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I can't find a formula for when the actual base of a $\log$ changes. I specifically mean this scenario.

$$ \lim_{h \to 0} \frac{\log_{a + h}[b] - \log_{a}[b]}{h} $$

The reason why I need it is that I am using an alternate notation that makes this problem obvious. Use up arrow for a power, and down arrow for $\log$.

$$ \underbrace{a \downarrow (b \uparrow c)}_{\log_a[b^c]} = \underbrace{(a \downarrow b)\ c}_{c\ \log_a[b]} $$

In this alternate notation, this is $a^b$ implicitly differentiated. It is it handles both $a$ and $b$ changing at the same time.

$$ d[a + b] = d[a] + d[b] $$

$$ d[a * b] = b\ d[a] + a\ d[b] $$

$$ d[a \uparrow b] = (a \uparrow b) ( \frac{b}{a} d[a] + \underbrace{(e \downarrow a)}_{ln[a]} d[b]) $$

$d[a \uparrow b]$ in more familiar notation looks like:

$$ d[a^b] = b\ a^{b-1} d[a] + a^b ln[a]\ d[b] $$

But this leaves the question of how to then define $d[a \downarrow b]$. It needs to handle the case where $a$ is not a constant, because this is a non-commutative and non-associative binary operator. I can't find anything anywhere that handles a changing base. I don't know what goes in place at the big star.

$$ d[a \downarrow b] = \bigstar d[a] + \frac{1}{b (e \downarrow a)}d[b] $$

What I did try was to just use change of base formula to get something like:

$$ (a + \delta a) \downarrow (b + \delta b) = \frac{e \downarrow (b + \delta b)}{e \downarrow (a + \delta a)} = \frac{\ln[b] + \ln[1 + \frac{\delta b}{b}]}{\ln[a] + \ln[1 + \frac{\delta a}{a}]} $$

Which doesn't look right, because I am expecting that when $\delta a = 0$, I get the normal formula for differential of $\log$ function.

C.F.G
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Rob
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    I didn't really quite understand actually, but from the first formula I guess you're trying to calculate $\frac{d\log_x(b)}{dx}$ – Fareed Abi Farraj Aug 21 '19 at 06:01
  • yes, implicitly differentiating $log_x[b]$ to be specific, where my notation would be $d[x \downarrow b]$, and both x and b are variables. – Rob Aug 21 '19 at 06:11

2 Answers2

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Using ordinary differentiation rules we get $$(\log_xb)'=\left(\frac{\ln b}{\ln x}\right)'=-\frac{\ln b}{x(\ln x)^2}=-\frac{\log_xb}{x\ln x}$$ which in your notation is $$-\frac{x\downarrow b}{x(e\downarrow x)}$$

Parcly Taxel
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  • I'm going to assume that the answer to this question is: $d[a \downarrow b] = -\frac{a \downarrow b}{a(e \downarrow a)}\ d[a] + \frac{1}{b(e \downarrow a)}\ d[b]$. – Rob Aug 21 '19 at 22:04
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$\displaystyle y = \log_x z $

or, $\displaystyle x^y = z $

or, $\displaystyle y \ln x = \ln z$

or, $\displaystyle dy ~ \ln x + y \frac{dx}{x} = \frac{dz}{z}$

or, $\displaystyle dz = z \ln x ~ dy + \frac{yz}{x} ~ dx$

or, $\displaystyle dz = x^y \ln x ~ dy + yx^{y-1} ~ dx $

Probably you are looking for the above result, right?

PTDS
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