g(x)=(6x^2-8x)ln(x)
2007-07-02 00:37:31 · 5 answers · asked by latina_babygirl002 1 in Science & Mathematics ➔ Mathematics
Use theProduct Rule
(uv)' = u'v + uv'
g(x)=(6x^2-8x)ln(x)
g'(x) = (12x - 8)ln(x) + (6x^2 - 8x)/x
= (12x - 8)ln(x) + (6x - 8)
2007-07-02 00:42:22 · answer #1 · answered by gudspeling 7 · 1⤊ 0⤋
to find the derivative you apply the product rule
g'(x) = u' v + u v'
u = (6x^2 - 8x)
u'=12x - 8
v=ln(x)
v'=1/x
therefore
g'(x)= (12x - 8)ln(x) + (6x^2 - 8x)/x
= (12x - 8)ln(x) + (6x - 8)
2007-07-02 07:50:02 · answer #2 · answered by Anonymous · 0⤊ 0⤋
g(x) = ( 6x^2 - 8 x ) ln x
g ' (x) = ( 6x^2 - 8 x ) 1 / x + ln x ( 12 x - 8 )
2007-07-02 07:52:25 · answer #3 · answered by CPUcate 6 · 0⤊ 0⤋
g `(x) = (12x - 8).ln x + (1/x).(6x² - 8x)
g `(x) = 4.(3x - 2).ln x + (1/x).(2x).(3x - 4)
g `(x) = 4.(3x - 2).ln x + 2.(3x - 4)
g `(x) = 2.[ 2.(3x - 2).ln x + 3x - 4 ]
2007-07-06 02:48:42 · answer #4 · answered by Como 7 · 0⤊ 0⤋
g`(x)=1/x*(6x^2-8x) + (12x-8)lnx
2007-07-02 09:11:04 · answer #5 · answered by Oanna 2 · 0⤊ 0⤋