设f(x)=(x-1)(x-2)...(x-10),求f'(10)的导数不好意思啊!是求f'(10)

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设f(x)=(x-1)(x-2)...(x-10),求f''(10)的导数不好意思啊!是求f''(10)设f(x)=(x-1)(x-2)...(x-10),求f''(10)的导数不好意思啊!是求f''(10)

设f(x)=(x-1)(x-2)...(x-10),求f'(10)的导数不好意思啊!是求f'(10)
设f(x)=(x-1)(x-2)...(x-10),求f'(10)的导数
不好意思啊!是求f'(10)

设f(x)=(x-1)(x-2)...(x-10),求f'(10)的导数不好意思啊!是求f'(10)
求f(10)的导数?楼主搞错了吧?是求f'(10)吧?
如果是求f(10)的导数,那就简单了,答案是0.原因是:f(10)是一个常数,而任何常数的导数都是0.
假设是求f'(10).
解法一:
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)]}'=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)
把x=10代入,就求出f'(10)了,会了吗?
解法二:
设y=(x-1)(x-2)……(x-9)
则:f(x)=y(x-10),
f'(x)=(x-10)y'+y(x-10)'=(10-x)y'+y
则:f'(x)|10=(10-10)y'+y=9!
即:f'(10)=9!
补充答案:
上面已经求出了f'(10)了呀!楼主没看到?还是没看出来?

0

观察它的导数结构,f'(10)只有(x-1)(x-2)...(x-9)这项不会等于0,带进去就是9!