Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.4 Partial Derivatives - 12.4 Exercises - Page 904: 35

Answer

$$\eqalign{ & {p_{uu}}\left( {u,v} \right) = \frac{{2{v^2} + 8 - 2{u^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}},\,\,\,\,\,\,\,\,{\text{ }}{p_{vv}}\left( {u,v} \right) = \frac{{2{u^2} + 8 - 2{v^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & {p_{uv}}\left( {u,v} \right) = {p_{vu}}\left( {u,v} \right) = - \frac{{4uv}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr} $$

Work Step by Step

$$\eqalign{ & p\left( {u,v} \right) = \ln \left( {{u^2} + {v^2} + 4} \right) \cr & {\text{Find the first partial derivatives }}{p_u}\left( {u,v} \right){\text{ and }}{p_v}\left( {u,v} \right){\text{ then}} \cr & {p_u}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {\ln \left( {{u^2} + {v^2} + 4} \right)} \right] \cr & {\text{treat }}v{\text{ as a constant}}{\text{, then}} \cr & {p_u}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\frac{\partial }{{\partial u}}\left[ {{u^2} + {v^2} + 4} \right] \cr & {p_u}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\left( {2u} \right) \cr & {p_u}\left( {u,v} \right) = \frac{{2u}}{{{u^2} + {v^2} + 4}} \cr & and \cr & {p_v}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {\ln \left( {{u^2} + {v^2} + 4} \right)} \right] \cr & {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr & {p_v}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\frac{\partial }{{\partial v}}\left[ {{u^2} + {v^2} + 4} \right] \cr & {p_v}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\left( {2v} \right) \cr & {p_v}\left( {u,v} \right) = \frac{{2v}}{{{u^2} + {v^2} + 4}} \cr & \cr & {\text{Find the second partial derivatives }}{p_{uv}}\left( {u,v} \right){\text{ and }}{p_{vu}}\left( {u,v} \right){\text{ then}} \cr & {p_{uv}}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {\frac{{2u}}{{{u^2} + {v^2} + 4}}} \right] \cr & {\text{use product rule}} \cr & {p_{uv}}\left( {u,v} \right) = 2u\frac{\partial }{{\partial v}}\left[ {\frac{1}{{{u^2} + {v^2} + 4}}} \right] \cr & {p_{uv}}\left( {u,v} \right) = 2u\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\frac{\partial }{{\partial v}}\left[ {{u^2} + {v^2} + 4} \right] \cr & {p_{uv}}\left( {u,v} \right) = 2u\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\left( {2v} \right) \cr & {p_{uv}}\left( {u,v} \right) = - \frac{{4uv}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & and \cr & {p_{vu}}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {\frac{{2v}}{{{u^2} + {v^2} + 4}}} \right] \cr & {\text{use product rule}} \cr & {p_{vu}}\left( {u,v} \right) = 2v\frac{\partial }{{\partial u}}\left[ {\frac{1}{{{u^2} + {v^2} + 4}}} \right] \cr & {p_{vu}}\left( {u,v} \right) = 2v\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\frac{\partial }{{\partial u}}\left[ {{u^2} + {v^2} + 4} \right] \cr & {p_{vu}}\left( {u,v} \right) = 2v\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\left( {2u} \right) \cr & {p_{vu}}\left( {u,v} \right) = - \frac{{4uv}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & \cr & {\text{Find the second partial derivatives }}{p_{uu}}\left( {u,v} \right){\text{ and }}{p_{vv}}\left( {u,v} \right){\text{ then}} \cr & {p_{uu}}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {\frac{{2u}}{{{u^2} + {v^2} + 4}}} \right] \cr & {\text{by using the quotient rule}} \cr & {p_{uu}}\left( {u,v} \right) = \frac{{\left( {{u^2} + {v^2} + 4} \right)\left( 2 \right) - 2u\left( {2u} \right)}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & {p_{uu}}\left( {u,v} \right) = \frac{{2{u^2} + 2{v^2} + 8 - 4{u^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & {p_{uu}}\left( {u,v} \right) = \frac{{2{v^2} + 8 - 2{u^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & and \cr & {\text{ }}{p_{vv}}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {\frac{{2v}}{{{u^2} + {v^2} + 4}}} \right] \cr & {\text{by using the quotient rule}} \cr & {p_{vv}}\left( {u,v} \right) = \frac{{\left( {{u^2} + {v^2} + 4} \right)\left( 2 \right) - 2v\left( {2v} \right)}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & {p_{vv}}\left( {u,v} \right) = \frac{{2{u^2} + 2{v^2} + 8 - 4{v^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr & {p_{vv}}\left( {u,v} \right) = \frac{{2{u^2} + 8 - 2{v^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr} $$
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