Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 36

$$\nabla z = \frac{1}{{{{\left( {x + 8y} \right)}^2}}}\left( {48y{e^{3y}}{\bf{i}} + \left( {18{x^2}{e^{3y}} + 144xy{e^{3y}} - 48x{e^{3y}}} \right){\bf{j}}} \right)$$

$$\eqalign{ & z = \frac{{6x{e^{3y}}}}{{x + 8y}} \cr & \cr & {\text{calculate the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr & {z_x} = \frac{\partial }{{\partial x}}\left[ {\frac{{6x{e^{3y}}}}{{x + 8y}}} \right] \cr & {\text{treat }}y{\text{ as a constant}}{\text{, and use the quotient rule for derivatives}} \cr & {z_x} = \frac{{\left( {x + 8y} \right)\frac{\partial }{{\partial x}}\left[ {6x{e^{3y}}} \right] - \left( {6x{e^{3y}}} \right)\frac{\partial }{{\partial x}}\left[ {x + 8y} \right]}}{{{{\left( {x + 8y} \right)}^2}}} \cr & {z_x} = \frac{{\left( {x + 8y} \right)\left( {6{e^{3y}}} \right) - \left( {6x{e^{3y}}} \right)\left( 1 \right)}}{{{{\left( {x + 8y} \right)}^2}}} \cr & {z_x} = \frac{{6x{e^{3y}} + 48y{e^{3y}} - 6x{e^{3y}}}}{{{{\left( {x + 8y} \right)}^2}}} \cr & {z_x} = \frac{{48y{e^{3y}}}}{{{{\left( {x + 8y} \right)}^2}}} \cr & and \cr & {z_y} = \frac{\partial }{{\partial y}}\left[ {\frac{{6x{e^{3y}}}}{{x + 8y}}} \right] \cr & {\text{treat }}y{\text{ as a constant}}{\text{, and use the quotient rule for derivatives}} \cr & {z_y} = \frac{{\left( {x + 8y} \right)\frac{\partial }{{\partial y}}\left[ {6x{e^{3y}}} \right] - 6x{e^{3y}}\frac{\partial }{{\partial y}}\left[ {x + 8y} \right]}}{{{{\left( {x + 8y} \right)}^2}}} \cr & {z_y} = \frac{{\left( {x + 8y} \right)\left( {18x{e^{3y}}} \right) - 6x{e^{3y}}\left( 8 \right)}}{{{{\left( {x + 8y} \right)}^2}}} \cr & {z_y} = \frac{{18{x^2}{e^{3y}} + 144xy{e^{3y}} - 48x{e^{3y}}}}{{{{\left( {x + 8y} \right)}^2}}} \cr & \cr & {\text{The gradient of the function }}z\left( {x,y} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr & \nabla z = {z_x}{\bf{i}} + {z_y}{\bf{j}} \cr & {\text{substituting the partial derivatives, we obtain}} \cr & \nabla z = \frac{{48y{e^{3y}}}}{{{{\left( {x + 8y} \right)}^2}}}{\bf{i}} + \frac{{18{x^2}{e^{3y}} + 144xy{e^{3y}} - 48x{e^{3y}}}}{{{{\left( {x + 8y} \right)}^2}}}{\bf{j}} \cr & {\text{factoring}} \cr & \nabla z = \frac{1}{{{{\left( {x + 8y} \right)}^2}}}\left( {48y{e^{3y}}{\bf{i}} + \left( {18{x^2}{e^{3y}} + 144xy{e^{3y}} - 48x{e^{3y}}} \right){\bf{j}}} \right) \cr} $$