y=3x^2*[(2x^2+5)]^1/3
du/dx=?
dy/du=?
dy/dx=?
請幫忙列出步驟
救命,十萬火急,微分題!!!!
2006-10-21 6:46 am
回答 (2)
2006-10-21 7:06 am
✔ 最佳答案
y=3x^2*[(2x^2+5)]^1/33x^2乘以[(2x^2+5)]^1/3
用product rule
留意[(2x^2+5)]^1/3為compostive function
dy/dx=3x^2(1/3(2x^2+5)^-2/3×4x)+(2x^2+5)^1/3×6x
=〔9x^3/((2x^2+5)^2/3)〕+6x(2x^2+5)^1/3
=3x(7x^2+10)/(2x^2+5)^2/3
2006-10-26 2:41 am
y=3x^2*[(2x^2+5)]^1/3
Let u=2x^2+5,w=u^(1/3)
Then w=(2x^2+5)^(1/3)
dw/dx
=(dw/du)(du/dx)
=[u^(-2/3)/3](4x)
=[(2x^2+5)^(-2/3)](4x)
=4x/(2x^2+5)^(2/3)
Use produce rule d(uv)/dx=u dv/dx+v du/dx
dy/dx
=(3x^2)[4x/(2x^2+5)^(2/3)]+(2x^2+5)^(1/3)(6x)
=12x^3/[4x/(2x^2+5)^(2/3)]+(6x)(2x^2+5)^(1/3)
=6x{2x^2/[4x/(2x^2+5)^(2/3)]+(2x^2+5)^(1/3)}
Let u=2x^2+5,w=u^(1/3)
Then w=(2x^2+5)^(1/3)
dw/dx
=(dw/du)(du/dx)
=[u^(-2/3)/3](4x)
=[(2x^2+5)^(-2/3)](4x)
=4x/(2x^2+5)^(2/3)
Use produce rule d(uv)/dx=u dv/dx+v du/dx
dy/dx
=(3x^2)[4x/(2x^2+5)^(2/3)]+(2x^2+5)^(1/3)(6x)
=12x^3/[4x/(2x^2+5)^(2/3)]+(6x)(2x^2+5)^(1/3)
=6x{2x^2/[4x/(2x^2+5)^(2/3)]+(2x^2+5)^(1/3)}
收錄日期: 2021-04-12 22:14:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061020000051KK04760