Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 10

$${f_x}\left( {x,y} \right) = 56{e^{7x - y}},\,\,\,\,{f_y}\left( {x,y} \right) = - 8{e^{7x - y}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = 56{e^{15}}{\text{ and }}{f_y}\left( { - 4,3} \right) = - 8{e^{ - 31}}$$

$$\eqalign{ & f\left( {x,y} \right) = 8{e^{7x - y}} \cr & {\text{find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {8{e^{7x - y}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = 8{e^{7x - y}}\frac{\partial }{{\partial x}}\left[ {7x - y} \right] \cr & {f_x}\left( {x,y} \right) = 8{e^{7x - y}}\left( 7 \right) \cr & {f_x}\left( {x,y} \right) = 56{e^{7x - y}} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = 56{e^{7\left( 2 \right) - \left( { - 1} \right)}} \cr & {f_x}\left( {2, - 1} \right) = 56{e^{15}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {8{e^{7x - y}}} \right] \cr & {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr & {f_y}\left( {x,y} \right) = 8{e^{7x - y}}\frac{\partial }{{\partial y}}\left[ {7x - y} \right] \cr & {f_y}\left( {x,y} \right) = 8{e^{7x - y}}\left( { - 1} \right) \cr & {f_y}\left( {x,y} \right) = - 8{e^{7x - y}} \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = - 8{e^{7\left( { - 4} \right) - 3}} \cr & {f_y}\left( { - 4,3} \right) = - 8{e^{ - 28 - 3}} \cr & {f_y}\left( { - 4,3} \right) = - 8{e^{ - 31}} \cr} $$