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: 39

Answer

$$\eqalign{ & {f_x}\left( {x,y,z} \right) = \frac{6}{{4z + 5}}{\text{,}}\,\,\,\,{f_y}\left( {x,y,z} \right) = - \frac{5}{{4z + 5}} \cr & {f_z}\left( {x,y,z} \right) = \frac{{4\left( {5y - 6x} \right)}}{{{{\left( {4z + 5} \right)}^2}}},\,\,\,\,{\text{ }}{f_{yz}}\left( {x,y,z} \right) = \frac{{20}}{{{{\left( {4z + 5} \right)}^2}}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( {x,y,z} \right) = \frac{{6x - 5y}}{{4z + 5}} \cr & {\text{Find }}{f_x}\left( {x,y,z} \right){\text{ treat }}y{\text{ and }}z{\text{ as constants}} \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{6x - 5y}}{{4z + 5}}} \right] \cr & {f_x}\left( {x,y,z} \right) = \frac{1}{{4z + 5}}\frac{\partial }{{\partial x}}\left[ {6x - 5y} \right] \cr & {\text{then}} \cr & {f_x}\left( {x,y,z} \right) = \frac{1}{{4z + 5}}\left( 6 \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{6}{{4z + 5}} \cr & \cr & {\text{Find }}{f_y}\left( {x,y,z} \right){\text{ treat }}x{\text{ and }}z{\text{ as constants}} \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{6x - 5y}}{{4z + 5}}} \right] \cr & {f_y}\left( {x,y,z} \right) = \frac{1}{{4z + 5}}\frac{\partial }{{\partial y}}\left[ {6x - 5y} \right] \cr & {\text{then}} \cr & {f_y}\left( {x,y,z} \right) = \frac{1}{{4z + 5}}\left( { - 5} \right) \cr & {f_y}\left( {x,y,z} \right) = - \frac{5}{{4z + 5}} \cr & \cr & {\text{Find }}{f_z}\left( {x,y,z} \right){\text{ treat }}x{\text{ and }}y{\text{ as constants}} \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {\frac{{6x - 5y}}{{4z + 5}}} \right] \cr & {f_z}\left( {x,y,z} \right) = \left( {6x - 5y} \right)\frac{\partial }{{\partial y}}\left[ {\frac{1}{{4z + 5}}} \right] \cr & {\text{then}} \cr & {f_z}\left( {x,y,z} \right) = \left( {6x - 5y} \right)\left( { - \frac{4}{{{{\left( {4z + 5} \right)}^2}}}} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{{4\left( {5y - 6x} \right)}}{{{{\left( {4z + 5} \right)}^2}}} \cr & \cr & {\text{Find }}{f_{yz}}\left( {x,y,z} \right){\text{ }} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ { - \frac{5}{{4z + 5}}} \right] \cr & {\text{then}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = - \frac{{ - 5\left( 4 \right)}}{{{{\left( {4z + 5} \right)}^2}}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = \frac{{20}}{{{{\left( {4z + 5} \right)}^2}}} \cr} $$
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