calyne
Differentiate the function y = ln(e^[x] + xe^[x])



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Arnab09
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apply chain rule


Arnab09
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who? me?

calyne
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.....

calyne
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help me

Arnab09
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u know how to apply chain rule? @calyne?

calyne
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yeah i know that pellet what's the derivative of e^[x]

Arnab09
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wait a while..

calyne
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what

calyne
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are you joking

Arnab09
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Arnab09
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that will be dy/dx, in the picture^^

apoorvk
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THAT WAS FAST!!! kudos @arnab09


Eyad
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@Arnab09 :have u just write that now and upload it now !??????

Arnab09
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yep..

Eyad
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WOW,that was Extremely fast !!!!!!

Arnab09
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maybe.. :/

Arnab09
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thanx, BTW

Arnab09
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u got it? @calyne??

Eyad
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your welcome :)

DHASHNI
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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)

DHASHNI
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and for ur info ln e =1

Arnab09
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idk how u got the second step, @dhashni

DHASHNI
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applying logarithm simplification
ln(e^(x))=xlne=x

Arnab09
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yeah, but the second part.. u cant get the term after + sign out of log

DHASHNI
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yeah i made a blunder mistake !!!!

DHASHNI
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dw:1334424165372:dw

DHASHNI
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dw:1334424204454:dw

Arnab09
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i.e: x+ln(1+x), nice..

DHASHNI
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dw:1334424237437:dw