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: 46

Answer

$$\eqalign{ & {g_x}\left( {x,y,z} \right) = 4xy - 3{z^4} \cr & {g_y}\left( {x,y,z} \right) = 2{x^2} + 20y{x^2} \cr & {g_z}\left( {x,y,z} \right) = - 12x{z^3} + 20{y^2}z \cr} $$

Work Step by Step

$$\eqalign{ & g\left( {x,y,z} \right) = 2{x^2}y - 3x{z^4} + 10{y^2}{z^2} \cr & {\text{Find the first partial derivative }}{g_x}\left( {x,y,z} \right) \cr & {g_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {2{x^2}y - 3x{z^4} + 10{y^2}{z^2}} \right] \cr & {\text{treat }}y{\text{ and }}z{\text{ as a constant}}{\text{, then }} \cr & {g_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {2{x^2}y} \right] - \frac{\partial }{{\partial x}}\left[ {3x{z^4}} \right] + \frac{\partial }{{\partial x}}\left[ {10{y^2}{z^2}} \right] \cr & {g_x}\left( {x,y,z} \right) = 2y\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] - 3{z^4}\frac{\partial }{{\partial x}}\left[ x \right] + 10{y^2}{z^2}\frac{\partial }{{\partial x}}\left[ 1 \right] \cr & {g_x}\left( {x,y,z} \right) = 4xy - 3{z^4} \cr & \cr & {\text{Find the first partial derivative }}{g_y}\left( {x,y,z} \right) \cr & {g_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {2{x^2}y - 3x{z^4} + 10{y^2}{z^2}} \right] \cr & {\text{treat }}x{\text{ and }}z{\text{ as a constant}}{\text{, then }} \cr & {g_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {2{x^2}y} \right] - \frac{\partial }{{\partial y}}\left[ {3x{z^4}} \right] + \frac{\partial }{{\partial y}}\left[ {10{y^2}{z^2}} \right] \cr & {g_y}\left( {x,y,z} \right) = 2{x^2}\frac{\partial }{{\partial y}}\left[ y \right] - 3{z^4}\frac{\partial }{{\partial y}}\left[ x \right] + 10{z^2}\frac{\partial }{{\partial y}}\left[ {{y^2}} \right] \cr & {g_y}\left( {x,y,z} \right) = 2{x^2} + 20y{x^2} \cr & \cr & {\text{Find the first partial derivative }}{g_z}\left( {x,y,z} \right) \cr & {g_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {2{x^2}y - 3x{z^4} + 10{y^2}{z^2}} \right] \cr & {\text{treat }}x{\text{ and }}y{\text{ as a constant}}{\text{, then }} \cr & {g_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {2{x^2}y} \right] - \frac{\partial }{{\partial z}}\left[ {3x{z^4}} \right] + \frac{\partial }{{\partial z}}\left[ {10{y^2}{z^2}} \right] \cr & {g_z}\left( {x,y,z} \right) = 2{x^2}y\frac{\partial }{{\partial z}}\left[ 1 \right] - 3x\frac{\partial }{{\partial z}}\left[ {{z^4}} \right] + 10{y^2}\frac{\partial }{{\partial z}}\left[ {{z^2}} \right] \cr & {g_z}\left( {x,y,z} \right) = 2{x^2}y\left( 0 \right) - 3x\left( {4{z^3}} \right) + 10{y^2}\left( {2z} \right) \cr & {g_z}\left( {x,y,z} \right) = - 12x{z^3} + 20{y^2}z \cr} $$
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