Calculus: Early Transcendentals (2nd Edition)

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

Chapter 12 - Functions of Several Veriables - 12.4 Partial Derivatives - 12.4 Exercises - Page 904: 37

Answer

$$\eqalign{ & {F_{rr}}\left( {r,s} \right) = 0,\,\,\,\,\,\,\,\,{\text{ }}{F_{ss}}\left( {r,s} \right) = r{e^s} \cr & {F_{sr}}\left( {r,s} \right) = {e^s}{\text{ and }}{F_{rs}}\left( {r,s} \right) = {e^s} \cr} $$

Work Step by Step

$$\eqalign{ & F\left( {r,s} \right) = r{e^s} \cr & {\text{Find the first partial derivatives }}{F_r}\left( {r,s} \right){\text{ and }}{F_s}\left( {r,s} \right){\text{ then}} \cr & {F_r}\left( {r,s} \right) = \frac{\partial }{{\partial r}}\left[ {r{e^s}} \right] \cr & {\text{treat }}s{\text{ as a constant}}{\text{, then}} \cr & {F_r}\left( {r,s} \right) = {e^s}\frac{\partial }{{\partial r}}\left[ r \right] \cr & {F_r}\left( {r,s} \right) = {e^s} \cr & and \cr & {F_s}\left( {r,s} \right) = \frac{\partial }{{\partial s}}\left[ {r{e^s}} \right] \cr & {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr & {F_s}\left( {r,s} \right) = r{e^s} \cr & \cr & {\text{Find the second partial derivatives }}{F_{rr}}\left( {r,s} \right){\text{ and }}{F_{ss}}\left( {r,s} \right){\text{ then}} \cr & {F_{rr}}\left( {r,s} \right) = \frac{\partial }{{\partial r}}\left[ {{e^s}} \right] \cr & {F_{rr}}\left( {r,s} \right) = 0 \cr & and \cr & {\text{ }}{F_{ss}}\left( {r,s} \right) = \frac{\partial }{{\partial s}}\left[ {r{e^s}} \right] \cr & {\text{ }}{F_{ss}}\left( {r,s} \right) = r{e^s} \cr & \cr & {\text{Find the second partial derivatives }}{F_{rs}}\left( {x,y} \right){\text{ and }}{F_{sr}}\left( {x,y} \right){\text{ then}} \cr & {F_{rs}}\left( {r,s} \right) = \frac{\partial }{{\partial s}}\left[ {{e^s}} \right] \cr & {F_{rs}}\left( {r,s} \right) = {e^s} \cr & and \cr & {F_{sr}}\left( {r,s} \right) = \frac{\partial }{{\partial r}}\left[ {r{e^s}} \right] \cr & {F_{sr}}\left( {r,s} \right) = {e^s} \cr} $$
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