First Derivative CoB Theorem

  • Thread starter Orion1
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  • #1
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Main Question or Discussion Point


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?
 

Answers and Replies

  • #2
lurflurf
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Orion1 said:

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?
It is true, and follows from
[tex]\log_v(u)=\frac{\log(u)}{\log(v)}[/tex]
 
  • #3
970
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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