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Nov 10, 2012 at 20:44 comment added chris I ve updated my answer thanks to the help of @rojo so that the output looks exactly as you had typeset.
Nov 10, 2012 at 19:43 history tweeted twitter.com/#!/StackMma/status/267351964059332608
Nov 10, 2012 at 18:58 comment added JohnD @VitaliyKaurov: I do not know why they make that line under the second box of code behave the way it does.
Nov 10, 2012 at 18:55 vote accept JohnD
Nov 10, 2012 at 18:46 answer added Rojo timeline score: 6
Nov 10, 2012 at 18:22 comment added Vitaliy Kaurov Related: Wolfram Blog By the way, do you understand why there is contrast in behavior?
Nov 10, 2012 at 17:55 comment added Jens You may find this question useful, too.
Nov 10, 2012 at 17:37 answer added chris timeline score: 17
Nov 10, 2012 at 17:29 comment added JohnD @chris: Indeed I think you have. I re-edited it to show what I do, leaving you room to formally answer the question with what you had. I just did not know that Derivative[...] functioned differently than using the partial symbol in the palette (at least in the sense here).
Nov 10, 2012 at 17:28 history edited JohnD CC BY-SA 3.0
dispayed partial notation as intended
Nov 10, 2012 at 17:21 comment added JohnD @J. M.: I see what you mean, but unlike the first answer in the link I don't want to appeal to limits since we aren't emphasizing what a partial derivative is but rather just need to evaluate partials at values to accomplish other things. The second answer does address the issue, but I'd say the code is more complicated than a replacement rule. YMMV of course.
Nov 10, 2012 at 17:20 comment added chris @texasAUtiger I have edited your question; but in the process may be answered it? In other words what's wrong with Derivative[1, 0][u][2, y]?
Nov 10, 2012 at 17:18 history edited chris CC BY-SA 3.0
deleted 66 characters in body
Nov 10, 2012 at 17:09 comment added J. M.'s missing motivation Seems to be a dupe of this; anyway, look up SeriesCoefficient[].
Nov 10, 2012 at 17:04 history asked JohnD CC BY-SA 3.0