If f(x)=sin(x³), find the value of its 2001st derivative, at x=0.
In a quick approach it seems that the only terms that could take a value different from 0 are those where the third derivate of x^3 (=6) appears alone.
If I haven't made mistakes, in pair derivatives we have 6*sin (x^3), and in impars we have 6*cos (x^3) (and with sign changing). For 2001st derivatives the sign would be negative and the value would be -6.
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Posted by armando
on 2005-03-30 19:27:14 |
solution?
