f'(x)=[x'(x-1)(x+2)(x-3)(x+4).(x+100)]+[x(x-1)'(x+2)(x-3)(x+4).(x+100)]+...+[x(x-1)(x+2)(x-3)(x+4).(x+100)']除开第2项外,其他各项均含有因式(x-1),当x=1时,这些项均为0故f'(1)=x(x-1)'(x+2)(x-3)(x+4).(x+100...