=(1/tan( x/y)) *sec^2( x/y)*dx/y
=2 /[y *sin(2x/y)]
dz/dy =d (ln tan x/y）/dy
=(1/tan( x/y)) *d (tan x/y)/dy
=(1/tan( x/y)) *sec^2( x/y)*d(x/y)
=(1/tan( x/y)) *sec^2( x/y)*(-x/y^2)
= - 2x/[y^2*sin(2x/y)]
