第1个回答 2017-12-30
y=1+xe^y
y'= e^y + (xe^y)y'
(1-xe^y)y' = e^y
y' =e^y/(1-xe^y)
y''
=[e^y/(1-xe^y)]y' - [ e^y/(1-xe^y)^2] .[ -e^y - xe^y.y']
=[e^y/(1-xe^y)].[e^y/(1-xe^y)] - [ e^y/(1-xe^y)^2] .{ -e^y - xe^y.[e^y/(1-xe^y)] }
=e^(2y)/(1-xe^y)^2 - [ e^y/(1-xe^y)^2] .[ -e^y /(1-xe^y)]
=e^(2y)/(1-xe^y)^2 + e^(2y)/(1-xe^y)^3