Thread: 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.

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)

Hello, zachb!

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

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

Guys did I solve this problem correctly?

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)

Originally Posted by qbkr21

Guys did I solve this problem correctly?

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

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

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

Yeah, Jhevon got it right first, but didn't simplify completely.


All of you, thanks for helping.

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

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)

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