Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 46

Answer

$${f_x}\left( {x,y,z} \right) = 4xyz\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)$$ $${f_y}\left( {x,y,z} \right) = 2{x^2}z\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)$$ $$\eqalign{ & {f_z}\left( {x,y,z} \right) = 2{x^2}y\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{1}{{2\sqrt z }}{\sinh ^2}\left( {{x^2}yz} \right)\sinh \left( {\sqrt z } \right) \cr} $$

Work Step by Step

$$\eqalign{ & f\left( {x,y,z} \right) = \cosh \left( {\sqrt z } \right){\sinh ^2}\left( {{x^2}yz} \right) \cr & {\text{Calculate }}{f_x}\left( {x,y,z} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {\cosh \left( {\sqrt z } \right){{\sinh }^2}\left( {{x^2}yz} \right)} \right] \cr & {f_x}\left( {x,y,z} \right) = \cosh \left( {\sqrt z } \right)\frac{\partial }{{\partial x}}\left[ {{{\sinh }^2}\left( {{x^2}yz} \right)} \right] \cr & {f_x}\left( {x,y,z} \right) = 2\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)\frac{\partial }{{\partial x}}\left[ {{x^2}yz} \right] \cr & {f_x}\left( {x,y,z} \right) = 2\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)\left( {2xyz} \right) \cr & {f_x}\left( {x,y,z} \right) = 4xyz\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right) \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y,z} \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {\cosh \left( {\sqrt z } \right){{\sinh }^2}\left( {{x^2}yz} \right)} \right] \cr & {f_y}\left( {x,y,z} \right) = \cosh \left( {\sqrt z } \right)\frac{\partial }{{\partial y}}\left[ {{{\sinh }^2}\left( {{x^2}yz} \right)} \right] \cr & {f_y}\left( {x,y,z} \right) = 2\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)\frac{\partial }{{\partial y}}\left[ {{x^2}yz} \right] \cr & {f_y}\left( {x,y,z} \right) = 2\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)\left( {{x^2}z} \right) \cr & {f_y}\left( {x,y,z} \right) = 2{x^2}z\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right) \cr & \cr & {\text{Calculate }}{f_z}\left( {x,y,z} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {\cosh \left( {\sqrt z } \right){{\sinh }^2}\left( {{x^2}yz} \right)} \right] \cr & {\text{use product rule}} \cr & {f_z}\left( {x,y,z} \right) = \cosh \left( {\sqrt z } \right)\frac{\partial }{{\partial z}}\left[ {{{\sinh }^2}\left( {{x^2}yz} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\sinh ^2}\left( {{x^2}yz} \right)\frac{\partial }{{\partial z}}\left[ {\cosh \left( {\sqrt z } \right)} \right] \cr & {f_z}\left( {x,y,z} \right) = 2\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right)\left( {{x^2}y} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\sinh ^2}\left( {{x^2}yz} \right)\sinh \left( {\sqrt z } \right)\left( {\frac{1}{{2\sqrt z }}} \right) \cr & {f_z}\left( {x,y,z} \right) = 2{x^2}y\cosh \left( {\sqrt z } \right)\sinh \left( {{x^2}yz} \right)\cosh \left( {{x^2}yz} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{1}{{2\sqrt z }}{\sinh ^2}\left( {{x^2}yz} \right)\sinh \left( {\sqrt z } \right) \cr} $$
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