Can someone help with this please. I dont know where to start and what rules I use to get through this question.
ln(1 + x/1 - x)
Thanks
0
Differentation
Started by AM4R, Dec 19 2006 09:00 PM
4 replies to this topic
#1
-
- Members
-
- 145 posts
Site Swot
- Gender:Male
- Gender:Male
Posted 19 December 2006 - 09:00 PM
::::::/\M/\R::::::
#2
-
- Members
-
- 390 posts
Top of the Class
- Location:Cambridge
- Interests:Muzak.
- Gender:Male
Posted 19 December 2006 - 09:36 PM
Let
, so you have 
. You can differentiate y with repsect to x using the chain rule now. Note that you will probably use the quotient rule to differentiate u with respect to x.
HMFC - Founded 1874, beefing the Cabbage since 1875
#3
-
- Members
-
- 1,955 posts
Fully Fledged Genius
- Gender:Male
Posted 20 December 2006 - 12:10 PM
You could algebraically simplify the:
(exclude the u, this is just cause I'm too lazy to rewrite the expression)
to 1/(1-x) which makes it a pretty straightforward integral, ln[1/(1-x)]
ps: ad_absurdum hope you don't mind me nicking your tex.
edit: sorry I'm pretty sure I read it was an integral question! Anyway you still could simplify and use the chain rule.
(exclude the u, this is just cause I'm too lazy to rewrite the expression)
to 1/(1-x) which makes it a pretty straightforward integral, ln[1/(1-x)]
ps: ad_absurdum hope you don't mind me nicking your tex.
edit: sorry I'm pretty sure I read it was an integral question! Anyway you still could simplify and use the chain rule.
#4
-
- Members
-
- 145 posts
Site Swot
- Gender:Male
- Gender:Male
Posted 20 December 2006 - 08:19 PM
whoops i didnt make that clear. Its differentiation and Its ln (1+X)/(1-X)
thanks
thanks
::::::/\M/\R::::::
#5
-
- Members
-
- 37 posts
Showing Improvement
- Gender:Male
Posted 20 December 2006 - 11:39 PM
just do:
ln((1+x)/(1-x)= ln(1+x)-ln(1-x)
that differentiates to:
1/(1+x)+1/(1-x)
ln((1+x)/(1-x)= ln(1+x)-ln(1-x)
that differentiates to:
1/(1+x)+1/(1-x)
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
