Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 23

Answer

$$\eqalign{ & {z_s} = \left( {t + 1} \right){e^{st + s + t}} \cr & {z_t} = \left( {s + 1} \right){e^{st + s + t}} \cr} $$

Work Step by Step

$$\eqalign{ & {\text{Let the functions }}z = {e^{x + y}},\,\,\,\,\,x = st{\text{ and }}y = s + t \cr & \cr & {\text{Calculate }}{z_s},{\text{ apply }}\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial s}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{x + y}}} \right] = {e^{x + y}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{e^{x + y}}} \right] = {e^{x + y}} \cr & \frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {st} \right] = t \cr & \frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {s + t} \right] = 1 \cr & {\text{Then}}{\text{,}} \cr & \frac{{\partial z}}{{\partial s}} = {e^{x + y}}\left( t \right) + {e^{x + y}}\left( 1 \right) \cr & \frac{{\partial z}}{{\partial s}} = t{e^{x + y}} + {e^{x + y}} \cr & \frac{{\partial z}}{{\partial s}} = \left( {t + 1} \right){e^{x + y}} \cr & {\text{Write in terms of }}s{\text{ and }}t \cr & \frac{{\partial z}}{{\partial s}} = \left( {t + 1} \right){e^{st + s + t}} \cr & \cr & {\text{Calculate }}{z_t},{\text{ apply }}\frac{{\partial z}}{{\partial t}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial t}} \cr & \frac{{\partial x}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {st} \right] = s \cr & \frac{{\partial y}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {s + t} \right] = 1 \cr & {\text{Then}}{\text{,}} \cr & \frac{{\partial z}}{{\partial t}} = {e^{x + y}}\left( s \right) + {e^{x + y}}\left( 1 \right) \cr & \frac{{\partial z}}{{\partial t}} = s{e^{x + y}} + {e^{x + y}} \cr & \frac{{\partial z}}{{\partial t}} = \left( {s + 1} \right){e^{x + y}} \cr & {\text{Write in terms of }}s{\text{ and }}t \cr & \frac{{\partial z}}{{\partial t}} = \left( {s + 1} \right){e^{st + s + t}} \cr} $$
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