If f(x)=sin(x³), find the value of its 2001st derivative, at x=0.
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