Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 10

$$\frac{{dz}}{{dt}} = 2t + 4{t^{ - 5}}$$

$$\eqalign{ & {\text{we have }}z = {x^2}y - x{y^3},{\text{ with }}x = {t^2}{\text{ and }}y = {t^{ - 2}} \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^2}y - x{y^3}} \right] \cr & \frac{{\partial z}}{{\partial x}} = y\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] - {y^3}\frac{\partial }{{\partial x}}\left[ x \right] \cr & \frac{{\partial z}}{{\partial x}} = 2xy - {y^3} \cr & and \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^2}y - x{y^3}} \right] \cr & \frac{{\partial z}}{{\partial x}} = {x^2}\frac{\partial }{{\partial y}}\left[ y \right] - x\frac{\partial }{{\partial x}}\left[ {{y^3}} \right] \cr & \frac{{\partial z}}{{\partial x}} = {x^2} - 3x{y^2} \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[ {{t^{ - 2}}} \right] = - 2{t^{ - 3}} \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( {2xy - {y^3}} \right)\left( {2t} \right) + \left( {{x^2} - 3x{y^2}} \right)\left( { - 2{t^{ - 3}}} \right) \cr & {\text{replace }}x = {t^2}{\text{ and }}y = {t^{ - 2}} \cr & \frac{{dz}}{{dt}} = \left( {2\left( {{t^2}} \right)\left( {{t^{ - 2}}} \right) - {{\left( {{t^{ - 2}}} \right)}^3}} \right)\left( {2t} \right) + \left( {{{\left( {{t^2}} \right)}^2} - 3\left( {{t^2}} \right){{\left( {{t^{ - 2}}} \right)}^2}} \right)\left( { - 2{t^{ - 3}}} \right) \cr & {\text{simplifying}} \cr & \frac{{dz}}{{dt}} = \left( {2 - {t^{ - 6}}} \right)\left( {2t} \right) + \left( {{t^4} - 3{t^{ - 2}}} \right)\left( { - 2{t^{ - 3}}} \right) \cr & \frac{{dz}}{{dt}} = 4t - 2{t^{ - 5}} - 2t + 6{t^{ - 5}} \cr & \frac{{dz}}{{dt}} = 2t + 4{t^{ - 5}} \cr} $$