Answer
$h'(x)=6x(x^2+3)^2$
Work Step by Step
$h(x)=(x^2+3)^3=(x^2+3)^2(x^2+3)=(x^4+6x^2+9)(x^2+3)$
Product Rule $h'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x^4+6x^2+9 ;u’(x)=4x^3+12x $
$v(x)=x^2+3 ;v’(x)= 2x$
$h'(x)=(4x^3+12x)(x^2+3)+(2x)(x^4+6x^2+9)$
$=2x^5+12x^3+18x+4x^5+12x^3+12x+36x$
$=6x^5+36x^3+54x=6x(x^4+6x^2+9)$
$=(6x)(x^2+3)^2$
