设f(x)=x(x^2+1)(x^2+2)…(x^2+2009)则f'(0)=
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![设f(x)=x(x^2+1)(x^2+2)…(x^2+2009)则f'(0)=](/uploads/image/z/8650154-2-4.jpg?t=%E8%AE%BEf%28x%29%3Dx%28x%5E2%2B1%29%28x%5E2%2B2%29%E2%80%A6%28x%5E2%2B2009%29%E5%88%99f%27%280%29%3D)
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!