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【1】f[x]=x[x+1][x+2].[x+100]
[2]f[x]=a0 x^n+a1 x^[n-1]+.a[n-1]x+ an
人气:454 ℃ 时间:2020-01-31 03:59:09
解答
(1)f(x)=x(x+1)(x+2)(x+3)......(x+100)
lnf(x)=lnx+ln(x+1)+ln(x+2)+...+ln(x+100)
[lnf(x)]'=f'(x)/f(x)=1/(x)+1/(x+1)+1/(x+2)+.+1/(x+100)
所以,
f'(x)=f(x)×[1/(x)+1/(x+1)+1/(x+2)+...+1/(x+100)]
(2)f(x)=a0 x^n+a1 x^[n-1]+.a[n-1]x+ an
f'(x)=n×a0×x^(n-1)+(n-1)×a1×x^(n-2)+……+a(n-1)
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