若f(x)=(x-1)(x-2)…(x-9)(x-10)则f'(10)=
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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.
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!
再问: 第一种解法中,若把10带入的话,f(x)不是等于0吗,那结果就得0吗,与第二种解法答案不一样啊
再答: 你不会带入计算,找个人帮忙看看计算吧。你肯定算错了。 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) =(x-2)……(x-10)+(x-1)(x-3)……(x-10)+(x-1)(x-2)(x-4)……(x-10)+(x-1)(x-2)……(x-9) 算到这里再带入会吧?
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!
再问: 第一种解法中,若把10带入的话,f(x)不是等于0吗,那结果就得0吗,与第二种解法答案不一样啊
再答: 你不会带入计算,找个人帮忙看看计算吧。你肯定算错了。 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) =(x-2)……(x-10)+(x-1)(x-3)……(x-10)+(x-1)(x-2)(x-4)……(x-10)+(x-1)(x-2)……(x-9) 算到这里再带入会吧?
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