f(x)=x(x+1)(x+2)...(x+2009),求f'(0)
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f(x)=x(x+1)(x+2)...(x+2009),求f'(0)
f(x)=x(x+1)(x+2)...(x+2009),求f'(0)
f(x)=x(x+1)(x+2)...(x+2009),求f'(0)
应该是2009的阶乘,用求导的定义来算f‘(0)=lim[f(X)-f(0)]/f(x)令x趋于零
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