设f(x)=x(x^2+1)(x^2+2)…(x^2+2009)则f'(0)=

设f(x)=x(x^2+1)(x^2+2)…(x^2+2009)则f'(0)=
设f(x)=x(x^2+1)(x^2+2)…(x^2+2009)则f'(0)=

设f(x)=x(x^2+1)(x^2+2)…(x^2+2009)则f'(0)=
设u（x）=(x^2+1)(x^2+2)…(x^2+2009)
所以f(x)=xu(x)
f'(x)=x'u(x)+xu'(x)=u(x)+xu'(x)
x=0时,xu'(x)=0*u‘（0）=0
所以f'(0)=u(0)=1*2*3*……*2009=2009!