Answer
$\displaystyle \frac{dy}{dx}=x(x^{2}-1)^{3}(11x^{3} -3x+16)$
Work Step by Step
Use the product rule
$\color{blue}{[u(x)\cdot v(x)]^{\prime}=u(x)\cdot v^{\prime}(x)+v(x)\cdot u^{\prime}(x)}$
For $v^{\prime}(x)$, we need the chain rule.
$\left[\begin{array}{llll}
u(x)=x^{3}+2 & , & v(x)=(x^{2}-1)^{4} & \\
u^{\prime}(t)=3x^{2} & & v^{\prime}(x)=[w(x)]^{4} & \\
& & v^{\prime}(x)=4\cdot[w(x)]^{3}\cdot w^{\prime}(x) & (\text{chain rule)})\\
& & v^{\prime}(x)=4(x^{2}-1)^{3}\cdot 2x &
\end{array}\right]$
$\displaystyle \frac{dy}{dx}=u(x)\cdot v^{\prime}(x)+v(x)\cdot u^{\prime}(x)$
$=(x^{3}+2)[4(x^{2}-1)^{3}\cdot 2x]+(x^{2}-1)^{4}(3x^{2})$
$=8x(x^{3}+2)(x^{2}-1)^{3}+3x^{2}(x^{2}-1)^{4}$
$=x(x^{2}-1)^{3}[8(x^{3}+2)+3x(x^{2}-1)]$
$=x(x^{2}-1)^{3}(8x^{3}+16+3x^{3}-3x)$
$=x(x^{2}-1)^{3}(11x^{3} -3x+16)$
