注意那里是-1的96次方。
f‘(2)=(x-1)(x-3)...(x-100)/[(x+1)...(x+100)]^2
=4*98!/(102!)^2
因为有x-2的都是0
f‘(x)=[(x-2)(x-3)…(x-100)+(x-1)(x-3)…(x-100)+…(x-1)(x-2)…(x-100)]*(x+1)(x+2)…(x+100)+(x-1)(x-2)…(x-100)*[(x+2)(x+3)…(x+100)+(x+1)(x+3)…(x+100)+…(x+1)(x+2)…(x+100)]/[(x+1)(x+2)…(x+100)]^2
当x=2时,所有的x-2=0,
f‘(2)=(x-1)(x-3)...(x-100)/[(x+1)...(x+100)]^2=2*98!/(102!)^2