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First Derivative CoB Theorem

  1. Aug 13, 2005 #1

    Any Calculus researchers interested in disproving this theorem with a simple base and function?

    Orion1 change of base theorem:
    [tex]\frac{d}{dx} (\log_v u) = \frac{1}{u \ln(v)} \frac{du}{dx} - \frac{\ln(u)}{v \ln^2 (v)} \frac{dv}{dx}[/tex]

    Is this theorem correct?

    Does this theorem accept functions in the base?
     
  2. jcsd
  3. Aug 13, 2005 #2

    lurflurf

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    It is true, and follows from
    [tex]\log_v(u)=\frac{\log(u)}{\log(v)}[/tex]
     
  4. Aug 14, 2005 #3
    Theorem Proof...


    First derivative Change of Base (proof 1):
    [tex]\frac{d}{dx} (\log_v u) = \frac{d}{dx} \left( \frac{\ln(u)}{\ln(v)} \right) = \frac{1}{u \ln(v)} \frac{du}{dx} - \frac{\ln(u)}{v \ln^2 (v)} \frac{dv}{dx}[/tex]

    First derivative Change of Base theorem:
    [tex]\boxed{\frac{d}{dx} (\log_v u) = \frac{1}{u \ln(v)} \frac{du}{dx} - \frac{\ln(u)}{v \ln^2 (v)} \frac{dv}{dx}}[/tex]
     
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