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

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Arnab09
 2 years ago
Best ResponseYou've already chosen the best response.3u know how to apply chain rule? @calyne?

calyne
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i know that pellet what's the derivative of e^[x]

Arnab09
 2 years ago
Best ResponseYou've already chosen the best response.3that will be dy/dx, in the picture^^

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.0THAT WAS FAST!!! kudos @arnab09

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0y'=x\(x+1),here is the steps :Dhttp://www4c.wolframalpha.com/Calculate/MSP/MSP15221a13hh36a69i0df100003gi12aabhd746hb7?MSPStoreType=image/gif&s=4&w=472&h=1028

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0@Arnab09 :have u just write that now and upload it now !??????

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0WOW,that was Extremely fast !!!!!!

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.1y = 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
 2 years ago
Best ResponseYou've already chosen the best response.1and for ur info ln e =1

Arnab09
 2 years ago
Best ResponseYou've already chosen the best response.3idk how u got the second step, @dhashni

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.1applying logarithm simplification ln(e^(x))=xlne=x

Arnab09
 2 years ago
Best ResponseYou've already chosen the best response.3yeah, but the second part.. u cant get the term after + sign out of log

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.1yeah i made a blunder mistake !!!!

Arnab09
 2 years ago
Best ResponseYou've already chosen the best response.3i.e: x+ln(1+x), nice..
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