# First Derivative CoB Theorem

Orion1

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

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

Is this theorem correct?

Does this theorem accept functions in the base?

Homework Helper
Orion1 said:

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

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

Is this theorem correct?

Does this theorem accept functions in the base?
It is true, and follows from
$$\log_v(u)=\frac{\log(u)}{\log(v)}$$

Orion1
Theorem Proof...

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

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