Question

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

Answer 1

apply chain rule

Answer 2

hello

Answer 3

who? me?

Answer 4

.....

Answer 5

help me

Answer 6

u know how to apply chain rule? @calyne?

Answer 7

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

Answer 8

wait a while..

Answer 9

what

Answer 10

are you joking

Answer 11

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

Answer 12

THAT WAS FAST!!! kudos @arnab09

Answer 13

y'=-x\(x+1),here is the steps :D http://www4c.wolframalpha.com/Calculate/MSP/MSP15221a13hh36a69i0df100003gi12aabhd746hb7?MSPStoreType=image/gif&s=4&w=472&h=1028

Answer 14

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

Answer 15

yep..

Answer 16

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

Answer 17

maybe.. :/

Answer 18

thanx, BTW

Answer 19

u got it? @calyne??

Answer 20

your welcome :)

Answer 21

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)

Answer 22

and for ur info ln e =1

Answer 23

idk how u got the second step, @dhashni

Answer 24

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

Answer 25

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

Answer 26

yeah i made a blunder mistake !!!!