y=tan(x+y)得y'=(1+y')/(cos(x+y))^2解得y'=-1/(sin(x+y))^2=-(sin(x+y))^(-2)y''=2(sin(x+y))^(-3)*cos(x+y)*(1+y') =2(sin(x+y))^(-3)*cos(x+y)*(1-1/(sin(x+y))^2) =-2(cos(x+y))^3(sin(x+y))^(-5)
