两边对x求导:y'=(1+y')cos(x+y)
得:y'=cos(x+y)/[1-cos(x+y)]
再对y'求导:y"=[-(1+y')sin(x+y)(1-cos(x+y))-cos(x+y)*(1+y')(sin(x+y)]/[1-cos(x+y)]^2
=-sin(x+y)(1+y')/[1-cos(x+y)]^2
再代入y',得:
y"=-sin(x+y)[1+cos(x+y)/(1-cos(x+y))]/[1-cos(x+y)]^2
=-sin(x+y)/[1-cos(x+y)]^3
追问谢了