2006-12-08 10:53:04 · 3 answers · asked by kondiii 1 in Science & Mathematics ➔ Mathematics
im looking for nth derivative
2006-12-08 11:26:59 · update #1
Thus y(x)=6*x^(-3)+0.5; y’=6*(-3)*x^(-4); y’’=6*(-3)*(-4)*x^(-5); y’’’=6*(-3)*(-4)*(-5)*x^(-6); enough for induction? Then y(n) = 3 * (-1)^n * (n+2)! * x^(-n-3) for n>=1; Check it!
2006-12-08 13:26:42 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I don't think there will be a "formula" to follow...
f'(x) = [3x^3(3x^2) - (12+x^3)6x^2]/4x^6
= [9x^5 - 72x^2 - 6x^5]/4x^6
= [3x^5 - 72x^2]/4x^6
= 3x^2[x^3 - 24]/4x^6
= 3[x^3 - 24]/4x^4
I don't see a great way to write that in terms of the original f(x), so I don't think there's gonna be a generic formula for f(n)(x).
Could be wrong though....
2006-12-08 12:01:00 · answer #2 · answered by Jim Burnell 6 · 0⤊ 0⤋
I'm not exactly sure what you're asking. Can you clarify?
2006-12-08 11:05:36 · answer #3 · answered by Puggy 7 · 0⤊ 0⤋