Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 970: 17

Answer

$g_x(x,y) = 5y(2xy + 1)(x + {x^2}y)^4$ $g_y(x,y) = (x + x^2y)^4 (6x^2y+x)$

Work Step by Step

$$\eqalign{ & g\left( {x,y} \right) = y{\left( {x + {x^2}y} \right)^5} \cr & \cr & {\text{Differentiate with respect to }}x \cr & {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {y{{\left( {x + {x^2}y} \right)}^5}} \right] \cr & {\text{Considering }}y{\text{ as a constant}} \cr & {g_x}\left( {x,y} \right) = y\frac{\partial }{{\partial x}}\left[ {{{\left( {x + {x^2}y} \right)}^5}} \right] \cr & {\text{Using the Chain Rule}} \cr & {g_x}\left( {x,y} \right) = 5y{\left( {x + {x^2}y} \right)^4}\frac{\partial }{{\partial x}}\left[ {x + {x^2}y} \right] \cr & {g_x}\left( {x,y} \right) = 5y{\left( {x + {x^2}y} \right)^4}\left( {1 + 2xy} \right) \cr & {g_x}\left( {x,y} \right) = 5y\left( {2xy + 1} \right){\left( {x + {x^2}y} \right)^4} \cr & \cr & {\text{Differentiate with respect to }}y \cr & {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {y{{\left( {x + {x^2}y} \right)}^5}} \right] \cr & {\text{Use the product rule together the chain rule}} \cr & {g_y}\left( {x,y} \right) = y\frac{\partial }{{\partial y}}\left[ {{{\left( {x + {x^2}y} \right)}^5}} \right] + {\left( {x + {x^2}y} \right)^5}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {g_y}\left( {x,y} \right) = 5y{\left( {x + {x^2}y} \right)^4}\left( {0 + {x^2}} \right) + {\left( {x + {x^2}y} \right)^5}\left( 1 \right) \cr & {\text{Simplify}} \cr & {g_y}\left( {x,y} \right) = 5{x^2}y{\left( {x + {x^2}y} \right)^4} + {\left( {x + {x^2}y} \right)^5} \cr &{g_y}\left( {x,y} \right) =(x+x^2y)^4(5x^2y+x+x^2y)\cr &{g_y}\left( {x,y} \right) =(x+x^2y)^4(6x^2y+x)} $$
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