f(x)=(x-1)(x+2)(x-3)(x+4)…(x+100),则f'(1)的值为多少?)

Question

f(x)=(x-1)(x+2)(x-3)(x+4)…(x+100),则f'(1)的值为多少?)

Answer 1

方法一
利用导数定义
注意到f(1)=0
f'(1)=lim(x->1)[f(x)-f(1)]/(x-1)]
=lim(x->1)f(x)/(x-1)
=lim(x->1)[(x+2)(x-3)(x+4)…(x+98)(x-99)(x+100)]
=3*(-2)*5*(-4)*7*...*99*(-98)*101
=-99!*101
方法2
记f(x)=(x-1)g(x),其中g(x)=(x+2)(x-3)(x+4)…(x+100)
f'(x)=g(x)+(x-1)g'(x)
则f'(1)=g(1)=3*(-2)*5*(-4)*7*...*99*(-98)*101=-99!*101