设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax
问题描述:
设函数z=x^2+y^2 ,而y=y(x)由方程(e^xy)-y=0所确定,求az/ax
答
(e^xy)-y=0两端对x求导得e^(xy)(y+xy')-y'=0y'=y*e^(xy)/[1-xe^(xy)]函数z=x^2+y^2 ∂z/∂x=2x+∂z/∂/y*dy/dx=2x+2y*y*e^(xy)/[1-xe^(xy)]=2x+2y^2*e^(xy)/[1-xe^(xy)]