Answer
$$\frac{{dw}}{{dt}} = {\mkern 1mu} 4{t^5}\cos \left( {t + 1} \right) + 20{t^4}\sin \left( {t + 1} \right)$$
Work Step by Step
$$\eqalign{
& w = xy\sin z,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x = {t^2},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} y = 4{t^3},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} z = t + 1 \cr
& {\text{Calculate the partial derivatives }}\frac{{\partial w}}{{\partial x}},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{\partial w}}{{\partial y}},{\mkern 1mu} {\mkern 1mu} \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 x}}\left[ {xy\sin z} \right] = x\sin z \cr
& \frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left[ {xy\sin z} \right] = xy\cos z \cr
& {\text{Calculate the derivatives }}\frac{{dx}}{{dt}},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{dy}}{{dt}},{\mkern 1mu} {\mkern 1mu} \frac{{dz}}{{dt}} \cr
& \frac{{dx}}{{dt}} = 2t,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{dy}}{{dt}} = 12{t^2},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{dz}}{{dt}} = 1 \cr
& {\text{Use the chain rule }}\frac{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + {\mkern 1mu} {\mkern 1mu} \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + {\mkern 1mu} {\mkern 1mu} \frac{{\partial w}}{{\partial z}}\frac{{dz}}{{dt}} \cr
& \frac{{dw}}{{dt}} = y\sin z\left( {2t} \right) + {\mkern 1mu} {\mkern 1mu} x\sin z\left( {12{t^2}} \right) + {\mkern 1mu} {\mkern 1mu} xy\cos z\left( 1 \right) \cr
& \frac{{dw}}{{dt}} = 2ty\sin z + {\mkern 1mu} {\mkern 1mu} 12x{t^2}\sin z + {\mkern 1mu} {\mkern 1mu} xy\cos z \cr
& {\text{Write in terms of }}t \cr
& \frac{{dw}}{{dt}} = 2t\left( {4{t^3}} \right)\sin \left( {t + 1} \right) + {\mkern 1mu} {\mkern 1mu} 12\left( {{t^2}} \right){t^2}\sin \left( {t + 1} \right) + {\mkern 1mu} {\mkern 1mu} \left( {{t^2}} \right)\left( {4{t^3}} \right)\cos \left( {t + 1} \right) \cr
& \frac{{dw}}{{dt}} = 8{t^4}\sin \left( {t + 1} \right) + {\mkern 1mu} 12{t^4}\sin \left( {t + 1} \right) + {\mkern 1mu} 4{t^5}\cos \left( {t + 1} \right) \cr
& \frac{{dw}}{{dt}} = {\mkern 1mu} 4{t^5}\cos \left( {t + 1} \right) + 20{t^4}\sin \left( {t + 1} \right) \cr} $$
