Question

y=1+xe^y,求二阶导,

Answer 1

两边求导

y'=e^y+y'xe^y

y'(1-xe^y)=e^y

y''(1-xe^y)-y'(e^y+y'xe^y)=y'e^y

y''(1-xe^y)-y'y'=y'e^y

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

y''=e^2y/(1-xe^y)^3+e^2y/(1-xe^y)^2