设f(x)=x(x-1)(x-2)……(x-2009),则f'(0)=

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设f(x)=x*(x-1)*(x-2)……(x-2009),则f''(0)=设f(x)=x*(x-1)*(x-2)……(x-2009),则f''(0)=设f(x)=x*(x-1)*(x-2)……(x-20

设f(x)=x*(x-1)*(x-2)……(x-2009),则f'(0)=
设f(x)=x*(x-1)*(x-2)……(x-2009),则f'(0)=

设f(x)=x*(x-1)*(x-2)……(x-2009),则f'(0)=
f(0)=0
所以：
f'(0)=lim[f（x）-f（0）]/(x-0)
=lim[x*(x-1)*(x-2)……(x-2009)]/x
=lim(x-1)*(x-2)……(x-2009)
=(-1)(-2)...(-2009)
=-2009!