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: 33

Answer

$${f_x}\left( {x,y} \right) = \frac{{{y^{ - 1/2}}}}{{{y^2} + {x^2}}},\,\,\,\,\,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{x{y^{ - 3/2}}}}{{{y^2} + {x^2}}} - \frac{3}{2}{y^{ - 5/2}}{\tan ^{ - 1}}\left( {x/y} \right)$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = {y^{ - 3/2}}{\tan ^{ - 1}}\left( {x/y} \right) \cr & {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr & {\text{use the quotient rule}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{y^{ - 3/2}}{{\tan }^{ - 1}}\left( {x/y} \right)} \right] \cr & {f_x}\left( {x,y} \right) = {y^{ - 3/2}}\frac{\partial }{{\partial x}}\left[ {{{\tan }^{ - 1}}\left( {x/y} \right)} \right] \cr & {\text{use the differentitation rule for }}\tan u \cr & {f_x}\left( {x,y} \right) = {y^{ - 3/2}}\left( {\frac{{1/y}}{{1 + {{\left( {x/y} \right)}^2}}}} \right) \cr & {\text{simplifying}} \cr & {f_x}\left( {x,y} \right) = {y^{ - 5/2}}\left( {\frac{{{y^2}}}{{{y^2} + {x^2}}}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{{y^{ - 1/2}}}}{{{y^2} + {x^2}}} \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr & {\text{use the quotient rule}} \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{y^{ - 3/2}}{{\tan }^{ - 1}}\left( {x/y} \right)} \right] \cr & {\text{by the product rule}} \cr & {f_y}\left( {x,y} \right) = {y^{ - 3/2}}\frac{\partial }{{\partial y}}\left[ {{{\tan }^{ - 1}}\left( {x/y} \right)} \right] + {\tan ^{ - 1}}\left( {x/y} \right)\frac{\partial }{{\partial y}}\left[ {{y^{ - 3/2}}} \right] \cr & {\text{use the differentitation rule for }}\tan u \cr & {f_y}\left( {x,y} \right) = {y^{ - 3/2}}\left( {\frac{{ - x/{y^2}}}{{1 + {{\left( {x/y} \right)}^2}}}} \right) + {\tan ^{ - 1}}\left( {x/y} \right)\left( { - \frac{3}{2}{y^{ - 5/2}}} \right) \cr & {\text{simplifying}} \cr & {f_y}\left( {x,y} \right) = \frac{{{y^{ - 7/2}}}}{x}\left( {\frac{{{y^2}}}{{{y^2} + {x^2}}}} \right) - \frac{3}{2}{y^{ - 5/2}}{\tan ^{ - 1}}\left( {x/y} \right) \cr & {f_y}\left( {x,y} \right) = \frac{{x{y^{ - 3/2}}}}{{{y^2} + {x^2}}} - \frac{3}{2}{y^{ - 5/2}}{\tan ^{ - 1}}\left( {x/y} \right) \cr} $$
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