f(x)=(x+1)(x+2)(x+3)......(x+100),求函数在0这一点的倒数值
是导数 不是倒数
人气:471 ℃ 时间:2020-02-03 18:27:28
解答
f(x)=(x+1)(x+2)(x+3)......(x+100),
lnf(x)=ln(x+1)+ln(x+2)+...+ln(x+100)
ln'f(x)=f'(x)/f(x)=1/(x+1)+1/(x+2)+.+1/(x+100)
f'(x)=f(x)*[1/(x+1)+1/(x+2)+...+1/(x+100)]
f'(0)=f(0)*[1/(0+1)+1/(0+2)+...+1/(0+100)]
f'(0)=100![1+1/2+1/3+...1/100]
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