Believe.

引用:

原帖由 W_a 於 2011-11-6 13:57 發表


應該係ln ... 我m1果邊教完


咩料... ...Who are you?



另外回樓主 , m2數黎? 我未教呢d野 , 想計都計唔到

...都係ln泥     系m2數;

W_a

引用:

原帖由 Believe. 於 2011-11-6 14:12 發表

...都係ln泥     系m2數;

咁應該係我ln果邊未教到咁深或者m1唔會教到咁深

(其實讀得m1就係唔想見到咁多代數~~~同埋配合番Econ)

花文.

我m1

Believe.

待破解    

A_c

我都要等你教我火西。

Imisstime

2.{ [(e^x)-(e^-x)]/[(e^x)+(e^-x)]} x (e^x)
give (e^2x-1)/(e^2x+1)
Since (e^2x+1) is 份母

Try d/dx {ln(e^2x+1)}
and u will get  (2e^2x)/(e^2x+1)

Different between 2e^2x/(e^2x+1) and [(e^x)-(e^-x)]/[(e^x)+(e^-x)]
is given as  [(e^x)-(e^-x)]/[(e^x)+(e^-x)] - {2e^2x/(e^2x+1)}
which is -(e^2x+1) / (e^2x+1) = -1
Therefore∫ [(e^x)-(e^-x)]/[(e^x)+(e^-x)] dx =  ln(e^2x+1) +∫ -1 dx

Believe.

引用:

原帖由 Imisstime 於 2011-11-6 22:51 發表
2.{ [(e^x)-(e^-x)]/[(e^x)+(e^-x)]} x (e^x)
give (e^2x-1)/(e^2x+1)
Since (e^2x+1) is 份母

Try d/dx {ln(e^2x+1)}
and u will get  (2e^2x)/(e^2x+1)

Different between 2e^2x/(e^2x+1) and [(e^x)-(e^-x)]/[( ...

thx  好在我理解到. . .
不過點解會咁念 btw 都係+1-1

A_c

樓上上高手，跪拜，，

Reeve.

different between

Imisstime

This is the way we are taught by the lesson...But they let u = something , and than du/dx
and work out the different between the du/dx and the question

sIuˇHanG

引用:

原帖由 A_c 於 2011-11-6 23:29 發表
樓上上高手，跪拜，，

同意 我完全Load唔到咩泥
我M1

R-果然翁

引用:

原帖由 Imisstime 於 2011-11-6 22:51 發表
2.{ [(e^x)-(e^-x)]/[(e^x)+(e^-x)]} x (e^x)
give (e^2x-1)/(e^2x+1)
Since (e^2x+1) is 份母

Try d/dx {ln(e^2x+1)}
and u will get  (2e^2x)/(e^2x+1)

Different between 2e^2x/(e^2x+1) and [(e^x)-(e^-x)]/[( ...

Professor特登開ac教人數

-iCeB.冰

完全唔知咩黎

bbwarrior

引用:

原帖由 Believe. 於 5-11-2011 10:00 PM 發表
1.∫(sec x)^5(csc x)^3dx

2.∫[(e^x)-(e^-x)]/[(e^x)+(e^-x)]dx


求高手 破解.

我擅長數學,不過ed似係附加數學