Chapter 12 - Functions of Several Veriables - 12.4 Partial Derivatives - 12.4 Exercises - Page 904: 54

$$\eqalign{ & {F_w}\left( {w,x,y,z} \right) = \sqrt {x + 2y + 3z} \cr & {F_x}\left( {w,x,y,z} \right) = \frac{w}{{2\sqrt {x + 2y + 3z} }} \cr & {F_y}\left( {w,x,y,z} \right) = \frac{w}{{\sqrt {x + 2y + 3z} }} \cr & {F_z}\left( {w,x,y,z} \right) = \frac{{3w}}{{2\sqrt {x + 2y + 3z} }} \cr} $$

$$\eqalign{ & F\left( {w,x,y,z} \right) = w\sqrt {x + 2y + 3z} \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\sqrt {x + 2y + 3z} } \right] \cr & {\text{treat }}x,y{\text{ and }}z{\text{ as a constants}} \cr & {F_w}\left( {w,x,y,z} \right) = \sqrt {x + 2y + 3z} \frac{\partial }{{\partial w}}\left[ w \right] \cr & {F_w}\left( {w,x,y,z} \right) = \sqrt {x + 2y + 3z} \left( 1 \right) \cr & {F_w}\left( {w,x,y,z} \right) = \sqrt {x + 2y + 3z} \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\sqrt {x + 2y + 3z} } \right] \cr & {\text{treat }}w,y{\text{ and }}z{\text{ as a constants}} \cr & {F_x}\left( {w,x,y,z} \right) = w\frac{\partial }{{\partial x}}\left[ {{{\left( {x + 2y + 3z} \right)}^{1/2}}} \right] \cr & {\text{use chain rule}} \cr & {F_x}\left( {w,x,y,z} \right) = w\left( {\frac{1}{2}} \right){\left( {x + 2y + 3z} \right)^{ - 1/2}}\frac{\partial }{{\partial x}}\left[ {x + 2y + 3z} \right] \cr & {F_x}\left( {w,x,y,z} \right) = w\left( {\frac{1}{2}} \right){\left( {x + 2y + 3z} \right)^{ - 1/2}}\left( 1 \right) \cr & {F_x}\left( {w,x,y,z} \right) = \frac{w}{{2\sqrt {x + 2y + 3z} }} \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\sqrt {x + 2y + 3z} } \right] \cr & {\text{treat }}w,x{\text{ and }}z{\text{ as a constants}} \cr & {F_y}\left( {w,x,y,z} \right) = w\frac{\partial }{{\partial y}}\left[ {{{\left( {x + 2y + 3z} \right)}^{1/2}}} \right] \cr & {\text{use chain rule}} \cr & {F_y}\left( {w,x,y,z} \right) = w\left( {\frac{1}{2}} \right){\left( {x + 2y + 3z} \right)^{ - 1/2}}\frac{\partial }{{\partial y}}\left[ {x + 2y + 3z} \right] \cr & {F_y}\left( {w,x,y,z} \right) = w\left( {\frac{1}{2}} \right){\left( {x + 2y + 3z} \right)^{ - 1/2}}\left( 2 \right) \cr & {F_y}\left( {w,x,y,z} \right) = \frac{w}{{\sqrt {x + 2y + 3z} }} \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\sqrt {x + 2y + 3z} } \right] \cr & {\text{treat }}w,x{\text{ and }}y{\text{ as a constants}} \cr & {F_z}\left( {w,x,y,z} \right) = w\frac{\partial }{{\partial z}}\left[ {{{\left( {x + 2y + 3z} \right)}^{1/2}}} \right] \cr & {\text{use chain rule}} \cr & {F_z}\left( {w,x,y,z} \right) = w\left( {\frac{1}{2}} \right){\left( {x + 2y + 3z} \right)^{ - 1/2}}\frac{\partial }{{\partial z}}\left[ {x + 2y + 3z} \right] \cr & {F_z}\left( {w,x,y,z} \right) = w\left( {\frac{1}{2}} \right){\left( {x + 2y + 3z} \right)^{ - 1/2}}\left( 3 \right) \cr & {F_z}\left( {w,x,y,z} \right) = \frac{{3w}}{{2\sqrt {x + 2y + 3z} }} \cr} $$