Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 41

$$\eqalign{ & {\text{a)}}{f_x}\left( {x,y,z} \right) = 2x{y^4}{z^3} + y,\,\,\,{\text{b) }}{f_y}\left( {x,y,z} \right) = 4{x^2}{y^3}{z^3} + x \cr & {\text{c)}}{f_z}\left( {x,y,z} \right) = 3{x^2}{z^2}{y^4} + 2z,\,\,\,{\text{d) }}{f_x}\left( {1,y,z} \right) = 2{y^4}{z^3} + y \cr & {\text{e) }}{f_y}\left( {1,2,z} \right) = 32{z^3} + 1,\,\,\,{\text{f)}}\,\,{f_z}\left( {1,2,3} \right) = 438 \cr} $$

$$\eqalign{ & {\text{Let }}f\left( {x,y,z} \right) = {x^2}{y^4}{z^3} + xy + {z^2} + 1 \cr & {\text{a) Calculate the partial derivative }}{f_x}\left( {x,y,z} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2}{y^4}{z^3} + xy + {z^2} + 1} \right] \cr & {f_x}\left( {x,y,z} \right) = {y^4}{z^3}\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] + y\frac{\partial }{{\partial x}}\left[ x \right] + \frac{\partial }{{\partial x}}\left[ {{z^2}} \right] + \frac{\partial }{{\partial x}}\left[ 1 \right] \cr & {f_x}\left( {x,y,z} \right) = 2x{y^4}{z^3} + y \cr & \cr & {\text{b) Calculate the partial derivative }}{f_y}\left( {x,y,z} \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2}{y^4}{z^3} + xy + {z^2} + 1} \right] \cr & {f_y}\left( {x,y,z} \right) = {z^3}{x^2}\frac{\partial }{{\partial y}}\left[ {{y^4}} \right] + x\frac{\partial }{{\partial y}}\left[ y \right] + \frac{\partial }{{\partial y}}\left[ {{z^2}} \right] + \frac{\partial }{{\partial y}}\left[ 1 \right] \cr & {f_y}\left( {x,y,z} \right) = {z^3}{x^2}\left( {4{y^3}} \right) + x\left( 1 \right) + 0 \cr & {f_y}\left( {x,y,z} \right) = 4{x^2}{y^3}{z^3} + x \cr & \cr & {\text{c) Calculate the partial derivative }}{f_z}\left( {x,y,z} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{x^2}{y^4}{z^3} + xy + {z^2} + 1} \right] \cr & {f_z}\left( {x,y,z} \right) = {x^2}{y^4}\frac{\partial }{{\partial z}}\left[ {{z^3}} \right] + \frac{\partial }{{\partial z}}\left[ {xy} \right] + \frac{\partial }{{\partial z}}\left[ {{z^2}} \right] + \frac{\partial }{{\partial z}}\left[ 1 \right] \cr & {f_z}\left( {x,y,z} \right) = {x^2}{y^4}\left( {3{z^2}} \right) + 0 + 2z + 0 \cr & {f_z}\left( {x,y,z} \right) = 3{x^2}{z^2}{y^4} + 2z \cr & \cr & {\text{d) Calculate the partial derivative }}f\left( {1,y,z} \right){\text{ }} \cr & {f_x}\left( {x,y,z} \right) = 2x{y^4}{z^3} + y \cr & {f_x}\left( {1,y,z} \right) = 2\left( 1 \right){y^4}{z^3} + y \cr & {f_x}\left( {1,y,z} \right) = 2{y^4}{z^3} + y \cr & \cr & {\text{e) Calculate the partial derivative }}{f_y}\left( {1,2,z} \right){\text{ }} \cr & {f_y}\left( {1,2,z} \right) = 4{x^2}{y^3}{z^3} + x \cr & {f_y}\left( {1,2,z} \right) = 4{\left( 1 \right)^2}{\left( 2 \right)^3}{z^3} + 1 \cr & {f_y}\left( {1,2,z} \right) = 32{z^3} + 1 \cr & \cr & {\text{f) Calculate the partial derivative }}{f_z}\left( {1,2,3} \right){\text{ }} \cr & {f_z}\left( {x,y,z} \right) = 3{x^2}{z^2}{y^4} + 2z \cr & {f_z}\left( {1,2,3} \right) = 3{\left( 1 \right)^2}{\left( 3 \right)^2}{\left( 2 \right)^4} + 2\left( 3 \right) \cr & {f_z}\left( {1,2,3} \right) = 438 \cr} $$