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So in differentiating any exponential function we find a common pattern. Namely, the derivative is equal to the function itself multiplied by a constant which does not depend on the input variable. How would one go about computing this constant? I guess I'm looking to understand how we deduce that this constant is equal to the natural logarithm of the exponential's base?
beginalign
fracddx2^x
&=lim_hto0frac2^x+h-2^xh\
&=2^xlim_hto0frac2^h-1h\[6px]
&=2^xln(2)
endalign

Your question assumes that you have some definition of $2^x$ which does not use logarithm. The problem of defining $a^x$ for $a>0,xinmathbb R $ without the use of logarithms is difficult and most textbooks avoid this approach.

Anyway assuming that this difficult part is taken care of, one can proceed as in your question and get $$(a^x) '=a^xlim_hto 0fraca^h-1htag1$$ From here we have to embark on the journey of exploring the limit $$lim_hto 0fraca^h-1htag2$$ by proving the following:

• the limit $(2)$ exists if $a>0$ and hence defines a function $F:mathbbR^+ tomathbb R $ via $F(a) =limlimits_hto 0dfraca^h-1h$.


• 
  $F$ is strictly increasing and continuous with $F(1)=0$.


• 
  $F(ab) =F(a) +F(b), a>0<b$.


• 
  $F(a^b) =bF(a), a>0,binmathbb R $.


• 
  $F$ is differentiable with derivative $F'(x) =1/x$.


• There is a positive number $e>1$ such that $F(e) =1$. And we have $F(e^x) =x, xinmathbb R $ and $e^F(x) =x, xinmathbb R ^+ $. Thus $F(x) $ and $e^ x$ are inverses to each other.

And once we have proved the above we can just say that $F(x) $ is the natural logarithm of $x$.

All of the above including the difficult definition of $a^x$ has been provided with details in my blog post.

Write $k=ln a$ so $fracddxa^x=fracddxe^kx=ke^kx$ by the chain rule. This is $a^xln a$.

Not a rigorous argument, but may help you understand.

Let $h=dfrac1n$. We have

$$frac2^h-1h=aiff(1+ah)^1/h=left(1+frac anright)^n=2$$

Taking the limit for $hto0$,

$$ln 2=aiff e^a=2.$$

The constant $e$ appears as the limit of

$$left(1+frac 1nright)^n.$$

The backwards answer.

Notice there is always a constant, call it $C_b$, where $frac db^xx = C_bcdot b^x$. We also notice that if $0 < b < c$ then $C_b < C_c$. And it appears that if we consider $C_x$ as a function of $x$ thatn $C_x$ is continuous, 1-1, and increasing. As $C_2 < 1$ and $C_3 > 1$ it seems there must by some number $E$ so that $C_E = 1$.

This means that as $b^x = E^xlog_E b$ that by the chain rule $frac db^xdx = (C_Ecdot E^xlog_E b)cdotlog_E b = log_E bcdot b^x$. But $frac db^xdx = C_bb^x$. So that means $C_b = log_E (b)$.

So we can use that as our definition. $e$ is the base where $frac de^xdx=e^x$. And as a result $frac db^xdx =frac de^xlog_e bdx =log_e bcdot b^x$.

All neat clean and accurate.

... but backwards.

Thing is .... we haven't actually defined what $b^x$ means if $x$ is irrational.

So most texts define $e$ and $ln x$ in other ways and derive these results.