Is it possible to find the derivative of $y = 2^x$ using Calculus Made Easy-style infinitesimals? I was wondering today...
Is it possible to find $dy/dx$ given $y=2^x$ using infinitesimals of the sort found in Thompson's Calculus Made Easy? 
For example, Thompson found the derivative of $x^2$ by letting $y+dy=(x+dx)^2$ and doing some algebra (plus letting some higher order differentials "vanish"). 
I tried to do something similar with my exponential function:
$y+dy=2^{x+dx}$
But I got stuck, and now I am not even sure it is possible to differentiate $2^x$ using this strategy. Any thoughts? 
 A: The key to all of this is the number $L(a) = (a^{dx}-1)/dx$. This is a new sort of function, not expressible as a rationally in terms of $a$. The crucial fact is that
$L$ is a logarithm
I say that $L$ is a logarithm, which requires proving that
$$ L(ab) = L(a) + L(b) \tag{1} $$
for $a,b>0$.
But
$$ L(ab) dx = a^{dx} b^{dx} - 1 = (1+L(a)dx)(1+L(a)dx) - 1 = (L(a)+L(b))dx $$
since $a^{dx} = 1 + L(a)dx $, and dividing by $dx$ gives the result. $(1)$ gives us a lot of information about $L$:
$$ L(a) = L(a \cdot 1) = L(a) + L(1) , $$
so $L(1)=0$, $0 = L(a/a) = L(a) + L(1/a) $, so $L(1/a)=-L(a)$, and
$$ L(a^n) = L(aa^{n-1}) = L(a) + L(a^{n-1}) = \dotsb = n L(a), \tag{2} $$
and eventually for any $b$,
$$ L(a^b) = bL(a) . $$
Derivative?
Which logarithm is it? Let's start with a seemingly more difficult question: what is the derivative of $L$?
$$ L(x+dx) = L(x(1+dx/x)) = L(x) + L(1+dx/x) = L(1) + L'(1) dx/x = L(x) + L'(1) dx/x $$
This is reasonably uncontroversial. The hairy part: what is $L'(1)$?
$$ L(x+dx) dy = (x+dx)^{dy}-1 = x^{dy}\left(1+\frac{dx}{x}\right)^{dy}-1 = (1+L(x)dy)\left(1+\frac{dx}{x}dy\right) -1 = \left(L(x) + \frac{dx}{x} \right) dy . $$
So this very suspicious manipulation gives $L'(x) = 1/x$. Hence $L$ is increasing, and in particular, $L(2)>0$, and for any positive $a,b$, 
$$L(a)=L(b) \implies L(a/b)=0 \quad \text{, so } a=b. $$ 
Moreover, $(2)$ implies that $L(x)$ is unbounded above as $x \to \infty$, unbounded below as $x \to 0$. Now we can ask
Which logarithm?
In particular, there is some number $E$ so that $L(E)=1$. But this means that
$$ L(E^x) = xL(E) = x, $$
for any real $x$. We can write $x=L(y)$, and then
$$ L(E^{L(y)}) = L(y) , $$
and so $(3)$ implies that
$$ E^{L(y)} = y , $$
for any positive $y$. So $E^x$ and $L(x)$ are inverse functions. We also have $E^{aL(y)} = E^{L(y^a)} = y^a$.
Wrapping up, $E^{dx} = 1 + L(E)dx = 1 + dx$, so $ (E^{dx}-1)/dx = 1 $, and using the definition $e = (1+dx)^{1/dx}$ (substituting the limit involving $n$ for a simple infinitesimal expression),
$$ L(e) = L((1+dx)^{1/dx}) = \frac{L(1+dx)}{dx} = \frac{L(1)+L'(1)dx}{dx} = L'(1) = 1 , $$
so $E=e$, and $L = \log$, pretty much whatever your definitions are.

Post mortem
Now, there are numerous problems with the above derivation: we haven't said how to define arbitrary real powers, but have used them liberally. The suspicious bit is some sort of nasty double limit that we'd be better off staying well clear of. Inverse functions have been used fairly glibly throughout. But most of this can actually be made rigorous!
A: Just because I amazed that it works, I will post for now using the power series (or actually just using Taylor polynome!) :
$$
2^{dx} = e^{dxln(2)} = 1 + \ln(2)dx + O((\ln(2)dx)^2) = 1 + \ln(2)dx,
$$
where $O(x^n)$ is the $O$ notation, that means all the terms are of order $n$ or higher.
So you actually get:
$$
y + dy = 2^x + \ln(2)dx2^x
$$
edit 
Maybe using Pascal triangle or the binomial theorem it sounds reasonable to you that
$$
\begin{aligned}
\left(1+\frac{y}{n}\right)^n = 1 + n \frac{y}{n} + \binom{n}{2}\frac{y^2}{n^2} +\dots
\end{aligned}
$$
so again you can infer
$$
e^{\ln(2)dx} = \lim_{n\to \infty} \left(1+\frac{\ln(2)dx}{n} \right)^n = \lim_{n\to \infty} \left(1 + \ln(2)dx + O\left(\frac{dx^2}{n^2}\right)\right) = 1 +\ln(2)dx
$$
