Answer
$g'(x)=\dfrac{6(3x^2-2)^2(3x^2+9x+2)}{(2x+3)^4}$
Work Step by Step
Using the quotient rule: $g’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(3x^2-2)^3; u'(x)=(3)(6x)(3x^2-2)^{3-1}=18x(3x^2-2)^2.$
$v(x)=(2x+3)^3; v'(x)=(3)(2)(2x+3)^{3-1}=6(2x+3)^2.$
$g'(x)=\dfrac{(18x(3x^2-2)^2)((2x+3)^3)-(6(2x+3)^2)((3x^2-2)^3)}{((2x+3)^3)^2}$
$=\dfrac{6(3x^2-2)^2(3x^2+9x+2)}{(2x+3)^4}$
