Question

已知{an}为公差是1的等差数列,函数f(x)=(x-a1)(x-a2).(x-an),则f'(a4)=__

Answer 1

f(x)=(x-a1)(x-a2).(x-an)

f(x)=(x-a4)*[(x-a1)(x-a2)(x-a3)(x-a5).(x-an)]

所以

f'(x)=(x-a1)(x-a2)(x-a3)(x-a5).(x-an)+(x-a4)*[(x-a1)(x-a2)(x-a3)(x-a5).(x-an)]'

所以

f'(a4)=(a4-a1)(a4-a2)(a4-a3)(a4-a5).(a4-an)+0

=3×2×1×（-1）（-2）×.×（-n+4）

=(-1)^(n-4)*6* (n-4)!