Chapter 12 - Functions of Several Veriables - 12.4 Partial Derivatives - 12.4 Exercises - Page 904: 51

$$\eqalign{ & {f_w}\left( {w,x,y,z} \right) = 2wx{y^2} \cr & {f_x}\left( {w,x,y,z} \right) = {w^2}{y^2} + {y^3}{z^2} \cr & {f_y}\left( {w,x,y,z} \right) = 2{w^2}xy + 3x{y^2}{z^2} \cr & {f_z}\left( {w,x,y,z} \right) = 2x{y^3}z \cr} $$

$$\eqalign{ & f\left( {w,x,y,z} \right) = {w^2}x{y^2} + x{y^3}{z^2} \cr & {\text{Find the first partial derivative }}{f_w}\left( {w,x,y,z} \right) \cr & {f_w}\left( {w,x,y,z} \right) = \frac{\partial }{{\partial w}}\left[ {{w^2}x{y^2} + x{y^3}{z^2}} \right] \cr & {\text{treat }}x,y{\text{ and }}z{\text{ as a constants}} \cr & {f_w}\left( {w,x,y,z} \right) = x{y^2}\frac{\partial }{{\partial w}}\left[ {{w^2}} \right] + \frac{\partial }{{\partial u}}\left[ {x{y^3}{z^2}} \right] \cr & {f_w}\left( {w,x,y,z} \right) = x{y^2}\left( {2w} \right) \cr & {f_w}\left( {w,x,y,z} \right) = 2wx{y^2} \cr & \cr & {\text{Find the first partial derivative }}{f_x}\left( {w,x,y,z} \right) \cr & {f_x}\left( {w,x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{w^2}x{y^2} + x{y^3}{z^2}} \right] \cr & {\text{treat }}w,y{\text{ and }}z{\text{ as a constants}} \cr & {f_x}\left( {w,x,y,z} \right) = {w^2}{y^2}\frac{\partial }{{\partial x}}\left[ x \right] + {y^3}{z^2}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {f_x}\left( {w,x,y,z} \right) = {w^2}{y^2}\left( 1 \right) + {y^3}{z^2}\left( 1 \right) \cr & {f_x}\left( {w,x,y,z} \right) = {w^2}{y^2} + {y^3}{z^2} \cr & \cr & {\text{Find the first partial derivative }}{f_y}\left( {w,x,y,z} \right) \cr & {f_y}\left( {w,x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{w^2}x{y^2} + x{y^3}{z^2}} \right] \cr & {\text{treat }}w,x{\text{ and }}z{\text{ as a constants}} \cr & {f_y}\left( {w,x,y,z} \right) = {w^2}x\frac{\partial }{{\partial y}}\left[ {{y^2}} \right] + x{z^2}\frac{\partial }{{\partial y}}\left[ {{y^3}} \right] \cr & {f_y}\left( {w,x,y,z} \right) = {w^2}x\left( {2y} \right) + x{z^2}\left( {3{y^2}} \right) \cr & {f_y}\left( {w,x,y,z} \right) = 2{w^2}xy + 3x{y^2}{z^2} \cr & \cr & {\text{Find the first partial derivative }}{f_z}\left( {w,x,y,z} \right) \cr & {f_z}\left( {w,x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{w^2}x{y^2} + x{y^3}{z^2}} \right] \cr & {\text{treat }}w,x{\text{ and }}y{\text{ as a constants}} \cr & {f_z}\left( {w,x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{w^2}x{y^2}} \right] + x{y^3}\frac{\partial }{{\partial z}}\left[ {{z^2}} \right] \cr & {f_z}\left( {w,x,y,z} \right) = x{y^3}\left( {2z} \right) \cr & {f_z}\left( {w,x,y,z} \right) = 2x{y^3}z \cr} $$