Question

求y-x（e的y次方）=1的二阶导数

Answer 1

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)^2

=[2e^(2y)-xe^(3y)]/(1-xe^y)^3.