*The quantity of (4x-1) is raised to the 1/2 power or squarerooted.
2007-12-30 09:09:36 · 4 answers · asked by x 1 in Science & Mathematics ➔ Mathematics
f (x) = (x)(4x - 1)^(1/2)
f `(x) = (4x - 1)^(1/2) + (1/2)(4x - 1)^(-1/2)(4x)
f `(x) = (4x - 1)^(1/2) + (2x) (4x - 1)^(- 1/2)
f `(x) = (4x - 1)^(-1/2)[ (4x - 1) + (2x) ]
f `(x) = (4x - 1)^(-1/2)[ (6x - 1) ]
f `(x) = (6x - 1) / (4x - 1)^(1/2)
2007-12-30 10:03:02 · answer #1 · answered by Como 7 · 1⤊ 0⤋
apply the product rule and chain rule
you get x(1/2((4x-1)^-1/2)(4)) + (4x-1)^1/2 (1)
=> 2x (4x-1)^-1/2) + (4x-1)^1/2
2007-12-30 09:17:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
use the product and the chain rule
f'(x) = 1(4x-1)^.5 + x[.5(4x-5)^(-.5)*4]
f'(x) = (4x-1)^5 + 2x(4x-5)^(-.5)
2007-12-30 09:14:51 · answer #3 · answered by Linda K 5 · 0⤊ 0⤋
(d/dx)[x(4x -- 1)^(1/2)]
= (4x -- 1)^(1/2)d(x)/dx + x[d/d(4x -- 1)](4x -- 1)^(1/2)*(d/dx)(4x -- 1)
= (4x -- 1)^(1/2) + x/2(4x -- 1)^(-- 1/2)*4
= (4x -- 1)^(1/2) [ 1 + 2x / (4x -- 1)]
= (4x -- 1)^(1/2) [ (6x -- 1) / (4x -- 1)]
2007-12-30 09:23:26 · answer #4 · answered by sv 7 · 0⤊ 0⤋