Question

设二元函数z=xe^(x+y)+(x+1)㏑(1+y),求全微分dz

Answer 1

z=xe^(x+y)+(x+1)㏑(1+y)

az/ax=e^(x+y)+xe^(x+y)+ln(1+y)

az/ay=xe^(x+y)+(x+1)/(1+y)

所以

dz=az/axdx+az/aydy

=[e^(x+y)+xe^(x+y)+ln(1+y)]dx+[xe^(x+y)+(x+1)/(1+y)]dy