Non-standard analysis way of proving that derivative of $e^x$ is $e^x$ What is the non-standard (infinitesimal) analysis way of proving that the derivative of $e^x$ is $e^x$? I tried to prove it myself, but I am having a hard time proving this without recourse to standard limit things.
 A: $$
\frac{e^{x + dx} -e^{x}}{dx}=  e^x \left( \frac{e^{dx} -1}{dx} \right)
$$
You transfer everything to non-standrad reals, so that $dx$ could become infinitesimal. 
Then you evaluate the exponential from its infinite series
$$
 \frac{e^{dx} -1}{dx} = 1+  \frac{dx}{2} + H.O.T (dx^2)
$$
the standard part of the last one  is  1.
It is exactly the same technique like working with approximations. 
A: The equation $\frac{dy}{dx}=y$ is usually solved informally by "separation of variables", which involves writing $\frac{dy}{y}=dx$ and then integrating both sides, producing $\log y= x$ or $y=e^x$. However, in the traditional approach this calculation is merely a heuristic device since the symbol $\frac{dy}{dx}$ does not mean a ratio of infinitesimals but is rather a formal indivisible symbol for the derivative.  Therefore justifying this calculation requires a separate proof.
Meanwhile, in the hyperreal framework where infinitesimals are part of the rigorous tool kit the calculation can be taken rather literally not as heuristics but, rather, as actual proof. Some details need to be worked out related to the formula $\frac{dy}{dx}=\text{st}\frac{\Delta y}{\Delta x}$ but it's basically a good proof.
