1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need formula or how to

  1. Nov 4, 2007 #1
    [SOLVED] need formula or how to...

    y= ln ex/ex (x is the power)
  2. jcsd
  3. Nov 4, 2007 #2


    User Avatar
    Homework Helper

    uhm use the "^" to represent powers, for example, x squared is written as x^2

    and please state your question better
  4. Nov 4, 2007 #3

    y= ln e^x/e^x

  5. Nov 4, 2007 #4


    User Avatar
    Homework Helper

    So the question is to differentiate

    [tex] y=\frac{ln e^x}{e^x}[/tex] ?

    if it is...what rule do you think you would use when you need to differentiate something of the form [itex]y=\frac{u}{v}[/itex] ?
  6. Nov 4, 2007 #5
    quotient rule.
    which is: first, derivative of second, minus second, derivative of first, divided by first squared. (but, i'm still stuck. could you work it out for me?)
    Last edited: Nov 4, 2007
  7. Nov 4, 2007 #6


    User Avatar
    Homework Helper

    Well firstly...lne^x can be simplified to xlne which is just x

    so you really want to find [tex]\frac{d}{dx}(\frac{x}{e^x})[/tex]

    so if u= x => [itex]\frac{du}{dx}=1[/itex]
    and v=e^x => [itex]\frac{dv}{dx}=e^x[/itex]

    the formula is
    [tex] \frac{d}{dx}(\frac{u}{v}) =\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]

    so then you'll have
    [tex]\frac{d}{dx}(\frac{x}{e^x}) = \frac{(e^x)(1) -(x)(e^x)}{(e^x)^2}[/tex]

    then you will simplify it
  8. Nov 5, 2007 #7
    thanks a lot. you made it very easy to understand. only part i didn't get is it says answer should be 1-x/e^x. what cancels out the power in the denominator?

    Thanks again!
  9. Nov 5, 2007 #8


    User Avatar
    Homework Helper

    ...well the e^x in the numerator can be factored out and it will cancel with an e^x in the denominator...so giving you your answer
  10. Nov 5, 2007 #9
    oh, i see. thank you very much.. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook