f(x)=(X-1)X(X-2).........X(X-101) 求f(x)的导数

两边同时取对数得到:
lny=ln(x-1)+ln(x-2)+...+ln(x-101)
两边同时求导得到
y'/y=1/(x-1)+1/(x-2)+...+1/(x-101)
所以y'=(x-1)(x-2)...(x-101)[1/(x-1)+..+1/(x-101)]

[lnf(x)]'=(ln[(X-1)X(X-2)…X(X-101))])'
={ln(x-1)+ln(x-2)+…+ln(x-101)}'
=1/(x-1)+1/(x-2)+…+1/(x-101)
即：
f'(x)/f(x)=1/(x-1)+1/(x-2)+…+1/(x-101)
所以f'(x)=f(x)*[1/(x-1)+1/(x-2)+…+1/(x-101)]
结果相信你会求了:)