Question

设f（X)=(X-1)(X-2)(X-3)...(X-100),求f′（1）

Answer 1

设g(x)=(X-2)(X-3)...(X-100)，则f(x)=(x-1)*g(x),所以f′(x)=g(x)+(f(x)-1)*g′(x)，所以f'(1)=(1-2)(1-3)(1-4)……(1-100)+（1-1）*g′(x)=-1*（-2）*（-3）*……（-99）

Answer 2

令M=f（x），f(x)=(x-1>M/(x-1)，
则f'(x)=(x-1)'M/(x-1) (x-1)[M/(x-1)]‘=(x-2)(x-3)…(x-100)十(x-1)[(x-2)(x-3)…(x-100)]‘=(x-2)(x-3)…(x-100) (x-1){(x-3)(x-4)…(x-100) (x-2)[(x-3)…(x-100)]'}=M/(x-1) M/(x-2) M/(x-3) …M/(x-100)
∴f'(1)=-1×-2×…-99=-99!

Answer 3

[1/（x-1）+1/(x-2)+...1/（x-100）]*f(x)

Answer 4

太难了 做不出啊