Question

y=1+xe∧y.求这个隐函数y的二阶导数.

Answer 1

y=1+xe^y方程两边求导
y'=e^y+xe^y*y'
y'(1-xe^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
=[e^(2y)*(2+x-xe^y)]/[(1-xe^y)^3]