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Math Help - Help with differentiation incl. ln.

  1. #1
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    Help with differentiation incl. ln.

    Hello MathHelp.
    So i'm trying to solve this problem below
    Help with differentiation incl. ln.-unavngivet.png
    And using the rules of differentiation that i know of i fail. Perhaps the book that i am reading is wrong. If anyone can help it would be greatly appreciated.

    - Simon.
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  2. #2
    Senior Member vincisonfire's Avatar
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    Re: Help with differentiation incl. ln.

    Hi,
    These is nothing special there. Just use the product rule together with the chain rule.
    Note that you have something like (x-1)ln(x-1)-x ln(x). If you differentiate you will have something of the form ln(x-1)-{x-1\over x-1}-ln(x)+{x\over x} = ln(x-1)-ln(x).
    It is true that all the constants makes it messy, but keep you head straight and just use the basics rules you know.
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  3. #3
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    Re: Help with differentiation incl. ln.

    First take out the constant

    \displaystyle \frac{d}{dE}\kappa_B\left( \left(1+\frac{E}{\epsilon}\right)\times \ln\left(1+ \frac{E}{\epsilon}\right)-\frac{E}{\epsilon}\ln\frac{E}{\epsilon}\right)

    \displaystyle = \kappa_B\frac{d}{dE}\left( \left(1+\frac{E}{\epsilon}\right)\times \ln\left(1+ \frac{E}{\epsilon}\right)-\frac{E}{\epsilon}\ln\frac{E}{\epsilon}\right)

    then try the product rule term by term i.e.

    \displaystyle  \frac{d}{dE}\left( \left(1+\frac{E}{\epsilon}\right)\times \ln\left(1+ \frac{E}{\epsilon}\right)\right)

    \displaystyle  =  \frac{d}{dE}\left(1+\frac{E}{\epsilon}\right) \times \ln\left(1+ \frac{E}{\epsilon}\right)+\left(1+\frac{E}{\epsilo  n}\right)\times \frac{d}{dE}\ln\left(1+ \frac{E}{\epsilon}\right)
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  4. #4
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    Re: Help with differentiation incl. ln.

    Thank you pickslides That's easier than expected.

    - Simon.
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