y=x^2-x√（2x+1）微分多少，算式給一下，我一直計算不出來謝謝?
回答 (1)
2020-10-23 2:02 am
y = x^2 - x√(2x+1)
y' = 2x - [ √(2x+1) + x．2/(2√(2x+1)) ] (*)
= 2x - [(2x+1) + x]/√(2x+1)
= 2x - (3x+1)/√(2x+1)
(*)
D(x√(2x+1)) = (Dx)√(2x+1) + x(D√(2x+1))
= √(2x+1) + x．1/(2√(2x+1)) D(2x+1)
= √(2x+1) + x．1/(2√(2x+1))．2
= √(2x+1) + x/√(2x+1)
y' = 2x - [ √(2x+1) + x．2/(2√(2x+1)) ] (*)
= 2x - [(2x+1) + x]/√(2x+1)
= 2x - (3x+1)/√(2x+1)
(*)
D(x√(2x+1)) = (Dx)√(2x+1) + x(D√(2x+1))
= √(2x+1) + x．1/(2√(2x+1)) D(2x+1)
= √(2x+1) + x．1/(2√(2x+1))．2
= √(2x+1) + x/√(2x+1)
收錄日期: 2021-05-04 02:30:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201022102720AAemrCd