Chapter 13 - Partial Derivatives - 13.4 Differentiability, Differentials, And Local Linearity - Exercises Set 13.4 - Page 947: 17

$$dw = 3{x^2}{y^2}zdx + 2{x^3}yzdy + {x^3}{y^2}dz$$

$$\eqalign{ & w = {x^3}{y^2}z \cr & {\text{Let }}w = f\left( {x,y,z} \right){\text{ }} \cr & {\text{Calculate the partial derivative }}{f_x}\left( {x,y,z} \right){\text{, treating }}y{\text{ and }}z{\text{ as constants}} \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{x^3}{y^2}z} \right] \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = {y^2}z\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = {y^2}z\left( {3{x^2}} \right) \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = 3{x^2}{y^2}z \cr & {\text{Calculate the partial derivative }}{f_y}\left( {x,y,z} \right){\text{, treating }}x{\text{ and }}z{\text{ as constants}} \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{x^3}{y^2}z} \right] \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = {x^3}z\frac{\partial }{{\partial y}}\left[ {{y^2}} \right] \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = {x^3}z\left( {2y} \right) \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = 2{x^3}yz \cr & {\text{Calculate the partial derivative }}{f_z}\left( {x,y,z} \right){\text{, treating }}x{\text{ and }}y{\text{ as constants}} \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{x^3}{y^2}z} \right] \cr & {\text{ }}{f_z}\left( {x,y,z} \right) = {x^3}{y^2}\frac{\partial }{{\partial z}}\left[ z \right] \cr & {\text{ }}{f_z}\left( {x,y,z} \right) = {x^3}{y^2}\left( 1 \right) \cr & {\text{ }}{f_z}\left( {x,y,z} \right) = {x^3}{y^2} \cr & \cr & {\text{The total differential of }}z{\text{ is given by}}:\,\,\left( {{\text{See page 944}}} \right) \cr & dw = {f_x}\left( {x,y,z} \right)dx + {f_y}\left( {x,y,z} \right)dy + {f_z}\left( {x,y,z} \right)dz \cr & {\text{Substitute the partial derivatives}} \cr & dw = 3{x^2}{y^2}zdx + 2{x^3}yzdy + {x^3}{y^2}dz \cr} $$