Answer
$160t^{5}(2t^{5}+3)^{3}+4(2t^{5}+3)^{4}$
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 $v^{\prime}(t)$, we need the chain rule.
$\left[\begin{array}{llll}
u(t)=4t & , & v(t)=(2t^{5}+3)^{4} & \\
u^{\prime}(t)=4 & & v^{\prime}(t)=[w(t)]^{4} & (\text{chain rule)})\\
& & v^{\prime}(t)=4\cdot[w(t)]^{3}\cdot w^{\prime}(t) & \\
& & v^{\prime}(t)=4(2t^{5}+3)^{3}\cdot(10t^{4}) &
\end{array}\right]$
$r^{\prime}(t)=(4t)[(2t^{5}+3)^{4}]^{\prime}+(2t^{5}+3)^{4}(4t)^{\prime}$
$=4t[4(2t^{5}+3)^{3}\cdot 10t^{4}]+(2t^{5}+3)^{4}\cdot 4$
$=160t^{5}(2t^{5}+3)^{3}+4(2t^{5}+3)^{4}$
