Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 39

$${z_x}\left( {1,2} \right) = \frac{1}{{\sqrt {17} }},\,\,\,{\text{and }}\,\,\,{z_y}\left( {1,2} \right) = \frac{8}{{\sqrt {17} }}$$

$$\eqalign{ & z = \sqrt {{x^2} + 4{y^2}} ;\,\,\,\,\,\,\,\partial z/\partial x\left( {1,2} \right),\,\,\,\,\,\,\partial f/\partial y\left( {1,2} \right) \cr & z = {\left( {{x^2} + 4{y^2}} \right)^{1/2}} \cr & \cr & {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr & {\text{and use the chain rule}} \cr & {z_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {{{\left( {{x^2} + 4{y^2}} \right)}^{1/2}}} \right) \cr & {z_x}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^2} + 4{y^2}} \right)^{ - 1/2}}\frac{\partial }{{\partial x}}\left( {{x^2} + 4{y^2}} \right) \cr & {z_x}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^2} + 4{y^2}} \right)^{ - 1/2}}\left( {2x} \right) \cr & {z_x}\left( {x,y} \right) = {\left( {{x^2} + 4{y^2}} \right)^{ - 1/2}}\left( x \right) \cr & {z_x}\left( {x,y} \right) = \frac{x}{{\sqrt {{x^2} + 4{y^2}} }} \cr & \cr & {\text{Evaluating }}{z_x}\left( {1,2} \right) \cr & {z_x}\left( {1,2} \right) = \frac{1}{{\sqrt {1 + 4{{\left( 2 \right)}^2}} }} \cr & {z_x}\left( {1,2} \right) = \frac{1}{{\sqrt {17} }} \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr & {\text{and use the chain rule}} \cr & {z_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{{\left( {{x^2} + 4{y^2}} \right)}^{1/2}}} \right) \cr & {z_y}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^2} + 4{y^2}} \right)^{ - 1/2}}\frac{\partial }{{\partial y}}\left( {{x^2} + 4{y^2}} \right) \cr & {z_y}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^2} + 4{y^2}} \right)^{ - 1/2}}\left( {8y} \right) \cr & {z_x}\left( {x,y} \right) = \frac{{4y}}{{\sqrt {{x^2} + 4{y^2}} }} \cr & \cr & {\text{Evaluating }}{z_y}\left( {1,2} \right) \cr & {z_y}\left( {1,2} \right) = \frac{{4\left( 2 \right)}}{{\sqrt {1 + 4{{\left( 2 \right)}^2}} }} \cr & {z_y}\left( {1,2} \right) = \frac{8}{{\sqrt {17} }} \cr} $$