Actually the question says to find the minimum value of the function. But for that i have to find the f'(x)=0.
Best answer up for grabs!
What's the derivative of 27^cosx + 81^sinx ?
2015-09-01 8:02 am
回答 (2)
2015-09-01 8:19 am
✔ 最佳答案
(d/dx) (27^cosx + 81^sinx)= [d(27^cosx)/dx] + [d(81^sinx)/dx]
= [d(27^cosx)/d(cosx)]•[d(cosx)/dx] + [d(81^sinx)/d(sinx)]•[d(sinx)/dx]
=(27^cosx) ln(27)•(-sinx) + (81^sinx) ln(81)•cosx
= cosx (81^sinx) ln(81) - sinx (27^cosx) ln(27)
2015-09-01 8:04 am
y = a^f(x) ---> y' = ln(a) f '(x) a^f(x)
收錄日期: 2021-04-30 12:26:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150901000233AAtIJbP