y=(x+1/x^2)^√ 7的微分?
回答 (1)
2015-11-11 9:27 am
y = (x + 1/x^2)^√7
y' = √7*(x + 1/x^2)^(√7-1)*(x + 1/x^2)'
= √7*(x + 1/x^2)^(√7-1)*(1 - 2/x^3)
= √7*[(x^3 + 1)/x^2]^(√7-1)*(x^3 - 2)/x^3
= √7*(x^3 + 1)^(√7-1)*(x^3 - 2)/x^(2√7-2+3)
= √7*(x^3 - 2)*(x^3 + 1)^(√7-1)/x^(2√7+1)
y' = √7*(x + 1/x^2)^(√7-1)*(x + 1/x^2)'
= √7*(x + 1/x^2)^(√7-1)*(1 - 2/x^3)
= √7*[(x^3 + 1)/x^2]^(√7-1)*(x^3 - 2)/x^3
= √7*(x^3 + 1)^(√7-1)*(x^3 - 2)/x^(2√7-2+3)
= √7*(x^3 - 2)*(x^3 + 1)^(√7-1)/x^(2√7+1)
收錄日期: 2021-04-30 19:58:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151110235857AAUnvG7