若f(x)=(x-1)(x-2)…(x-9)(x-10)则f'(10)=对于此题,有9!和0两种答案.能否给我一个清楚的过程,让我明白到底哪种对.
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![若f(x)=(x-1)(x-2)…(x-9)(x-10)则f'(10)=对于此题,有9!和0两种答案.能否给我一个清楚的过程,让我明白到底哪种对.](/uploads/image/z/14787981-45-1.jpg?t=%E8%8B%A5f%28x%29%3D%28x-1%29%28x-2%29%E2%80%A6%28x-9%29%28x-10%29%E5%88%99f%27%2810%29%3D%E5%AF%B9%E4%BA%8E%E6%AD%A4%E9%A2%98%2C%E6%9C%899%21%E5%92%8C0%E4%B8%A4%E7%A7%8D%E7%AD%94%E6%A1%88.%E8%83%BD%E5%90%A6%E7%BB%99%E6%88%91%E4%B8%80%E4%B8%AA%E6%B8%85%E6%A5%9A%E7%9A%84%E8%BF%87%E7%A8%8B%2C%E8%AE%A9%E6%88%91%E6%98%8E%E7%99%BD%E5%88%B0%E5%BA%95%E5%93%AA%E7%A7%8D%E5%AF%B9.)
若f(x)=(x-1)(x-2)…(x-9)(x-10)则f'(10)=对于此题,有9!和0两种答案.能否给我一个清楚的过程,让我明白到底哪种对.
若f(x)=(x-1)(x-2)…(x-9)(x-10)则f'(10)=
对于此题,有9!和0两种答案.能否给我一个清楚的过程,让我明白到底哪种对.
若f(x)=(x-1)(x-2)…(x-9)(x-10)则f'(10)=对于此题,有9!和0两种答案.能否给我一个清楚的过程,让我明白到底哪种对.
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.
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!
f(x)=(x-1)(x-2)…(x-9)(x-10)
设g(x)=(x-1)(x-2)………(x-9)
f(x)=g(x)(x-10)
f'(x)=g'(x)(x-10)+g(x)
f'(10)=0+g(10)
所以
f'(10)=(10-1)(10-2)…………(10-9)
=9!
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