Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 25

$$\eqalign{ & {r_{xx}}\left( {x,y} \right) = \frac{{12y}}{{{{\left( {x + y} \right)}^3}}},\,\,\,\,{r_{yy}}\left( {x,y} \right) = - \frac{{12x}}{{{{\left( {x + y} \right)}^3}}} \cr & {r_{xy}}\left( {x,y} \right) = {r_{yx}}\left( {x,y} \right) = \frac{{6\left( {y - x} \right)}}{{{{\left( {x + y} \right)}^3}}} \cr} $$

$$\eqalign{ & r\left( {x,y} \right) = \frac{{6y}}{{x + y}} \cr & {\text{Find }}{r_x}\left( {x,y} \right){\text{ and }}{r_y}\left( {x,y} \right) \cr & {r_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{6y}}{{x + y}}} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {r_x}\left( {x,y} \right) = 6y\frac{\partial }{{\partial x}}\left[ {\frac{1}{{x + y}}} \right] \cr & {r_x}\left( {x,y} \right) = 6y\frac{\partial }{{\partial x}}\left[ {{{\left( {x + y} \right)}^{ - 1}}} \right] \cr & {r_x}\left( {x,y} \right) = - 6y{\left( {x + y} \right)^{ - 2}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr & {r_x}\left( {x,y} \right) = - 6y{\left( {x + y} \right)^{ - 2}} \cr & \cr & {r_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{6y}}{{x + y}}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then using the quotient rule}} \cr & {r_y}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( 6 \right) - 6y\left( 1 \right)}}{{{{\left( {x + y} \right)}^2}}} \cr & {r_y}\left( {x,y} \right) = \frac{{6x + 6y - 6y}}{{{{\left( {x + y} \right)}^2}}} \cr & {r_y}\left( {x,y} \right) = \frac{{6x}}{{{{\left( {x + y} \right)}^2}}} \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {r_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 6y{{\left( {x + y} \right)}^{ - 2}}} \right] \cr & {r_{xy}}\left( {x,y} \right) = - 6y\frac{\partial }{{\partial y}}\left[ {{{\left( {x + y} \right)}^{ - 2}}} \right] - 6{\left( {x + y} \right)^{ - 2}}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {r_{xy}}\left( {x,y} \right) = 12y{\left( {x + y} \right)^{ - 3}} - 6{\left( {x + y} \right)^{ - 2}} \cr & {r_{xy}}\left( {x,y} \right) = 6{\left( {x + y} \right)^{ - 3}}\left( {2y - \left( {x + y} \right)} \right) \cr & {r_{xy}}\left( {x,y} \right) = 6{\left( {x + y} \right)^{ - 3}}\left( {y - x} \right) \cr & {r_{xy}}\left( {x,y} \right) = \frac{{6\left( {y - x} \right)}}{{{{\left( {x + y} \right)}^3}}} \cr & \cr & {r_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{6x}}{{{{\left( {x + y} \right)}^2}}}} \right] \cr & {\text{by using quotient rule}} \cr & {r_{yx}}\left( {x,y} \right) = \frac{{{{\left( {x + y} \right)}^2}\left( 6 \right) - 6x\left( 2 \right)\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^4}}} \cr & {r_{yx}}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( 6 \right) - 6x\left( 2 \right)}}{{{{\left( {x + y} \right)}^3}}} \cr & {r_{yx}}\left( {x,y} \right) = \frac{{6x + 6y - 12x}}{{{{\left( {x + y} \right)}^3}}} \cr & {r_{yx}}\left( {x,y} \right) = \frac{{6\left( {y - x} \right)}}{{{{\left( {x + y} \right)}^3}}} \cr & \cr & {r_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 6y{{\left( {x + y} \right)}^{ - 2}}} \right] \cr & {r_{xx}}\left( {x,y} \right) = - 6y\frac{\partial }{{\partial x}}\left[ {{{\left( {x + y} \right)}^{ - 2}}} \right] \cr & {r_{xx}}\left( {x,y} \right) = - 6y\left( { - 2} \right){\left( {x + y} \right)^{ - 3}} \cr & {r_{xx}}\left( {x,y} \right) = \frac{{12y}}{{{{\left( {x + y} \right)}^3}}} \cr & {\text{and}} \cr & {r_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{6x}}{{{{\left( {x + y} \right)}^2}}}} \right] \cr & {r_{yy}}\left( {x,y} \right) = 6x\frac{\partial }{{\partial y}}\left[ {{{\left( {x + y} \right)}^{ - 2}}} \right] \cr & {r_{yy}}\left( {x,y} \right) = 6x\left( { - 2} \right){\left( {x + y} \right)^{ - 3}} \cr & {r_{yy}}\left( {x,y} \right) = - \frac{{12x}}{{{{\left( {x + y} \right)}^3}}} \cr} $$