Question

高数题求解答过程

f(x)=x(x-1)(x+2)(x-3)(x+4)....(x+100)
求f'(1)
要过程，答案我有是-101！/100

Answer 1

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)|x=1
=1*3*(-2)*5*(-4)*...*(-98)*101
=-101!/100

Answer 2

由求导法则知f'(x)中除了第一项x(x+2)(x+3)...(x+100)外，其余加的项均含有(x-1)项，故当x=1时其余加的项的值均为0,所以就有
f'(1)=1*3*4*...*101=101!/2.

Answer 3

f(x)=(x-1)g(x)
g(x)=x(x+2)(x-3)(x+4)...(x+100)

f'(x)=g(x)+(x-1)g'(x)
f'(1)=g(1)+0*g'(1)=g(1)

g(1)=1*3*(-2)*5*(-4)*7*(-6)*9*...*(-98)*101
=-101!/100即为所求.