anonymous
  • anonymous
Differentiate the function y = ln(e^[-x] + xe^[-x])
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
apply chain rule
anonymous
  • anonymous
hello
anonymous
  • anonymous
who? me?

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anonymous
  • anonymous
.....
anonymous
  • anonymous
help me
anonymous
  • anonymous
u know how to apply chain rule? @calyne?
anonymous
  • anonymous
yeah i know that pellet what's the derivative of e^[-x]
anonymous
  • anonymous
wait a while..
anonymous
  • anonymous
what
anonymous
  • anonymous
are you joking
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
that will be dy/dx, in the picture^^
apoorvk
  • apoorvk
THAT WAS FAST!!! kudos @arnab09
anonymous
  • anonymous
y'=-x\(x+1),here is the steps :Dhttp://www4c.wolframalpha.com/Calculate/MSP/MSP15221a13hh36a69i0df100003gi12aabhd746hb7?MSPStoreType=image/gif&s=4&w=472&h=1028
anonymous
  • anonymous
@Arnab09 :have u just write that now and upload it now !??????
anonymous
  • anonymous
yep..
anonymous
  • anonymous
WOW,that was Extremely fast !!!!!!
anonymous
  • anonymous
maybe.. :/
anonymous
  • anonymous
thanx, BTW
anonymous
  • anonymous
u got it? @calyne??
anonymous
  • anonymous
your welcome :)
anonymous
  • anonymous
y = ln(e^[-x] + xe^[-x]) y=-x( ln (e))+xe^(-x)=-x+xe^(-x) =x(e(-x)-1) dy/dx = x(-e(-x))+(e(-x)-1)
anonymous
  • anonymous
and for ur info ln e =1
anonymous
  • anonymous
idk how u got the second step, @dhashni
anonymous
  • anonymous
applying logarithm simplification ln(e^(-x))=-xlne=-x
anonymous
  • anonymous
yeah, but the second part.. u cant get the term after + sign out of log
anonymous
  • anonymous
yeah i made a blunder mistake !!!!
anonymous
  • anonymous
|dw:1334424165372:dw|
anonymous
  • anonymous
|dw:1334424204454:dw|
anonymous
  • anonymous
i.e: -x+ln(1+x), nice..
anonymous
  • anonymous
|dw:1334424237437:dw|

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