Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 33

$$\nabla z = \cos \left( {7{y^2} - 7xy} \right)\left( { - 7y{\bf{i}} + \left( {14y - 7x} \right){\bf{j}}} \right)$$

$$\eqalign{ & z = \sin \left( {7{y^2} - 7xy} \right) \cr & {\text{calculate the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr & {z_x} = \frac{\partial }{{\partial x}}\left[ {\sin \left( {7{y^2} - 7xy} \right)} \right] \cr & {\text{treat }}y{\text{ as a constant}} \cr & {z_x} = \cos \left( {7{y^2} - 7xy} \right)\frac{\partial }{{\partial x}}\left[ {7{y^2} - 7xy} \right] \cr & {z_x} = \cos \left( {7{y^2} - 7xy} \right)\left( { - 7y} \right) \cr & {z_x} = - 7y\cos \left( {7{y^2} - 7xy} \right) \cr & and \cr & {z_y} = \cos \left( {7{y^2} - 7xy} \right)\frac{\partial }{{\partial y}}\left[ {7{y^2} - 7xy} \right] \cr & {\text{treat }}x{\text{ as a constant}} \cr & {z_y} = \cos \left( {7{y^2} - 7xy} \right)\left( {14y - 7x} \right) \cr & {z_y} = \left( {14y - 7x} \right)\cos \left( {7{y^2} - 7xy} \right) \cr & \cr & {\text{The gradient of the function }}z\left( {x,y} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr & \nabla z = {z_x}{\bf{i}} + {z_y}{\bf{j}} \cr & {\text{}}{\text{ substituting the partial derivatives, we obtain}} \cr & \nabla z = - 7y\cos \left( {7{y^2} - 7xy} \right){\bf{i}} + \left( {14y - 7x} \right)\cos \left( {7{y^2} - 7xy} \right){\bf{j}} \cr & {\text{factoring}} \cr & \nabla z = \cos \left( {7{y^2} - 7xy} \right)\left( { - 7y{\bf{i}} + \left( {14y - 7x} \right){\bf{j}}} \right) \cr} $$