設f(x)=(2x^3-4x^2)^6，求f'(1)?
回答 (2)
2020-04-16 6:57 pm
✔ 最佳答案
只是基本練習題: 連鎖律的練習.設 u(x) = 2x^3-4x^2,
則 f(x) = [u(x)]^6
∴ f'(x) = 6[u(x)]^5u'(x)
= 6(2x^3-4x^2)^5(6x^2-8x)
f'(1) = 6(2-4)^5(6-8) = 6(-2)^5(-2) = 384
2020-04-16 7:55 pm
f '(x)=6[(2x^3-4x^2)^5](6x^2-8x)
代入1後就能得到答案囉~
代入1後就能得到答案囉~
收錄日期: 2021-05-04 02:33:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200416102133AAirM2a