给这方程所确定的隐函数的二阶导数,

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
再问： л��