大一微积分求导,过程最好详细点啦,谢谢哟,很好采纳的!

2025-05-19 23:17:18
推荐回答(1个)
回答1:

y'=1/[ln²(ln³x)] ×[ln²(ln³x)]'
=1/[ln²(ln³x)] ×[2ln(ln³x)]×1/ln³x×(ln³x)'
=1/[ln²(ln³x)] ×[2ln(ln³x)]×1/ln³x×3ln²x·1/x