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: 8

Answer

$${f_x}\left( {x,y} \right) = 12{e^{3x + 2y}},\,\,\,\,{f_y}\left( {x,y} \right) = 8{e^{3x + 2y}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = 12{e^4}{\text{ and }}{f_y}\left( { - 4,3} \right) = 8{e^{ - 6}}$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = 4{e^{3x + 2y}} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{e^{3x + 2y}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = 4{e^{3x + 2y}}\frac{\partial }{{\partial x}}\left[ {3x + 2y} \right] \cr & {f_x}\left( {x,y} \right) = 4{e^{3x + 2y}}\left( 3 \right) \cr & {f_x}\left( {x,y} \right) = 12{e^{3x + 2y}} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = 12{e^{3\left( 2 \right) + 2\left( { - 1} \right)}} \cr & {f_x}\left( {2, - 1} \right) = 12{e^{6 - 2}} \cr & {f_x}\left( {2, - 1} \right) = 12{e^4} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{e^{3x + 2y}}} \right] \cr & {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr & {f_y}\left( {x,y} \right) = 4{e^{3x + 2y}}\frac{\partial }{{\partial y}}\left[ {3x + 2y} \right] \cr & {f_y}\left( {x,y} \right) = 4{e^{3x + 2y}}\left( 2 \right) \cr & {f_y}\left( {x,y} \right) = 8{e^{3x + 2y}} \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = 8{e^{3\left( { - 4} \right) + 2\left( 3 \right)}} \cr & {f_y}\left( { - 4,3} \right) = 8{e^{ - 12 + 6}} \cr & {f_y}\left( { - 4,3} \right) = 8{e^{ - 6}} \cr} $$
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