Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 20

\[{z_s} = 6s\cos \left( {3{s^2} - {t^2}} \right){\text{ and }}{z_t} = 6s\cos \left( {3{s^2} - {t^2}} \right)\]

$$\eqalign{ & {\text{Let the functions }}z = \sin \left( {2x + y} \right),{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x = {s^2} - {t^2}{\text{ and }}y = {s^2} + {t^2} \cr & {\text{Calculate }}{z_s},{\text{ apply }}\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial s}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\sin \left( {2x + y} \right)} \right] = 2\cos \left( {2x + y} \right) \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\sin \left( {2x + y} \right)} \right] = \cos \left( {2x + y} \right) \cr & \frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {{s^2} - {t^2}} \right] = 2s \cr & \frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {{s^2} + {t^2}} \right] = 2s \cr & {\text{Then,}} \cr & \frac{{\partial z}}{{\partial s}} = 2\cos \left( {2x + y} \right)\left( {2s} \right) + \cos \left( {2x + y} \right)\left( {2s} \right) \cr & \frac{{\partial z}}{{\partial s}} = 6s\cos \left( {2x + y} \right) \cr & {\text{Write in terms of }}s{\text{ and }}t \cr & \frac{{\partial z}}{{\partial s}} = 6s\cos \left( {2\left( {{s^2} - {t^2}} \right) + {s^2} + {t^2}} \right) \cr & \frac{{\partial z}}{{\partial s}} = 6s\cos \left( {2{s^2} - 2{t^2} + {s^2} + {t^2}} \right) \cr & \frac{{\partial z}}{{\partial s}} = 6s\cos \left( {3{s^2} - {t^2}} \right) \cr & or \cr & {z_s} = 6s\cos \left( {3{s^2} - {t^2}} \right) \cr & {\text{Calculate }}{z_t},{\text{ apply }}\frac{{\partial z}}{{\partial t}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial t}} \cr & \frac{{\partial x}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {{s^2} - {t^2}} \right] = - 2t \cr & \frac{{\partial y}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {{s^2} + {t^2}} \right] = 2t \cr & {\text{Then,}} \cr & \frac{{\partial z}}{{\partial t}} = 2\cos \left( {2x + y} \right)\left( { - 2t} \right) + \cos \left( {2x + y} \right)\left( {2t} \right) \cr & \frac{{\partial z}}{{\partial t}} = - 4t\cos \left( {2x + y} \right) + 2t\cos \left( {2x + y} \right) \cr & \frac{{\partial z}}{{\partial t}} = - 2t\cos \left( {2x + y} \right) \cr & {\text{Write in terms of }}s{\text{ and }}t \cr & \frac{{\partial z}}{{\partial t}} = - 2t\cos \left( {2{s^2} - 2{t^2} + {s^2} + {t^2}} \right) \cr & \frac{{\partial z}}{{\partial t}} = - 2t\cos \left( {3{s^2} - {t^2}} \right) \cr & {z_t} = 6s\cos \left( {3{s^2} - {t^2}} \right) \cr} $$