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.