答：[(-1)^(n+1)]/[n(n+1)]
方法一
记f(x)=(x-1)g(x),其中g(x)=[(x-2)...(x-n)]/[(x+1)(x+2)...(x+n)],
当n为奇数时g(1)=(n-1)!/(n+1)!=1/[n(n+1)],当n为偶数时g(1)=-(n-1)!/(n+1)!=-1/[n(n+1)],
求导f'(x)=g(x)+(x-1)g'(x)
则f'(1)=g(1).
当n为奇数时f'(1)=g(1)=(n-1)!/(n+1)!=1/[n(n+1)],
当n为偶数时f'(1)=g(1)=-(n-1)!/(n+1)!=-1/[n(n+1)].
则f'(1)=[(-1)^(n+1)]/[n(n+1)]
方法二
利用导数定义
注意到f(1)=0
f'(1)=(x->1)lim[f(x)-f(1)]/(x-1)=(x->1)f(x)/(x-1)
=(x->1)lim[(x-2)(x-3)...(x-n)]/[(x+1)(x+2)...(x+n)]
=[(-1)*(-2)*...*(1-n)]/[2*3*...*(n-1)*n*(n+1)],
当n为奇数时f'(1)=1/[n(n+1)],当n为偶数时f'(1)=-1/[n(n+1)].
则f'(1)=[(-1)^(n+1)]/[n(n+1)]