f(x) = (x^4 +1) / x^2 . find f '(x) .
2006-09-07 08:58:52 · 8 answers · asked by Asker 1 in Science & Mathematics ➔ Mathematics
if f(x) = u(x) / v(x)
f'(x) = ( u'(x) (v(x) - v'(x) u(x) / [ v(x^2)]
f'(x) = (x^4+1)' x^2 - (x^2)' (x^4+1) =
( 4x^3 + 0) x^2 - (2x)(x^4+1) / ((x^2)^2)=
(4x^5 - 2x^5 - 2x) / x^4 =
2x^5 - 2x / x^4 =
x(2x^4 - 2) / x^4 =
(2x^4 - 2) / x^3
Its Our correct answer, hope i could help you.
Good Luck.
2006-09-07 09:18:13 · answer #1 · answered by sweetie 5 · 1⤊ 0⤋
f(x) = (x^4+1) / x^2 = (x^4/x^2) + (1/x^2) = x^2 + x^-2
f'(x) = 2x - 2x^-3
2006-09-07 09:10:37 · answer #2 · answered by Kyrix 6 · 0⤊ 0⤋
I wouldn't use the quotient rule here. I would just simplify f(x) to x^2+x^-2 and use the power rule, giving you f'(x) = 2x-(2/x^3).
2006-09-07 09:10:07 · answer #3 · answered by sgp19 2 · 1⤊ 0⤋
use the quotient rule. d/dx(u/v) = u'v - uv' / v^2.
[x^3(x^2) - (x^4+1)x ] / x^4
[x^5 - (x^5 + x)] / x^4.
-x / x^4.
-1/x^3.
2006-09-07 09:03:00 · answer #4 · answered by the redcuber 6 · 0⤊ 0⤋
f(x) = (x^4 + 1)/(x^2)
f'(x) = (((x^2)(x^4 + 1)') - ((x^4 + 1)(x^2)'))/((x^2)^2)
f'(x) = (((x^2)(4x^3)) - ((x^4 + 1)(2x))/(x^4)
f'(x) = (4x^5 - (2x^4 + 2x))/(x^4)
f'(x) = (4x^5 - 2x^4 - 2x)/(x^4)
f'(x) = (x(4x^4 - 2x^3 - 2))/(x(x^3))
f'(x) = (4x^4 - 2x^3 - 2)/(x^3)
ANS : (4x^4 - 2x^3 - 2)/(x^3)
2006-09-07 14:00:22 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
f(x) = (x^4 +1) / x^2 . find f '(x) .
f(x) = (x^2 +x^-2
f '(x) = (2x -2x^-3.
.
2006-09-07 09:08:02 · answer #6 · answered by Amar Soni 7 · 0⤊ 0⤋
[2x(x^4+1)-x^2*4x^3]/x^4
(2x^5+2x-4x^5)/x^4
(-2x^5+2x)/x^4
-2(x^4-1)/x^3
2006-09-07 09:08:35 · answer #7 · answered by raj 7 · 0⤊ 0⤋
quotient rule
2006-09-07 09:00:11 · answer #8 · answered by jasonalwaysready 4 · 0⤊ 0⤋