设函数y=f(x+y),其中f具有二阶导数,且f'不等于1,求二? ,简单的描述是,答案是设函数y=f(x+y) ,其中f具有二阶导数,且f'不等于1,求二阶导数

y=f(x+y)
则：
y'=f'(x+y)*(1+y')=f'(x+y)+f'(x+y)*y'
===> [1-f'(x+y)]*y'=f'(x+y)
===> y'=f'(x+y)/[1-f'(x+y)]
所以：
y''={f''(x+y)*(1+y')*[1-f'(x+y)]-f'(x+y)*[-f''(x+y)*(1+y')]}/[1-f'(x+y)]^2
=f''(x+y)*(1+y')/[1-f'(x+y)]^2
=f''(x+y)*[1/1-f'(x+y)]/[1-f'(x+y)]^2
=f''(x+y)/[1-f'(x+y)]^3.