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Jul 16 at 9:44 answer added Sjoerd Smit timeline score: 4
Jul 16 at 8:35 history edited Prem CC BY-SA 4.0
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Jul 16 at 8:30 history edited Prem CC BY-SA 4.0
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Jul 16 at 8:21 comment added Prem @BobHanlon You are right. I forgot to mention in the question the ReplaceRepeated works.
Jul 16 at 8:18 comment added Prem @creidhne Do you mean that Replace will only work once for one of the rule that is found to apply? I tested this using following code 1 + x + y /. {x -> x1, y -> y1}. The result is 1 + x1 + y1. Both the replacements happened here without any problem.
Jul 15 at 18:12 answer added Roland F timeline score: 2
Jul 15 at 16:46 history edited creidhne
The bugs tag is reserved for questions where the problem has been vetted by this community and the observed behavior is confirmed to be a bug. Please do not use this tag for new questions
Jul 15 at 16:43 comment added creidhne These results are not a bug. Replace documentation explains that when a list of rules is given, the result with the first rule that applies is returned. Your expected result is given by Replace Repeated, which repeatedly performs replacements until the expression no longer changes.
Jul 15 at 16:06 history edited user64494 CC BY-SA 4.0
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Jul 15 at 15:17 comment added edinorog2196 A workaround is to replace g[x] -> x D[g[x]/x, x]. In my experience it is better to just substitute a single thing rather than an expression g[x]/x which could not be found in the code (you have (Cos[x]g[x])/x)
Jul 15 at 14:39 comment added AsukaMinato 1/x // FullForm gives Power[x,-1]
Jul 15 at 14:38 comment added Bob Hanlon I don't know why it doesn't work, but you can use ReplaceRepeated (//.) instead of ReplaceAll (/.)
S Jul 15 at 14:20 review First questions
Jul 15 at 15:00
S Jul 15 at 14:20 history asked Prem CC BY-SA 4.0