小卒資訊論壇 - Powered by Discuz! Board

標題: M1 [打印本頁]

作者: lovecake 時間: 2011-5-5 07:11 PM 標題: M1

[attach]33690[/attach]

作者: Simon 時間: 2011-5-5 07:42 PM

Just use chain rule to do it.

作者: lovecake 時間: 2011-5-5 08:44 PM

次方有個X我唔識做

作者: Simon 時間: 2011-5-5 08:56 PM

h(x) = ln(x^(2x) + 1) = lnu

dh　 dh　　du
----- = ------ ．------
dx　　du　　dx

你可以做左y = x^x 呢個先

作者: 夏娜醬大好 時間: 2011-5-5 09:11 PM

唔識做 x^(2x) ？
d/dx x^(2x)
= d/dx e^(2x ln x)
= (2+2 ln x)e^(2x ln x)
= (2+2 ln x)x^(2x)

作者: 桃子 時間: 2011-5-5 09:59 PM

u = x^(2x) + 1
u-1 = x^(2x)
ln(u-1) = 2x ln x
Diff. both sides w.r.t. x,
1/(u-1) *(du/dx)= 2(1+ln x)

du/dx = 2(u-1)(1+ln x)

h'(x)
=dh/du * du/dx
=(1/u) * 2(u-1)(1+ln x)
=2(u-1)(1+ln x) / u

When x=2, u=2^4 + 1=17

h'(2) = 2*16 (1+ln 2) / 17
=(32/17) (1+ln 2)

I hope that I didn't make careless mistakes.

[ 本帖最後由桃子於 2011-5-5 10:02 PM 編輯 ]

歡迎光臨小卒資訊論壇 (http://lsforum.net/board/)