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Homework Help: Differentiate y=\ln(e^{-x} + xe^{-x})

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Differentiate

    [tex]y=\ln(e^{-x} + xe^{-x})[/tex]


    2. Relevant equations
    - Logarithmic differentiation?


    3. The attempt at a solution
    Here goes:

    [tex]y=\ln(e^{-x} + xe^{-x})[/tex]

    [tex]y'=\frac{e^{-x} + (x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x)}{e^{-x}+xe^{-x}})[/tex]

    [tex]\frac{e^{-x}+xe^{-x}+e^{-x}}{e^{-x}+xe^{-x}}[/tex]

    I must be missing something huge here because the book's solution is

    [tex]y'=\frac{-x}{1+x}[/tex]
     
  2. jcsd
  3. Jul 29, 2008 #2
    Ok I looked at this a second time and got a bit further.. I'm still missing the solution though.

    [tex]\frac{e^{-x}+xe^{-x}+e^{-x}}{e^{-x}+xe^{-x}}=\frac{2e^{-x}+xe^{-x}}{e^{-x}+xe^{-x}}[/tex]

    [tex]\frac{\frac{1}{e^{x}}+\frac{x}{e^{x}}+\frac{1}{e^{x}}}{\frac{1}{e^{x}}+\frac{x}{e^{x}}}[/tex]

    [tex]\frac{\frac{2}{e^{x}}+\frac{x}{e^{x}}}{\frac{1}{e^{x}}+\frac{x}{e^{x}}}[/tex]

    [tex]\frac{\frac{2+x}{e^{x}}}{\frac{1+x}{e^{x}}}[/tex]

    [tex]\frac{2+x}{e^{x}}\frac{e^{x}}{1+x}[/tex]

    [tex]\frac{e^{x}(2+x)}{e^{x}(1+x)}[/tex]

    [tex]\frac{2+x}{1+x}[/tex]

    ?
     
  4. Jul 29, 2008 #3
    The derivative of e^x is itself, but you have "e^(-x)" which is not itself.
     
  5. Jul 29, 2008 #4
    You should first try simplifying the initial equation (common factoring). You can then use the distributive ln properties to make your work easier.
     
  6. Jul 29, 2008 #5
    Oops.. I assumed that if the derivative of e^x is itself, the x stood as a placeholder for, well, pretty much anything, including -x.

    So um.. what does e^(-x) come out to be? Ah wait.. just saw it.

    If y = e^(-x), then y' = e^(-x)ln(e)

    Hmm.. is that correct? This is the rule for differentiating exponential functions. e IS a constant, so it's valid. Right?
     
  7. Jul 29, 2008 #6
    y = e^(-x) is then y' = e^(-x) * -1

    The reason why you multiply the answer by -1 is that you still have to take the derivative of the exponent. In this case the exponent was -x.

    Also you should read my previous post.

    Another note you should take into account is that text book answers are always simplified down.
     
  8. Jul 29, 2008 #7
    Ok I tried the following:

    [tex]y=\ln(e^{-x}+xe^{-x})[/tex]

    [tex]y'=\ln e^{-x}+\ln(xe^{-x})[/tex]

    [tex]-e^{-x}\ln e+\ln x+\ln e^{-x}[/tex]

    [tex]-e^{-x}+\ln x-e^{-x}\ln e[/tex]

    [tex]-e^{-x}+\ln x-e^{-x}[/tex]

    [tex]-2e^{-x}+\frac{1}{x}[/tex]

    [tex]-2\frac{1}{e^x}+\frac{1}{x}=-\frac{2}{e^x}+\frac{1}{x}[/tex]

    But frankly, I don't know if this is correct. I might have not followed proper rules there :/
     
  9. Jul 29, 2008 #8
    You misused the Logarithmic property.

    [tex]\ln{ab}=\ln a+\ln b[/tex]

    [tex]\ln{(a+b)}\neq\ln a+\ln b[/tex]

    Your first post is more correct than your recent ones. All you have to do is go back to e^(-x) and put a negative in the front.
     
  10. Jul 29, 2008 #9
    Got it! Thanks for the tip :)

    [tex]y'=\frac{e^{-x} + (x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x)}{e^{-x}+xe^{-x}})[/tex]

    [tex]\frac{-e^{-x}-xe^{-x}+e^{-x}}{e^{-x}+xe^{-x}}[/tex]

    [tex]\frac{e^{-x}(1-1-x)}{e^{-x}+xe^{-x}}[/tex]

    [tex]\frac{e^{-x}(-x)}{e^{-x}(x+1)}[/tex]

    [tex]\frac{-x}{x+1}[/tex]

    ! :)
     
  11. Sep 16, 2009 #10
    Hi, I know this is kind of old, but I came across this same problem in my book - and I don't get how to get from step a to step b - like this:

    [tex]
    y=\ln(e^{-x} + xe^{-x})
    [/tex]


    [tex]
    y'=\frac{e^{-x} + (x\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}x)}{e^{-x}+xe^{-x}})
    [/tex]


    can anyone help? What rules are you using to go from the first one to the second one?
    thanks a lot,

    Aniella
     
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