# More Differentiating w/ LN

Printable View

• Mar 17th 2007, 08:25 PM
zachb
More 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.
• Mar 17th 2007, 08:40 PM
Jhevon
Quote:

Originally Posted by zachb
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.

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)
• Mar 17th 2007, 08:44 PM
Soroban
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

• Mar 17th 2007, 08:56 PM
qbkr21
Re:
Guys did I solve this problem correctly?

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

;)
• Mar 17th 2007, 08:59 PM
Jhevon
Quote:

Originally Posted by Soroban
Hello, zachb!

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

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

Are you tired Soroban? -1 + 1/(1 + x) = -x/(1 + x)
• Mar 17th 2007, 09:00 PM
Jhevon
Quote:

Originally Posted by qbkr21
Guys did I solve this problem correctly?

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

;)

no, almost, you have to use the product rule to find the derivative of xe^-x, and you had to multiply the first e^-x in the last set of brackets by -1 not 1
• Mar 17th 2007, 09:03 PM
Jhevon
Quote:

Originally Posted by Jhevon
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)

don't worry zachb that Soroban's answer and my answer looks different, they are the same thing. if you factor out an e^-x out of the top and bottom of my fraction, you can cancel them and end up with -x/(1 + x), which looks like Soroban's answer
• Mar 17th 2007, 09:04 PM
qbkr21
Re:
I am confused Jhevon. I did use the product rule for the last step...I am trying to understand what I am missing?
• Mar 17th 2007, 09:05 PM
zachb
Yeah, Jhevon got it right first, but didn't simplify completely.

All of you, thanks for helping. :)
• Mar 17th 2007, 09:09 PM
Jhevon
Quote:

Originally Posted by qbkr21
I am confused Jhevon. I did use the product rule for the last step...I am trying to understand what I am missing?

oh, my bad, false alarm, you are correct qbkr21. sorry about that! still friends right? :)

i seem to have selective vision today. it's like the third time i said someone say something that they didnt say, and when i reread, i realize i was wrong
• Mar 17th 2007, 09:11 PM
qbkr21
Re:
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)
• Mar 17th 2007, 09:19 PM
Jhevon
Quote:

Originally Posted by qbkr21
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)

that's alright man. as long as i know its good between us, no smilies necessary. i guess it's best to play it safe for the time being