Question

求由方程y-1=xe^y所确定的隐函数的二阶导数y".

Answer 1

y-1=xe^y

两边同时对x求导得

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

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

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

=e^y/(2-y)

y''=(e^y*y'+e^y*y')/(2-y)²

=(2e^y)e^y/(2-y)³

=2e^2y/(2-y)³