Calculus: Early Transcendentals (2nd Edition)

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

Chapter 12 - Functions of Several Veriables - Review Exercises - Page 960: 40

Answer

$$\eqalign{ & {g_x}\left( {x,y,z} \right) = 4y{z^2} - \frac{3}{y} \cr & {g_y}\left( {x,y,z} \right) = 4x{z^2} + \frac{{3x}}{{{y^2}}} \cr & {g_z}\left( {x,y,z} \right) = 8xyz \cr} $$

Work Step by Step

$$\eqalign{ & g\left( {x,y,z} \right) = 4xy{z^2} - \frac{{3x}}{y} \cr & g\left( {x,y,z} \right) = 4xy{z^2} - 3x{y^{ - 1}} \cr & {\text{Find the partial derivatives }}{g_x}\left( {x,y,z} \right){\text{,}}\,\,{g_y}\left( {x,y,z} \right){\text{ and }}{g_z}\left( {x,y,z} \right){\text{ then}} \cr & {g_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {4xy{z^2} - 3x{y^{ - 1}}} \right] \cr & {\text{treat }}y{\text{ and }}z{\text{ as a constant}} \cr & {g_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {4xy{z^2}} \right] - \frac{\partial }{{\partial x}}\left[ {3x{y^{ - 1}}} \right] \cr & {g_x}\left( {x,y,z} \right) = 4y{z^2}\frac{\partial }{{\partial x}}\left[ x \right] - 3{y^{ - 1}}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {g_x}\left( {x,y,z} \right) = 4y{z^2} - 3{y^{ - 1}} \cr & {g_x}\left( {x,y,z} \right) = 4y{z^2} - \frac{3}{y} \cr & \cr & {g_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {4xy{z^2} - 3x{y^{ - 1}}} \right] \cr & {\text{treat }}x{\text{ and }}z{\text{ as a constant}} \cr & {g_y}\left( {x,y,z} \right) = 4x{z^2}\frac{\partial }{{\partial y}}\left[ y \right] - 3x\frac{\partial }{{\partial y}}\left[ {{y^{ - 1}}} \right] \cr & {g_y}\left( {x,y,z} \right) = 4x{z^2}\left( 1 \right) - 3x\left( { - {y^{ - 2}}} \right) \cr & {g_y}\left( {x,y,z} \right) = 4x{z^2} + 3x{y^{ - 2}} \cr & {g_y}\left( {x,y,z} \right) = 4x{z^2} + \frac{{3x}}{{{y^2}}} \cr & \cr & and \cr & \cr & {g_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {4xy{z^2} - 3x{y^{ - 1}}} \right] \cr & {\text{treat }}y{\text{ and }}x{\text{ as a constant}} \cr & {g_z}\left( {x,y,z} \right) = 4xy\frac{\partial }{{\partial z}}\left[ {{z^2}} \right] + 3x{y^{ - 1}}\frac{\partial }{{\partial z}}\left[ 1 \right] \cr & {g_z}\left( {x,y,z} \right) = 4xy\left( {2z} \right) \cr & {g_z}\left( {x,y,z} \right) = 8xyz \cr} $$
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