2
$\begingroup$

I am trying to take he derivative of mod(remainder).

In:= D[Mod[x, y], x]
Out = Mod^(1,0)[x,y]

I cannot understand the meaning of Mod^(1,0). What does the exponential actually mean ?

$\endgroup$

closed as off-topic by MarcoB, Cassini, march, Simon Woods, m_goldberg May 25 '16 at 22:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Cassini, march, Simon Woods, m_goldberg
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ It means the derivative with respect to the first argument. Look up Derivative[]. $\endgroup$ – J. M. will be back soon May 25 '16 at 17:26
  • $\begingroup$ Look at the FullForm of your result or evaluate D[Mod[x, y], x] // FullForm $\endgroup$ – Bob Hanlon May 25 '16 at 19:16
  • $\begingroup$ I'm not sure about the "easily found in the documentation" part :) The notation is described explicitly in Derivatives of Unknown Functions. A broader discussion can be found in The Representation of Derivatives. $\endgroup$ – WReach May 26 '16 at 14:22