## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 40

#### Answer

$${f_{xx}}\left( {x,y} \right) = \frac{{2\left( {y + 3} \right)}}{{{{\left( {x - 1} \right)}^3}}}{\text{ and }}{f_{xy}}\left( {x,y} \right) = - \frac{1}{{{{\left( {x - 1} \right)}^2}}}$$

#### Work Step by Step

\eqalign{ & f\left( {x,y} \right) = \frac{{3x + y}}{{x - 1}} \cr & \cr & {\text{find }}{f_x}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{3x + y}}{{x - 1}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then by using the quotient rule}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {x - 1} \right)\frac{\partial }{{\partial x}}\left[ {3x + y} \right] - \left( {3x + y} \right)\frac{\partial }{{\partial x}}\left[ {x - 1} \right]}}{{{{\left( {x - 1} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {x - 1} \right)\left( 3 \right) - \left( {3x + y} \right)\left( 1 \right)}}{{{{\left( {x - 1} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_x}\left( {x,y} \right) = \frac{{3x - 3 - 3x - y}}{{{{\left( {x - 1} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{ - 3 - y}}{{{{\left( {x - 1} \right)}^2}}} \cr & \cr & {\text{find }}{f_{xx}}\left( {x,y} \right) \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{ - 3 - y}}{{{{\left( {x - 1} \right)}^2}}}} \right] \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\left( { - 3 - y} \right){{\left( {x - 1} \right)}^{ - 2}}} \right] \cr & {\text{by using the chain rule}} \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = - 2\left( { - 3 - y} \right){\left( {x - 1} \right)^{ - 3}}\frac{\partial }{{\partial x}}\left[ {x - 1} \right] \cr & {\text{solving derivatives}} \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = 2\left( {y + 3} \right){\left( {x - 1} \right)^{ - 3}} \cr & {f_{xx}}\left( {x,y} \right) = \frac{{2\left( {y + 3} \right)}}{{{{\left( {x - 1} \right)}^3}}} \cr & \cr & {\text{find }}{f_{xy}}\left( {x,y} \right) \cr & {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{ - 3 - y}}{{{{\left( {x - 1} \right)}^2}}}} \right] \cr & {\text{treating }}x{\text{ as a constant}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}}\frac{\partial }{{\partial y}}\left[ { - 3 - y} \right] \cr & {f_{xy}}\left( {x,y} \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}}\left( { - 1} \right) \cr & {f_{xy}}\left( {x,y} \right) = - \frac{1}{{{{\left( {x - 1} \right)}^2}}} \cr}

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