Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 970: 12

Answer

${g_x}\left( {x,t} \right) = t{e^{xt}}{\text{ and }}{g_t}\left( {x,t} \right) = x{e^{xt}}$

Work Step by Step

$$\eqalign{ & g\left( {x,t} \right) = {e^{xt}} \cr & \cr & {\text{Differentiate with respect to }}x \cr & {g_x}\left( {x,t} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{xt}}} \right] \cr & {\text{Using the Chain Rule}} \cr & {g_x}\left( {x,t} \right) = {e^{xt}}\frac{\partial }{{\partial x}}\left[ {xt} \right] \cr & {\text{Considering }}t{\text{ as a constant}} \cr & {g_x}\left( {x,t} \right) = t{e^{xt}}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {g_x}\left( {x,t} \right) = t{e^{xt}}\left( 1 \right) \cr & {g_x}\left( {x,t} \right) = t{e^{xt}} \cr & \cr & {\text{Differentiate with respect to }}t \cr & {g_t}\left( {x,t} \right) = \frac{\partial }{{\partial t}}\left[ {{e^{xt}}} \right] \cr & {\text{Using the Chain Rule}} \cr & {g_t}\left( {x,t} \right) = {e^{xt}}\frac{\partial }{{\partial t}}\left[ {xt} \right] \cr & {\text{Considering }}x{\text{ as a constant}} \cr & {g_t}\left( {x,t} \right) = x{e^{xt}}\frac{\partial }{{\partial t}}\left[ t \right] \cr & {g_t}\left( {x,t} \right) = x{e^{xt}}\left( 1 \right) \cr & {g_t}\left( {x,t} \right) = x{e^{xt}}} $$
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