Answer
$$\,\,\,\,\,\frac{{dw}}{{dt}} = 165{t^{32}}$$
Work Step by Step
$$\eqalign{
& w = 5{x^2}{y^3}{z^4};\,\,\,\,\,x = {t^2},\,\,\,y = {t^3},\,\,\,z = {t^5} \cr
& {\text{Calculate 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( {5{x^2}{y^3}{z^4}} \right) \cr
& \,\,\,\,{\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {5{x^2}{y^3}{z^4}} \right) \cr
& \,\,\,\,\frac{{\partial w}}{{\partial x}} = 10x{y^3}{z^4} \cr
& \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {5{x^2}{y^3}{z^4}} \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ and }}z{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = 5{x^2}{z^4}\left( {3{y^2}} \right) \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = 15{x^2}{y^2}{z^4} \cr
& \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left( {5{x^2}{y^3}{z^4}} \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ and }}y{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = 5{x^2}{y^3}\left( {4{z^3}} \right) \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = 20{x^2}{y^3}{z^3} \cr
& \cr
& {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{,}}\,\,{\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( {{t^3}} \right) = 3{t^2} \cr
& \,\,\,\,\frac{{dz}}{{dt}} = \frac{d}{{dt}}\left( {{t^5}} \right) = 5{t^4} \cr
& \cr
& {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr
& \,\,\,\,\frac{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dz}}{{dt}} \cr
& \,\,\,\,\,{\text{Substitute the derivatives}} \cr
& \,\,\,\,\frac{{dw}}{{dt}} = \left( {10x{y^3}{z^4}} \right)\left( {2t} \right) + \left( {15{x^2}{y^2}{z^4}} \right)\left( {3{t^2}} \right) + \left( {20{x^2}{y^3}{z^3}} \right)\left( {5{t^4}} \right) \cr
& \,\,\,\,{\text{Where }}x = {t^2},\,\,\,y = {t^3},\,\,\,z = {t^5} \cr
& \,\,\,\,\frac{{dw}}{{dt}} = \left( {10\left( {{t^2}} \right){{\left( {{t^3}} \right)}^3}{{\left( {{t^5}} \right)}^4}} \right)\left( {2t} \right) + \left( {15{{\left( {{t^2}} \right)}^2}{{\left( {{t^3}} \right)}^2}{{\left( {{t^5}} \right)}^4}} \right)\left( {3{t^2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {20{{\left( {{t^2}} \right)}^2}{{\left( {{t^3}} \right)}^3}{{\left( {{t^5}} \right)}^3}} \right)\left( {5{t^4}} \right) \cr
& \,\,\,\,\,{\text{simplifying}} \cr
& \,\,\,\,\,\frac{{dw}}{{dt}} = 20{t^{4 + 9 + 15 + 4}} + 45{t^{4 + 6 + 20 + 2}} + 100{t^{4 + 9 + 15 + 4}} \cr
& \,\,\,\,\,\frac{{dw}}{{dt}} = 165{t^{32}} \cr} $$
