求y-x（e的y次方）=1的二阶导数

对y-x*e^y=1求导,得y'-e^y-xe^y*y'=0,∴(1-xe^y)y'=e^y,∴y'=e^y/(1-xe^y),∴y''=[e^y*y'*(1-xe^y)-e^y*(-e^y-xe^y*y')]/(1-xe^y)^2={y'[e^y-xe^(2y)+xe^(2y)]+e^(2y)}/(1-xe^y)^2=[e^(2y)/(1-xe^y)+e^(2y)]/(1-xe^y...