设f(x)=(x-1)(x-2)...(x-10),求f'(a)
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解决时间 2021-11-13 04:18
- 提问者网友：疯子也有疯子的情调
- 2021-11-13 00:03
设f(x)=(x-1)(x-2)...(x-10),求f'(a)
最佳答案
- 五星知识达人网友：酒醒三更
- 2021-11-13 01:10
f(x)=(x-1)(x-2)...(x-10)
lnf(x)=ln[(x-1)(x-2)...(x-10)]
lnf(x)=ln(x-1)+ln(x-2)+...+ln(x-10)
[lnf(x)]'=1/(x-1)+1/(x-2)+...+1/(x-10)
[lnf(x)]'=[1/f(x)]*f'(x)
[ln(x-1)+ln(x-2)+...+ln(x-10)]'=1/(x-1)+1/(x-2)+...+1/(x-10)
f'(x)/f(x)=1/(x-1)+1/(x-2)+...+1/(x-10)
f'(x)=[1/(x-1)+1/(x-2)+...+1/(x-10)]f(x)=[1/(x-1)+1/(x-2)+...+1/(x-10)]×[(x-1)(x-2)...(x-10)]
∴f'(a)=[1/(a-1)+1/(a-2)+...+1/(a-10)]×(a-1)(a-2)...(a-10)
lnf(x)=ln[(x-1)(x-2)...(x-10)]
lnf(x)=ln(x-1)+ln(x-2)+...+ln(x-10)
[lnf(x)]'=1/(x-1)+1/(x-2)+...+1/(x-10)
[lnf(x)]'=[1/f(x)]*f'(x)
[ln(x-1)+ln(x-2)+...+ln(x-10)]'=1/(x-1)+1/(x-2)+...+1/(x-10)
f'(x)/f(x)=1/(x-1)+1/(x-2)+...+1/(x-10)
f'(x)=[1/(x-1)+1/(x-2)+...+1/(x-10)]f(x)=[1/(x-1)+1/(x-2)+...+1/(x-10)]×[(x-1)(x-2)...(x-10)]
∴f'(a)=[1/(a-1)+1/(a-2)+...+1/(a-10)]×(a-1)(a-2)...(a-10)
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