calyne 3 years ago Differentiate the function y = ln(e^[-x] + xe^[-x])

1. Arnab09

apply chain rule

2. BlowjobBlowjobBlowjob

hello

3. Arnab09

who? me?

4. calyne

.....

5. calyne

help me

6. Arnab09

u know how to apply chain rule? @calyne?

7. calyne

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

8. Arnab09

wait a while..

9. calyne

what

10. calyne

are you joking

11. Arnab09

12. Arnab09

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

13. apoorvk

THAT WAS FAST!!! kudos @arnab09

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

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

16. Arnab09

yep..

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

18. Arnab09

maybe.. :/

19. Arnab09

thanx, BTW

20. Arnab09

u got it? @calyne??

22. DHASHNI

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)

23. DHASHNI

and for ur info ln e =1

24. Arnab09

idk how u got the second step, @dhashni

25. DHASHNI

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

26. Arnab09

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

27. DHASHNI

yeah i made a blunder mistake !!!!

28. DHASHNI

|dw:1334424165372:dw|

29. DHASHNI

|dw:1334424204454:dw|

30. Arnab09

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

31. DHASHNI

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