For standard analysis, an answer (in the concept of limit) is here.

However, when dealing with infinitesimals in non-standard analysis, that limit technique will not work.

So in non standard analysis, should we necessarily consider derivative as slope between two infinitesimally apart points? Or should we still consider it as slope at "a point in hyper real number line $(\varepsilon)$"?

If the latter is true, please give a reason.


Actually, every technique of standard analysis also works in nonstandard analysis; this is formalized by the transfer principle.

In particular, the notions of limit for (internal) functions on the hyperreals is meaningful, and can even be computed by the usual $\epsilon$-$\delta$ approach.

And the notion of derivative for (internal) functions on the hyperreals is also meaningful, and satisfies the identity

$$ f'(a) = \lim_{\epsilon \to 0} \frac{f(a+\epsilon) - f(a)}{\epsilon} $$

in the sense that if either side exists, then so does the either and both sides are equal.

What nonstandard analysis does is allow for new techniques by comparing the standard and nonstandard models. In particular, if $f$ is a standard function and $a$ is a standard real number, then

$$ f'(a) = \mathrm{std} \left( \frac{f(a+\epsilon) - f(a)}{\epsilon} \right) $$

in the sense that there is a canonical way to interpret $a$ as a hyperreal and $f$ as a function on the hyperreals, and if the right side has the same value for every nonzero infinitesimal $\epsilon$, then the left hand side exists and equality holds. Conversely, if the left hand side exists, then the right hand side does as well for every (nonstandard) nonzero infinitesimal $\epsilon$, and they are equal.

And note that this isn't just the difference quotient; one has to do the additional operation of taking the standard part.

Just like the standard formulation of derivatives, one doesn't define the derivative simply by taking the different quotient at two points — one has to do something extra in order to pass from that idea to the derivative.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.