Question

请问这道高数，这题求导怎么出来的？

Answer 1

运用了乘积的导
f=xg(x)[其中g=(x一1)…(x+100)
f'=g(x)+xg(x)'
所以f(o)'=g(o)=1x…100=1001
望采纳

Answer 2

f(x) =x(x-1)(x+2)(x-3)....(x+100)
lnf(x) =lnx +ln(x-1) +....+ln(x+100)
f'(x)/f(x) =1/x+1/(x-1)+1/(x+2)+....+1/(x+100)
f'(x) =[1/x+1/(x-1)+1/(x+2)+....+1/(x+100)] . [x(x-1)(x+2)(x-3)....(x+100)]
f'(0)
=(0-1)(0+2)(0-3)....(0+100)
=100!

Answer 3

令u=x，v=（x-1）（x+2）（x-3）...(x+100),由f’(x）=u’v+uv’得：f’(x)=v+uv’=（x-1）（x+2）（x-3）...(x+100)+uv’，显然，u（0）=0，f’(0）=v（0）=（-1）x2x（-3）x...x100=100！

Answer 4

根据乘积的求导法则：(uvw)'=u'vw+uv'w+uvw'
同理：(uvwz)'=u'vwz+uv'wz+uvw'z+uvwz'
…………
f(x)=x(x-1)(x+2)(x-3)(x+4)……(x+100)
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)]+……+[x(x-1)(x+2)(x-3)(x+4)……(x+100)']
=[(x-1)(x+2)(x-3)(x+4)……(x+100)]+[x(x+2)(x-3)(x+4)……(x+100)]+[x(x-1)(x-3)(x+4)……(x+100)]+……+[x(x-1)(x+2)(x-3)(x+4)……(x-99)']
f'(0)=(0-1)(0+2)(0-3)(0+4)……(0+100)
=(-1)×2×(-3)×4×……×（-99）×100
=1×2×3×4×……×100
=100！