Answer
$(x^{2}-3)(6x^{2}+1)+ 2x^{2}(2x^{2}+1)$
Work Step by Step
Following example 1b, a more general product rule was introduced:
$(f\cdot g\cdot h)^{\prime}=f^{\prime}gh+fg^{\prime}h+fgh^{\prime}$
$f(x)=x,\qquad f^{\prime}(x)=1$
$g(x)=x^{2}-3,\qquad g^{\prime}(x)=2x$
$h(x)=2x^{2}+1,\qquad h^{\prime}(x)=2(2x)=4x$
$\displaystyle \frac{dy}{dx}=(f\cdot g\cdot h)^{\prime}(x)=$
$=(1)(x^{2}-3)(2x^{2}+1)+ x(2x)(2x^{2}+1)+ x(x^{2}-3)(4x)$
$=(x^{2}-3)(2x^{2}+1)+ 2x^{2}(2x^{2}+1)+ 4x^{2}(x^{2}-3)$
$=(x^{2}-3)(2x^{2}+1+4x^{2})+ 2x^{2}(2x^{2}+1)$
$=(x^{2}-3)(6x^{2}+1)+ 2x^{2}(2x^{2}+1)$
... we do not need to expand the answer..
