Chapter 13 - Functions of Several Variables - Review Exercises - Page 960: 24

$$\eqalign{ & {h_{xy}}\left( {x,y} \right) = \frac{{{x^2} - {y^2}}}{{{{\left( {x + y} \right)}^4}}} \cr & {h_{yx}}\left( {x,y} \right) = \frac{{{x^2} - {y^2}}}{{{{\left( {x + y} \right)}^4}}} \cr} $$

$$\eqalign{ & h\left( {x,y} \right) = \frac{x}{{x + y}} \cr & {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr & {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{x}{{x + y}}} \right] \cr & {\text{By the quotient rule}} \cr & {h_x}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( 1 \right) - x\left( 1 \right)}}{{{{\left( {x + y} \right)}^2}}} \cr & {h_x}\left( {x,y} \right) = \frac{{x + y - x}}{{{{\left( {x + y} \right)}^2}}} \cr & {h_x}\left( {x,y} \right) = \frac{y}{{{{\left( {x + y} \right)}^2}}} \cr & {\text{Calculate }}{h_{xy}}\left( {x,y} \right){\text{ differentiating }}{h_x}\left( {x,y} \right){\text{ with respect to }}y \cr & {\text{ and treating }}y{\text{ as a constant}}{\text{.}} \cr & {h_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{y}{{{{\left( {x + y} \right)}^2}}}} \right] \cr & {h_{xy}}\left( {x,y} \right) = \frac{{{{\left( {x + y} \right)}^2} - 2y\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^4}}} \cr & {h_{xy}}\left( {x,y} \right) = \frac{{{x^2} + 2xy + {y^2} - 2xy - 2{y^2}}}{{{{\left( {x + y} \right)}^4}}} \cr & {h_{xy}}\left( {x,y} \right) = \frac{{{x^2} - {y^2}}}{{{{\left( {x + y} \right)}^4}}} \cr & \cr & {\text{Calculate }}{h_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr & {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{x}{{x + y}}} \right] \cr & {h_y}\left( {x,y} \right) = - \frac{x}{{{{\left( {x + y} \right)}^2}}} \cr & {\text{Calculate }}{h_{yx}}\left( {x,y} \right){\text{ differentiating }}{h_y}\left( {x,y} \right){\text{ with respect to }}x \cr & {\text{ and treating }}y{\text{ as a constant}}{\text{.}} \cr & {h_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - \frac{x}{{{{\left( {x + y} \right)}^2}}}} \right] \cr & {h_{yx}}\left( {x,y} \right) = - \frac{{{{\left( {x + y} \right)}^2} - 2x\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^4}}} \cr & {h_{yx}}\left( {x,y} \right) = - \frac{{{x^2} + 2xy + {y^2} - 2{x^2} - 2xy}}{{{{\left( {x + y} \right)}^4}}} \cr & {h_{yx}}\left( {x,y} \right) = - \frac{{{y^2} - {x^2}}}{{{{\left( {x + y} \right)}^4}}} \cr & {h_{yx}}\left( {x,y} \right) = \frac{{{x^2} - {y^2}}}{{{{\left( {x + y} \right)}^4}}} \cr} $$