Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 19

$$\eqalign{ & {z_s} = 2\left( {s - t} \right)\sin {t^2} \cr & {z_t} = 2\left( {s - t} \right)\left( {t\left( {s - t} \right)\cos {t^2} - \sin {t^2}} \right) \cr} $$

$$\eqalign{ & {\text{Let the functions }}z = {x^2}\sin y,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x = s - t{\text{ and }}y = {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[ {{x^2}\sin y} \right] = 2x\sin y \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^2}\sin y} \right] = {x^2}\cos y \cr & \frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {s - t} \right] = 1 \cr & \frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {{t^2}} \right] = 0 \cr & {\text{Then,}} \cr & \frac{{\partial z}}{{\partial s}} = \left( {2x\sin y} \right)\left( 1 \right) + \left( {{x^2}\cos y} \right)\left( 0 \right) \cr & \frac{{\partial z}}{{\partial s}} = 2x\sin y \cr & {\text{Write in terms of }}s{\text{ and }}t \cr & \frac{{\partial z}}{{\partial s}} = 2\left( {s - t} \right)\sin {t^2} \cr & or \cr & {z_s} = 2\left( {s - t} \right)\sin {t^2} \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 - t} \right] = - 1 \cr & \frac{{\partial y}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {{t^2}} \right] = 2t \cr & {\text{Then,}} \cr & \frac{{\partial z}}{{\partial t}} = \left( {2x\sin y} \right)\left( { - 1} \right) + \left( {{x^2}\cos y} \right)\left( {2t} \right) \cr & \frac{{\partial z}}{{\partial t}} = - 2x\sin y + 2t{x^2}\cos y \cr & \frac{{\partial z}}{{\partial t}} = 2x\left( {tx\cos y - \sin y} \right) \cr & {\text{Write in terms of }}s{\text{ and }}t \cr & \frac{{\partial z}}{{\partial t}} = 2\left( {s - t} \right)\left( {t\left( {s - t} \right)\cos {t^2} - \sin {t^2}} \right) \cr & or \cr & {z_t} = 2\left( {s - t} \right)\left( {t\left( {s - t} \right)\cos {t^2} - \sin {t^2}} \right) \cr} $$