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: 50

Answer

$$\eqalign{ & \frac{{\partial w}}{{\partial x}} = 2{y^3}{e^{2x + 3z}} \cr & \frac{{\partial w}}{{\partial y}} = 3{y^2}{e^{2x + 3z}} \cr & \frac{{\partial w}}{{\partial z}} = 3{y^3}{e^{2x + 3z}} \cr} $$

Work Step by Step

$$\eqalign{ & w = {y^3}{e^{2x + 3z}} \cr & \cr & {\text{Calculate }}\frac{{\partial w}}{{\partial x}} \cr & \frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{y^3}{e^{2x + 3z}}} \right] \cr & \frac{{\partial w}}{{\partial x}} = {y^3}\frac{\partial }{{\partial x}}\left[ {{e^{2x + 3z}}} \right] \cr & \frac{{\partial w}}{{\partial x}} = {y^3}\left( {2{e^{2x + 3z}}} \right) \cr & \frac{{\partial w}}{{\partial x}} = 2{y^3}{e^{2x + 3z}} \cr & \cr & {\text{Calculate }}\frac{{\partial w}}{{\partial y}} \cr & \frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{y^3}{e^{2x + 3z}}} \right] \cr & \frac{{\partial w}}{{\partial y}} = {e^{2x + 3z}}\frac{\partial }{{\partial y}}\left[ {{y^3}} \right] \cr & \frac{{\partial w}}{{\partial y}} = 3{y^2}{e^{2x + 3z}} \cr & \cr & {\text{Calculate }}\frac{{\partial w}}{{\partial z}} \cr & \frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left[ {{y^3}{e^{2x + 3z}}} \right] \cr & \frac{{\partial w}}{{\partial z}} = {y^3}\frac{\partial }{{\partial z}}\left[ {{e^{2x + 3z}}} \right] \cr & \frac{{\partial w}}{{\partial z}} = {y^3}\left( {3{e^{2x + 3z}}} \right) \cr & \frac{{\partial w}}{{\partial z}} = 3{y^3}{e^{2x + 3z}} \cr} $$
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