Answer
$$\frac{{dz}}{{dt}} = 2t\sin \left( {4{t^3}} \right) + 12{t^4}\cos \left( {4{t^3}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{we have }}z = x\sin y,{\text{ with }}x = {t^2}{\text{ and }}y = 4{t^3} \cr
& {\text{find }}\frac{{dz}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr
& \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& {\text{find the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {x\sin y} \right] = \sin y \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {x\sin y} \right] = x\cos y \cr
& {\text{find the partial derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{t^2}} \right] = 2t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {4{t^3}} \right] = 12{t^2} \cr
& \cr
& {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr
& \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr
& \frac{{dz}}{{dt}} = \left( {\sin y} \right)\left( {2t} \right) + \left( {x\cos y} \right)\left( {12{t^2}} \right) \cr
& \frac{{dz}}{{dt}} = 2t\sin y + 12x{t^2}\cos y \cr
& {\text{replace }}x = {t^2}{\text{ and }}y = 4{t^3} \cr
& \frac{{dz}}{{dt}} = 2t\sin \left( {4{t^3}} \right) + 12\left( {{t^2}} \right){t^2}\cos \left( {4{t^3}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dz}}{{dt}} = 2t\sin \left( {4{t^3}} \right) + 12{t^4}\cos \left( {4{t^3}} \right) \cr} $$
