设隐函数z=f(x,y)由方程x/z=lnz/y所确定,求az/ax,az/ay

Question

如题所述

Answer 1

化简，得

x/z=lnz-lny
x=zlnz-zlny
令F(x,y,z)=zlnz-zlny-x
Fx=-1
Fy=-z/y
Fz=lnz+1-lny
所以
az/ax=-Fx/Fz
=1/(lnz+1-lny)
az/ay=-Fy/Fz
=(z/y)/(lnz+1-lny)本回答被提问者采纳