Basic question about nonstandard derivative I'm trying to understand how the nonstandard derivative works.
For instance, consider the function
$f(x) = \frac{1}{2} x^2$
The derivative is
$f'(x) = st \left( \frac{\frac{1}{2}(x + \epsilon)^2 - \frac{1}{2}x^2}{\epsilon} \right)$
for some infinitesimal $\epsilon$. This works out to
$f'(x) = st(x + \frac{\epsilon}{2}) = st(x)$
So the derivative of $\frac{1}{2} x^2$ is the standard part function.
Is this correct? I was expecting the derivative to be $f'(x) = x$ instead. $st(x)$ doesn't even have the same domain as the original function. Or am I misunderstanding something?
 A: The non-standard definition of derivatives that you quote in your question defines the ordinary derivative function, mapping real numbers $x$ to real numbers $f'(x)$.  Here $f$ itself should also be an ordinary function mapping real numbers to real numbers.  Where the definition appears to apply $f$ to a non-standard input, $x+\varepsilon$, it ought to apply not the ordinary $f$ but rather its canonical extension, usually called ${}^*\!f$, which takes hyperreals to hyperreals.
A: $$f'(x) = st \left( \frac{\frac{1}{2}(x + \epsilon)^2 - \frac{1}{2}x^2}{\epsilon} \right)= st \left( \frac{\frac{1}{2}x^2 + \epsilon x + \frac{1}{2}\epsilon^2 - \frac{1}{2}x^2}{\epsilon} \right) \\ = st \left( \frac{\epsilon x + \frac{1}{2}\epsilon^2}{\epsilon} \right) = st \left( x + \frac{1}{2}\epsilon\right)=x$$
Note that the standard part function is a function from the hyperreals to the reals. We can just evaluate it, since $x$ is real. Since $\epsilon$ is extremely small, $x$ is the nearest real number. (I assume that you want the derivative with domain $\mathbb R$, not with the hyperreals as domain)
