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

See The Solution Submitted by Old Original Oskar!    
Rating: 3.8000 (5 votes)

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Solution Simple solution (spoiler alert!) | Comment 3 of 9 |
What a nice problem!. Afer spending some time taking derivatives I got a big headache. I took a step back and thought series!.

We all know that sin z = z - z^3/3! + z^5/5! + ... + (-1)^(n+1) z^(2n-1)/(2n-1)!

If z = x^3, the only important term is the one with exponent that satisfies,

3(2n-1) - 2001 = 0.

Which gives n = 334. Therefore the solution is given by

(-1)^(334+1) z^2001/667! derived 2001 times.

This gives -(2001)!/667!

  Posted by ajosin on 2005-03-30 22:31:43
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