Question

设f（x）=x（x-1）（x-2）……（x-100）求f'(0)及f（x）的101次幂

Answer 1

将f(x)拆成两项的乘积
即f(x)=X [(x-1)(x-2)……（x-100)]
那f'(x)=x' [(x-1)(x-2)……（x-100)]+x [(x-1)(x-2)……（x-100)]'
=(x-1)(x-2)……（x-100)+x [(x-1)(x-2)……（x-100)]'
后面这个导数不用求出来
所以f'(0)=(-1)(-2)……(-100)+0
=100*99*98*.....*1=100！

Answer 2

2边取对数有
lnf(x)=lnx+ln(x-1)+...+ln(x-100) 2边求导得
1/f(x)*f(x)'=[1/x+1/(x-1)+...+1/(x-100)] suoyi:
f(x)'=f(x)*[1/x+1/(x-1)+...+1/(x-100)] suoyi:
f(0)'=100!
f（x）的101次幂是设么意思，不明白要求社么。

Answer 3

f'(0)=100!