设cos（x+y）=y，求d²y/dx²

cos(x+y)=y
两边对 x求导-sin(x+y)(1+y')=y'
解出y'=-(sin(x+y))/[1+sin(x+y)]
再对x求导y''=[sin(x+y)-cos(x+y)](1+y')/[1+sin(x+y)]^2
把y'代入得
y''=[sin(x+y)-cos(x+y)]/[1+sin(x+y)]^3

cos（x+y）=y
∴cos（x+y）′=y′
得-sin(x+y) (x+y)′=y′
-sin(x+y) (1+y′)=y′
∴y′=-sin(x+y)/[1+sin(x+y)]
再次求导得y′′=[sin(x+y)-cos(x+y)]/[1+sin(x+y)]^3

-Cos[x + y]