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高等数学求偏导数这题怎么做: x z=ln tan----- y
题目是z=ln tanx/y
希望能把这道题的步骤给我看
人气:408 ℃ 时间:2020-05-20 23:07:15
解答
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)]
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