微积分-求导

求该函数的导数,谢谢大佬了
2025-06-28 07:53:38
推荐回答(4个)
回答1:

如下

回答2:

y=(cosx)^x
lny = xlncosx
(1/y)y' =lncosx - xtanx
y' =[lncosx - xtanx] .(cosx)^x

回答3:

y = (cosx)^x = e^[xlncosx]
y' = e^[xlncosx] (lncosx - xsinx/cosx) = (cosx)^x (lncosx - xsinx/cosx)

回答4:


😁