Answer
$$\frac{{dw}}{{dt}} = 20{t^4}\sin \left( {t + 1} \right) + 4{t^5}\cos \left( {t + 1} \right)$$
Work Step by Step
$$\eqalign{
& {\text{we have }}w = xy\sin z,{\text{ with }}x = {t^2}{\text{, }}y = 4{t^3}{\text{ and }}z = t + 1 \cr
& {\text{find }}\frac{{dw}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr
& \frac{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial w}}{{\partial z}}\frac{{dz}}{{dt}} \cr
& {\text{find the partial derivatives }}\frac{{\partial w}}{{\partial x}}{\text{,}}\frac{{\partial w}}{{\partial y}}{\text{ and }}\frac{{\partial w}}{{\partial z}} \cr
& \frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {xy\sin z} \right] = y\sin z \cr
& \frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {xy\sin z} \right] = x\sin z \cr
& and \cr
& \frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left[ {xy\sin z} \right] = xy\cos z \cr
& \cr
& {\text{find the partial derivatives }}\frac{{dx}}{{dt}}{\text{,}}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dz}}{{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
& \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {t + 1} \right] = 1 \cr
& \cr
& {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr
& \frac{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial w}}{{\partial z}}\frac{{dz}}{{dt}} \cr
& \frac{{dw}}{{dt}} = \left( {y\sin z} \right)\left( {2t} \right) + \left( {x\sin z} \right)\left( {12{t^2}} \right) + \left( {xy\cos z} \right)\left( 1 \right) \cr
& \frac{{dw}}{{dt}} = 2ty\sin z + 12{t^2}x\sin z + xy\cos z \cr
& {\text{replace }}x = {t^2}{\text{, }}y = 4{t^3}{\text{ and }}z = t + 1 \cr
& \frac{{dw}}{{dt}} = 2t\left( {4{t^3}} \right)\sin \left( {t + 1} \right) + 12{t^2}\left( {{t^2}} \right)\sin \left( {t + 1} \right) + \left( {{t^2}} \right)\left( {4{t^3}} \right)\cos \left( {t + 1} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dw}}{{dt}} = 8{t^4}\sin \left( {t + 1} \right) + 12{t^4}\sin \left( {t + 1} \right) + 4{t^5}\cos \left( {t + 1} \right) \cr
& \frac{{dw}}{{dt}} = 20{t^4}\sin \left( {t + 1} \right) + 4{t^5}\cos \left( {t + 1} \right) \cr} $$
