Answer
$\displaystyle \frac{dy}{dx}=3x^{2}(3x^{4}+1)^{3}(19x^{4}+64x+1)$
Work Step by Step
Use the product rule
$\color{blue}{[u(t)\cdot v(t)]^{\prime}=u(t)\cdot v^{\prime}(t)+v(t)\cdot u^{\prime}(t)}$
For $u^{\prime}(x)$, we need the chain rule.
$\left[\begin{array}{lllll}
u(x)=(3x^{4}+1)^{4} & , & & v(t)=3x^{4}+1 & \\
u(x)=[w(x)]^{4} & & & v^{\prime}(t)=12x^{3} & \\
u^{\prime}(x)=4\cdot[w(x)]^{3}\cdot w^{\prime}(x) & ,(\text{chain rule)}) & & & \\
u^{\prime}(x)=(3x^{4}+1)^{3}\cdot 12x^{3} & & & &
\end{array}\right]$
$\displaystyle \frac{dy}{dx}=(3x^{4}+1)^{4}[x^{3}+4]^{\prime}+(x^{3}+4)[(3x^{4}+1)^{4}]^{\prime}$
$=(3x^{4}+1)^{4}(3x^{2})+(x^{3}+4)[4(3x^{4}+1)^{3}\cdot 12x^{3}]$
$=3x^{2}(3x^{4}+1)^{4}+48x^{3}(x^{3}+4)(3x^{4}+1)^{3}$
$=3x^{2}(3x^{4}+1)^{3}[3x^{4}+1+16x(x^{3}+4)]$
$=3x^{2}(3x^{4}+1)^{3}(3x^{4}+1+16x^{4}+64)$
$=3x^{2}(3x^{4}+1)^{3}(19x^{4}+64x+1)$
