Question

一道关于导数的选择题

若函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),且f'(x)是函数f(x)的导函数，则f'(1)=( )
A.24 B.-24 C.10 D.-10
请解释原因，谢谢啦！

Answer 1

f(1)=0
f'(1)=[f(1+deta)-f(1]/deta=f(1+deta)/deta
代入f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),得
f'(1)=deta*(-1+deta)*(-2+deta)*(-3+deta)*(-4+deta)/deta=(-1+deta)*(-2+deta)*(-3+deta)*(-4+deta)
deta约等于0
所以结果是24

Answer 2

f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)
f'(x)=(x-2)(x-3)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5)+(x-1)(x-2)(x-4)(x-5)+ (x-1)(x-2)(x-3)(x-5)+(x-1)(x-2)(x-3)(x-4)
f'(1)=(1-2)(1-3)(1-4)(1-5)+0+0+0+0
=24
所以选A

Answer 3

g(x)=(x-2)(x-3)(x-4)(x-5)
f(x)=(x-1)g(x)
f'(x)=g(x)+(x-1)g'(x)

f'(1)=g(1)=4!*(-1)^4=24

选A

Answer 4

A.因为如果带1进去就是0，所以剩f'(x）=（-1）*（-2）*（-3）*（-4）=24