设z=f(x,y)由z+x+y-(e∧x+y+z)+2=0求dz

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设z=f(x,y)由z+x+y-(e∧x+y+z)+2=0求dz

设z=f(x,y)由z+x+y-(e∧x+y+z)+2=0求dz
设z=f(x,y)由z+x+y-(e∧x+y+z)+2=0求dz

设z=f(x,y)由z+x+y-(e∧x+y+z)+2=0求dz
z+x+y-e^(x+y+z)+2=0,x+y+z=0 显然不满足,即 x+y+z=0≠0.
两边对 x 求偏导数,注意 z 是 x,y 的函数,得
z‘+1-(1+z')e^(x+y+z)=0,解得 z' = -1,
同理 z'= -1.得 dz=-dx-dy.