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Math Help - Differentiating log of a rational function

  1. #1
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    Differentiating log of a rational function

    Differentiate y=log \frac{x+1}{x+4}
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    Jen
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    Last edited by Jen; May 17th 2009 at 11:50 AM.
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  3. #3
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    Wait. I think I got it

    Is it:

    y=\log x+1 - \log x+4
    = \frac{1}{x+1} - \frac{1}{x+4}
    = 5

    ??
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  4. #4
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    Quote Originally Posted by nerdzor View Post
    Wait. I think I got it

    Is it:

    y=\log x+1 - \log x+4
    = \frac{1}{x+1} - \frac{1}{x+4}
    = 5

    ??
    1. Is \log meant to be the same as \ln ?

    If so

    y=\log (x+1) - \log( x+4) ~\implies~ y'=\dfrac1{x+1} - \dfrac1{x+4} = \dfrac{(x+4)-(x+1)}{(x+4)(x+1)}= \dfrac3{(x+4)(x+1)}

    2. If log means \log_{10} then you have transform these logarithms into natural logarithms using the base change formula.
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