Question

已知f(x)=(x-1)(x-2)(x-3)(x-4)....(x-2011),求f'(2011)=?

Answer 1

f(x)=(x-1)(x-2)(x-3)(x-4)....(x-2011)
=[(x-1)(x-2)(x-3)(x-4)....(x-2010)](x-2011)
所以
f'(x)=[(x-1)(x-2)(x-3)(x-4)....(x-2010)]'(x-2011)+[(x-1)(x-2)(x-3)(x-4)....(x-2010)](x-2011)'
=[(x-1)(x-2)(x-3)(x-4)....(x-2010)]'(x-2011)+[(x-1)(x-2)(x-3)(x-4)....(x-2010)]
f'(2011)=[(x-1)(x-2)(x-3)(x-4)....(x-2010)]' (2011-2011)+[2010*2009*...*1]
=2010!

追问

额、我都被绕糊涂了、

追答

哪一部有问题？

Answer 2

f'(x)=(x-2)(x-3)(x-4)...(x-2011) + (x-1)(x-3)(x-4)...(x-2011) +... + (x-1)(x-2)(x-3)(x-4)...(x-2010)
f'(2011)=(2010×2009×...×1)=2010!

追问

答案为 2010

追答

f(x)=g(x)h(x)p(x)q(x)
f'(x)=g'(x)h(x)p(x)q(x)+g(x)h'(x)p(x)q(x)+g(x)h(x)p'(x)q(x)+g(x)h(x)p(x)q'(x)
答案为 2010的阶乘，即：2010！

Answer 3

方法1
let f(x)=(x-2011)g(x),g(x)=(x-1)(x-2)...(x-2010),g(2011)=2010!
then
f'(x)=g(x)+(x-2011)g'(x)
so f'(2011)=g(2011)=2010!

方法2
按导数定义
f'(2011)=lim(x->2011)[f(x)-f(2011)]/(x-2011)=lim(x->2011)【(x-1)(x-2)...(x-2010)】=2010!
其中f(2011)=0

Answer 4

f(2011)=(x-1)(x-2)(x-3)(x-4)....(x-2011)=常数
f'(2011)=0