# Chapter 3: Derivatives - Practice Exercises - Page 178: 57

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

#### 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}

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