Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 29

Answer

$$\eqalign{ & {r_{xx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}},\,\,\,\,\,\,{r_{yy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr & {r_{xy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}},\,\,\,\,\,\,{r_{yx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr} $$

Work Step by Step

$$\eqalign{ & r = \ln \left| {x + y} \right| \cr & {\text{Find }}{r_x}{\text{ and }}{r_y} \cr & {r_x} = \frac{\partial }{{\partial x}}\left[ {\ln \left| {x + y} \right|} \right] \cr & {\text{treat }}y{\text{ as a constant and use }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr & {r_x} = \frac{1}{{x + y}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr & {\text{then}} \cr & {r_x} = \frac{1}{{x + y}}\left( 1 \right) \cr & {r_x} = \frac{1}{{x + y}} \cr & \cr & {r_y} = \frac{\partial }{{\partial y}}\left[ {\ln \left| {x + y} \right|} \right] \cr & {\text{treat }}x{\text{ as a constant and use }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr & {r_y} = \frac{1}{{x + y}}\frac{\partial }{{\partial y}}\left[ {x + y} \right] \cr & {\text{then}} \cr & {r_y} = \frac{1}{{x + y}}\left( 1 \right) \cr & {r_y} = \frac{1}{{x + y}} \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {r_{xy}} = \frac{\partial }{{\partial y}}\left[ {\frac{1}{{x + y}}} \right] \cr & {r_{xy}} = \frac{\partial }{{\partial y}}\left[ {{{\left( {x + y} \right)}^{ - 1}}} \right] \cr & {r_{xy}} = - {\left( {x + y} \right)^{ - 2}}\frac{\partial }{{\partial y}}\left[ {x + y} \right] \cr & {r_{xy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr & and \cr & {r_{yx}} = \frac{\partial }{{\partial x}}\left[ {\frac{1}{{x + y}}} \right] \cr & {r_{yx}} = \frac{\partial }{{\partial x}}\left[ {{{\left( {x + y} \right)}^{ - 1}}} \right] \cr & {r_{yx}} = - {\left( {x + y} \right)^{ - 2}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr & {r_{yx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr & \cr & {r_{xx}} = \frac{\partial }{{\partial x}}\left( {\frac{1}{{x + y}}} \right) \cr & {r_{xx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr & and \cr & {r_{yy}} = \frac{\partial }{{\partial x}}\left( {\frac{1}{{x + y}}} \right) \cr & {r_{yy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr} $$
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