两边对x求导
y'*sinx+ycosx- [-sin(x+y)*(1+y')]=0
y'(sinx+sin(x+y))=y(1-cosx)
y'=[1-cosx]/[sinx+sin(x+y)]
0/0所以需要洛必达
先关于x
y'=[sinx]/[cosx+cos(x+y)(1+y')]
所以y'[cosx+cos(x+y)(1+y')]=sinx
令x->0
y'[1+cosy *(1+y')]=0
令y->pi
y'[1-(1+y')]=0
(y')^2=0,y'=0