高数对数求导
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发布时间:2024-10-13 07:28
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热心网友
时间:2024-10-28 00:18
用反证法
设{an+bn}收敛
根据收敛的定义,an数列和an+bn数列都有极限
所以可以设lim(n→∞)an=c
lim(n→∞)(an+bn)=d
那么根据极限是四则运算,有
lim(n→∞)bn=lim(n→∞)[(an+bn)-an]
=lim(n→∞)(an+bn)-lim(n→∞)an
=d-c
所以bn也有极限,bn也收敛
这和题目规定bn发散矛盾
所以an+bn也发散。追问are you foolish?nonsense bullshit in your mouth make you have to jet them out sb
热心网友
时间:2024-10-28 00:19
解:∵y={(x-3)^5*(x+1)^(1/3)}/(x-2)
==>lny=5ln(x-3)+ln(x+1)/3-ln(x-2) (等式两端取对数)
==>y'/y=5/(x-3)+(1/3)/(x+1)-1/(x-2) (等式两端对x求导)
==>y'=[5/(x-3)+(1/3)/(x+1)-1/(x-2)]y (等式两端同乘y)
∴y'=[5/(x-3)+(1/3)/(x+1)-1/(x-2)]y
=[5/(x-3)+(1/3)/(x+1)-1/(x-2)]*{(x-3)^5*(x+1)^(1/3)}/(x-2)。