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Desperately Difficult Derivative (Posted on 2005-03-30) Difficulty: 4 of 5
If f(x)=sin(x³), find the value of its 2001st derivative, at x=0.

  Submitted by Old Original Oskar!    
Rating: 3.8000 (5 votes)
Solution: (Hide)
As sin(x)=x-x^3/3!+x^5/5!-x^7/7!..., f(x)=x^3-x^9/3!+x^15/5!-x^21/7!...= (terms x^K for K<2001) -x^2001/667! + (terms x^K for K>2001).

Differentiating 2001 times, we get -2001!/667! + (terms x^K for K>0) so at 0, the result is -2001!/667!.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsPuzzle Thoughts K Sengupta2023-03-26 08:49:40
QuestionNo SubjectVee-Liem Veefessional2010-04-19 18:06:51
re(3): solution?Old Original Oskar!2005-04-04 18:54:39
re(4): solution?ajosin2005-04-01 23:03:53
re(3): solution?armando2005-04-01 10:29:19
re(2): solution?ajosin2005-03-31 19:11:55
Solutionre: Simple solution (spoiler alert!)Charlie2005-03-31 15:09:28
SolutionSimple solution (spoiler alert!)ajosin2005-03-30 22:31:43
Some Thoughtsre: solution?Old Original Oskar!2005-03-30 20:50:05
solution?armando2005-03-30 19:27:14
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