Question

帮忙求这个函数的导数

要详细过程！！！！！！！！！

Answer 1

F(x)=be^(x-1)/x= (b/e)e^x/x
F'(x) = (b/e) (xe^x-e^x)/x^2 = (b/e)(x-1)e^x/x^2.

Answer 2

解由分式函数求导法则
知f'(x)=[be^(x-1)/x]'
={[be^(x-1)]'x-[be^(x-1)]x'}/x^2
={[be^(x-1)(x-1)']x-[be^(x-1)]}/x^2
={[be^(x-1)]x-[be^(x-1)]}/x^2
=be^(x-1)（x-1）/x^2