I need to differentiate y = ln(e^-x + xe^-x)

so,

y' = lne^-x + lnex^-x

y' = lne^-x(1 + x)

If this is correct so far, what am I supposed to do next? And if I'm completely wrong, what's the correct way to do this?

Thanks.

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- March 17th 2007, 09:25 PMzachbMore Differentiating w/ LN
I need to differentiate y = ln(e^-x + xe^-x)

so,

y' = lne^-x + lnex^-x

y' = lne^-x(1 + x)

If this is correct so far, what am I supposed to do next? And if I'm completely wrong, what's the correct way to do this?

Thanks. - March 17th 2007, 09:40 PMJhevon
you're completely wrong (sorry for being so blunt, but you asked). first you didn't even differentiate, second, you didn't split up the log correctly. the correct way to do this is by the chain rule.

the chain rule says, if we have a composite function f(g(x)), its derivative is given by f ' (g(x))*g'(x). in this case we have a composite function, we can think of ln(x) as f(x) and e^-x + xe^-x as g(x)

y = ln(e^-x + xe^-x)

=> y' = 1/(e^-x + xe^-x) * (-e^-x + e^-x -xe^-x) ..........we have to use the product rule to differentiate the xe^-x

=> y' = (-e^-x + e^-x -xe^-x)/(e^-x + xe^-x)

=> y' = (-xe^-x)/(e^-x + xe^-x) - March 17th 2007, 09:44 PMSoroban
Hello, zachb!

You can't "split" logs like that . . . and you missed a product.

Quote:

Differentiate: .y .= .ln(e^{-x} + x·e^{-x})

But you can simplify first: . y .= .ln[e^{-x}(1 + x)] .= .ln(e^{-x)) + ln(1 + x)

. . and we have: .y .= .-x + ln(1 + x)

. . . . . . . . . . . . . . . . . . . .1 . . . . . -1

Differentiate: . y' .= .-1 + ------ .= .-------

. . . . . . . . . . . . . . - - . . 1 + x . - .1 + x

- March 17th 2007, 09:56 PMqbkr21Re:
Guys did I solve this problem correctly?

http://item.slide.com/r/1/88/i/KvWKl...WK5csj2LaOOPs/

;) - March 17th 2007, 09:59 PMJhevon
- March 17th 2007, 10:00 PMJhevon
- March 17th 2007, 10:03 PMJhevon
- March 17th 2007, 10:04 PMqbkr21Re:
I am confused Jhevon. I did use the product rule for the last step...I am trying to understand what I am missing?

- March 17th 2007, 10:05 PMzachb
Yeah, Jhevon got it right first, but didn't simplify completely.

All of you, thanks for helping. :) - March 17th 2007, 10:09 PMJhevon
- March 17th 2007, 10:11 PMqbkr21Re:
Jhevon remember what I told you: YOU ARE THE MAN!!!!!!!!!!!!!!!!!!!!!!!

(P.S. I am not using smilies because I do no wish to receive another 5 pt. infraction) - March 17th 2007, 10:19 PMJhevon