Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 44

$${f_{xx}}\left( {x,y} \right) = - \frac{{9{y^4}}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}}{\text{ and }}{f_{xy}}\left( {x,y} \right) = \frac{{6y}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}}$$

$$\eqalign{ & f\left( {x,y} \right) = \ln \left| {1 + 3x{y^2}} \right| \cr & {\text{find }}{f_x}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left| {1 + 3x{y^2}} \right|} \right] \cr & {\text{use the rule }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + 3x{y^2}}}\frac{\partial }{{\partial x}}\left[ {1 + 3x{y^2}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + 3x{y^2}}}\left( {3{y^2}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{3{y^2}}}{{1 + 3x{y^2}}} \cr & \cr & {\text{find }}{f_{xx}}\left( {x,y} \right) \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{3{y^2}}}{{1 + 3x{y^2}}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then }} \cr & {f_{xx}}\left( {x,y} \right) = 3{y^2}\frac{\partial }{{\partial x}}\left[ {{{\left( {1 + 3x{y^2}} \right)}^{ - 1}}} \right] \cr & {f_{xx}}\left( {x,y} \right) = - 3{y^2}{\left( {1 + 3x{y^2}} \right)^{ - 2}}\frac{\partial }{{\partial x}}\left[ {1 + 3x{y^2}} \right] \cr & {f_{xx}}\left( {x,y} \right) = - 3{y^2}{\left( {1 + 3x{y^2}} \right)^{ - 2}}\left( {3{y^2}} \right) \cr & {f_{xx}}\left( {x,y} \right) = - 9{y^4}{\left( {1 + 3x{y^2}} \right)^{ - 2}} \cr & {f_{xx}}\left( {x,y} \right) = - \frac{{9{y^4}}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}} \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^2}}}{{1 + 3x{y^2}}}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then by using the quotient rule}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{\left( {1 + 3x{y^2}} \right)\frac{\partial }{{\partial y}}\left[ {3{y^2}} \right] - 3{y^2}\frac{\partial }{{\partial y}}\left[ {1 + 3x{y^2}} \right]}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{\left( {1 + 3x{y^2}} \right)\left( {6y} \right) - 3{y^2}\left( {6xy} \right)}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}} \cr & {\text{multiply}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{6y + 18x{y^3} - 18x{y^3}}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{6y}}{{{{\left( {1 + 3x{y^2}} \right)}^2}}} \cr} $$