Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 956: 1

$$\,\,\,\,\frac{{dz}}{{dt}} = 42{t^{13}}$$

$$\eqalign{ & z = 3{x^2}{y^3};\,\,\,\,\,\,\,\,\,x = {t^4},\,\,\,\,\,\,\,y = {t^2} \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial y}}{{\partial x}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {3{x^2}{y^3}} \right) \cr & \,\,\,\,{\text{Treat }}y{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = 3{y^3}\frac{\partial }{{\partial x}}\left( {{x^2}} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = 3{y^3}\left( {2x} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = 6x{y^3} \cr & and \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {3{x^2}{y^3}} \right) \cr & \,\,\,\,{\text{Treat }}x{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = 3{x^2}\frac{\partial }{{\partial y}}\left( {{y^3}} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = 3{x^2}\left( {3{y^2}} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = 9{x^2}{y^2} \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {{t^4}} \right) = 4{t^3} \cr & \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^2}} \right) = 2t \cr & \cr & {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr & \,\,\,\,\,{\text{Substitute the derivatives}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {6x{y^3}} \right)\left( {4{t^3}} \right) + \left( {9{x^2}{y^2}} \right)\left( {2t} \right) \cr & \,\,\,\,{\text{Where }}x = {t^4}{\text{ and}}\,\,\,\,\,y = {t^2} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {6\left( {{t^4}} \right){{\left( {{t^2}} \right)}^3}} \right)\left( {4{t^3}} \right) + \left( {9{{\left( {{t^4}} \right)}^2}{{\left( {{t^2}} \right)}^2}} \right)\left( {2t} \right) \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {6{t^{10}}} \right)\left( {4{t^3}} \right) + \left( {9{t^{12}}} \right)\left( {2t} \right) \cr & \,\,\,\,\frac{{dz}}{{dt}} = 24{t^{13}} + 18{t^{13}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = 42{t^{13}} \cr} $$