#### Answer

$${\left. {\frac{{dy}}{{dt}}} \right|_{t = 0}} = 0$$

#### Work Step by Step

$$\eqalign{
& y = 3\sin 2x{\text{ and }}x = {t^2} + \pi \cr
& {\text{then substituting }}{t^2} + \pi {\text{ for }}y \cr
& y = 3\sin 2\left( {{t^2} + \pi } \right) \cr
& y = 3\sin \left( {2{t^2} + 2\pi } \right) \cr
& {\text{differentiate both sides with respect to }}t \cr
& \frac{{dy}}{{dt}} = 3\frac{d}{{dt}}\left( {\sin \left( {2{t^2} + 2\pi } \right)} \right) \cr
& \frac{{dy}}{{dt}} = 3\cos \left( {2{t^2} + 2\pi } \right)\frac{d}{{dt}}\left( {2{t^2} + 2\pi } \right) \cr
& \frac{{dy}}{{dt}} = 3\cos \left( {2{t^2} + 2\pi } \right)\left( {4t} \right) \cr
& \frac{{dy}}{{dt}} = 12t\cos \left( {2{t^2} + 2\pi } \right) \cr
& {\text{Evaluate at }}t = 0 \cr
& {\left. {\frac{{dy}}{{dt}}} \right|_{t = 0}} = 12\left( 0 \right)\cos \left( {2{{\left( 0 \right)}^2} + 2\pi } \right) \cr
& {\left. {\frac{{dy}}{{dt}}} \right|_{t = 0}} = 0 \cr} $$