Differentiate the functions: f(x)= 1/(x to the 2nd + 2)to the 3rd
2007-04-07 08:52:49 · 7 answers · asked by m s 1 in Science & Mathematics ➔ Mathematics
if you mean simplify then
f(x)= 1/(x to the 2nd + 2)to the 3rd
f(x)=1/(x^6+2^3) distribute the third
f(x)=1/(x^6+8) solve the 2^3 final answer
if not sorry
2007-04-07 08:58:48 · answer #1 · answered by Kelsey 3 · 0⤊ 0⤋
1/(x^2+2)^3
1/(x^2+^3+2^3)
1/(x^5+8)
f(x)= 1/ x to the 5th + 8
2007-04-07 09:44:04 · answer #2 · answered by bernel1403 5 · 0⤊ 0⤋
1/x,-1/x^2,+2/x^3,-6/x^4
2007-04-07 09:04:48 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
f(x) = 1/ (x^2+2)^3
that makes
(x^2 + 2) ^-3 so its derivate is
-3(x^2 + 2)^-4 (2x)
or -6x / (x^2 + 2)^4
2007-04-07 09:01:13 · answer #4 · answered by w1ckeds1ck312121 3 · 0⤊ 0⤋
y = 1/(x²+2)³
let u=1, v=(x²+2)³
Now use the quotient rule:
dy/dx = (vdu/dx - udv/dx) / v²
2007-04-07 08:59:43 · answer #5 · answered by Anonymous · 0⤊ 0⤋
f(x) = 1/(x^2 +2)^3
f(x) = (x^2+2)-3
f'(x) =(-3(x^2 +2)-4)(2x)
= -6x/(x^2+2)^4
2007-04-07 09:07:05 · answer #6 · answered by bignose68 4 · 1⤊ 0⤋
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2007-04-07 09:13:04 · answer #7 · answered by Pam 5 · 0⤊ 0⤋