Chapter 15 - Review - Review Exercises - Page 1139: 16

$${f_x} = - \frac{6}{{{x^2}y}} + \frac{y}{6},{\text{ }}{f_y} = - \frac{6}{{x{y^2}}} + \frac{x}{6},{\text{ and }}{f_{yy}} = \frac{{12}}{{x{y^3}}}$$

$$\eqalign{ & f\left( {x,y} \right) = \frac{6}{{xy}} + \frac{{xy}}{6} \cr & {\text{Find }}{f_x} \cr & {f_x} = {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{6}{{xy}} + \frac{{xy}}{6}} \right] \cr & {f_x} = \frac{\partial }{{\partial x}}\left[ {\frac{6}{{xy}} + \frac{{xy}}{6}} \right] \cr & {f_x} = \frac{\partial }{{\partial x}}\left[ {\frac{6}{{xy}}} \right] + \frac{\partial }{{\partial x}}\left[ {\frac{{xy}}{6}} \right] \cr & {\text{Consider }}y{\text{ as a constant}} \cr & {f_x} = \frac{6}{y}\frac{\partial }{{\partial x}}\left[ {\frac{1}{x}} \right] + \frac{y}{6}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {f_x} = \frac{6}{y}\left( { - \frac{1}{{{x^2}}}} \right) + \frac{y}{6} \cr & {f_x} = - \frac{6}{{{x^2}y}} + \frac{y}{6} \cr & \cr & {\text{Find }}{f_y} \cr & {f_y} = {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{6}{{xy}} + \frac{{xy}}{6}} \right] \cr & {f_y} = \frac{\partial }{{\partial y}}\left[ {\frac{6}{{xy}} + \frac{{xy}}{6}} \right] \cr & {f_y} = \frac{\partial }{{\partial y}}\left[ {\frac{6}{{xy}}} \right] + \frac{\partial }{{\partial y}}\left[ {\frac{{xy}}{6}} \right] \cr & {\text{Consider }}x{\text{ as a constant}} \cr & {f_y} = \frac{6}{x}\frac{\partial }{{\partial y}}\left[ {\frac{1}{y}} \right] + \frac{x}{6}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {f_y} = \frac{6}{x}\left( { - \frac{1}{{{y^2}}}} \right) + \frac{x}{6} \cr & {f_y} = - \frac{6}{{x{y^2}}} + \frac{x}{6} \cr & \cr & {\text{Find }}{f_{yy}} \cr & {f_{yy}} = \frac{\partial }{{\partial y}}\left[ {{f_y}} \right] \cr & {f_{yy}} = \frac{\partial }{{\partial y}}\left[ { - \frac{6}{{x{y^2}}} + \frac{x}{6}} \right] \cr & {f_{yy}} = \frac{\partial }{{\partial y}}\left[ { - \frac{{6{y^{ - 2}}}}{x} + \frac{x}{6}} \right] \cr & {f_{yy}} = - \frac{{ - 12{y^{ - 3}}}}{x} + 0 \cr & {f_{yy}} = \frac{{12}}{{x{y^3}}} \cr} $$