设函数y=f(x+y),其中f具有二阶导数,且f'不等于1,求二?
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.