Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 897: 82

Answer

$$\eqalign{ & {z_{xx}} = \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},{\text{ }}{z_{xy}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{yy}} = - \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},{\text{ }}{z_{yx}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr} $$

Work Step by Step

$$\eqalign{ & z = \arctan \frac{y}{x} \cr & {\text{Calculate the second partial derivatives }}{z_{xx}},{\text{ }}{z_{yy}},{\text{ }}{z_{xy}},{\text{ }}{x_{yx}} \cr & {z_x} = \frac{\partial }{{\partial x}}\left[ {\arctan \frac{y}{x}} \right] \cr & {z_x} = \frac{{\frac{\partial }{{\partial x}}\left[ {\frac{y}{x}} \right]}}{{1 + {{\left( {\frac{y}{x}} \right)}^2}}} \cr & {z_x} = \frac{{ - \frac{y}{{{x^2}}}}}{{1 + \frac{{{y^2}}}{{{x^2}}}}} = \frac{{ - \frac{y}{{{x^2}}}}}{{\frac{{{x^2} + {y^2}}}{{{x^2}}}}} \cr & {z_x} = - \frac{y}{{{x^2} + {y^2}}} \cr & {z_{xx}} = \frac{\partial }{{\partial x}}\left[ { - \frac{y}{{{x^2} + {y^2}}}} \right] \cr & {z_{xx}} = - \frac{{y\left( { - 2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{xx}} = \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{xy}} = - \underbrace {\frac{\partial }{{\partial y}}\left[ {\frac{y}{{{x^2} + {y^2}}}} \right]}_{{\text{Quotient rule}}} \cr & {z_{xy}} = - \frac{{\left( {{x^2} + {y^2}} \right)\left( 1 \right) - y\left( {2y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{xy}} = - \frac{{{x^2} + {y^2} - 2{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{xy}} = - \frac{{{x^2} - {y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{xy}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & and \cr & {z_y} = \frac{\partial }{{\partial y}}\left[ {\arctan \frac{y}{x}} \right] \cr & {z_y} = \frac{{\frac{\partial }{{\partial y}}\left[ {\frac{y}{x}} \right]}}{{1 + {{\left( {\frac{y}{x}} \right)}^2}}} = \frac{{\frac{1}{x}}}{{1 + \frac{{{y^2}}}{{{x^2}}}}} = \frac{{\frac{1}{x}}}{{\frac{{{x^2} + {y^2}}}{{{x^2}}}}} = \frac{x}{{{x^2} + {y^2}}} \cr & {z_{yy}} = \frac{\partial }{{\partial y}}\left[ {\frac{x}{{{x^2} + {y^2}}}} \right] = - \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{yx}} = \underbrace {\frac{\partial }{{\partial x}}\left[ {\frac{x}{{{x^2} + {y^2}}}} \right]}_{{\text{Quotient rule}}} \cr & {z_{yx}} = \frac{{\left( {{x^2} + {y^2}} \right) - x\left( {2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{yx}} = \frac{{{x^2} + {y^2} - 2{x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{yx}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{Summary:}} \cr & {z_{xx}} = \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},{\text{ }}{z_{xy}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {z_{yy}} = - \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},{\text{ }}{z_{yx}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr} $$
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