求y=(x-1)(x-2)(x-3)...（x-10)（x大于10）的导数.f(x)=(x-1)(x-2)……(x-10),ln[f(x)]=ln[(x-1)(x-2)……(x-10)]ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10){ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(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)我想问下[ln(x-1)+ln(x-2)+……+ln(x-10)]'为何是1/(x-1)+1/(x-2)+……+1/(x-10)不是要进行二次求导么?ln(x-2)'应该是1/(2-x)啊
问题描述：
求y=(x-1)(x-2)(x-3)...（x-10)（x大于10）的导数.
f(x)=(x-1)(x-2)……(x-10),
ln[f(x)]=ln[(x-1)(x-2)……(x-10)]
ln[f(x)]=ln(x-1)+ln(x-2)+……+ln(x-10)
{ln[f(x)]}'=[ln(x-1)+ln(x-2)+……+ln(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)
我想问下[ln(x-1)+ln(x-2)+……+ln(x-10)]'为何是1/(x-1)+1/(x-2)+……+1/(x-10)
不是要进行二次求导么?ln(x-2)'应该是1/(2-x)啊
答
[ln(x-2)]'=[1/(x-2)](x-2)'=1/(x-2)
你的那个负号是哪来的?你做错了.
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