Chapter 14 - Section 14.6 - Directional Derivatives and the Gradient Vector - 14.6 Exercise - Page 1005: 7

$\frac{{5\sqrt 2 }}{{74}}$

$$\eqalign{ & f\left( {x,y} \right) = \arctan \left( {xy} \right),{\text{ }}\left( {2, - 3} \right),{\text{ }}\theta = \frac{{3\pi }}{4} \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\arctan \left( {xy} \right)} \right] \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {xy} \right)}^2}}}\frac{\partial }{{\partial x}}\left[ {xy} \right] \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + {x^2}{y^2}}}\left( y \right) \cr & {f_x}\left( {x,y} \right) = \frac{y}{{1 + {x^2}{y^2}}} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\arctan \left( {xy} \right)} \right] \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {xy} \right)}^2}}}\frac{\partial }{{\partial y}}\left[ {xy} \right] \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + {x^2}{y^2}}}\left( x \right) \cr & {f_y}\left( {x,y} \right) = \frac{x}{{1 + {x^2}{y^2}}} \cr & \cr & {\text{The gradient of the function is given by }} \cr & \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr & {\text{Substituting the partial derivatives}}{\text{, we obtain}} \cr & \nabla f\left( {x,y} \right) = \frac{y}{{1 + {x^2}{y^2}}}{\bf{i}} + \frac{x}{{1 + {x^2}{y^2}}}{\bf{j}} \cr & {\text{Calculate the gradient at the point }}\left( {2, - 3} \right) \cr & \nabla f\left( {2, - 3} \right) = \frac{{ - 3}}{{1 + {{\left( 2 \right)}^2}{{\left( { - 3} \right)}^2}}}{\bf{i}} + \frac{2}{{1 + {{\left( 2 \right)}^2}{{\left( { - 3} \right)}^2}}}{\bf{j}} \cr & \nabla f\left( {2, - 3} \right) = - \frac{3}{{37}}{\bf{i}} + \frac{2}{{37}}{\bf{j}} \cr & \cr & {\text{The directional derivative of }}f\left( {x,y} \right){\text{ at }}\left( {{x_0},{y_0}} \right){\text{ in the}} \cr & {\text{direction of a unit vector }}{\bf{u}} = \cos \theta {\bf{i}} + \sin \theta {\bf{j}}{\text{ is}} \cr & {D_u}f\left( {x,y} \right) = {f_x}\left( {x,y} \right)\cos \theta + {f_y}\left( {x,y} \right)\sin \theta \cr & {\text{At the point }}\left( {2, - 3} \right){\text{ and }}\theta = \frac{{3\pi }}{4}{\text{ we obtain}} \cr & {D_u}f\left( {2, - 3} \right) = {f_x}\left( {2, - 3} \right)\cos \left( {\frac{{3\pi }}{4}} \right) + {f_y}\left( {2, - 3} \right)\sin \left( {\frac{{3\pi }}{4}} \right) \cr & {D_u}f\left( {2, - 3} \right) = \left( { - \frac{3}{{37}}} \right)\cos \left( {\frac{{3\pi }}{4}} \right) + \left( {\frac{2}{{37}}} \right)\sin \left( {\frac{{3\pi }}{4}} \right) \cr & {\text{Simplifying}} \cr & {D_u}f\left( {2, - 3} \right) = \frac{{3\sqrt 2 }}{{74}} + \frac{{\sqrt 2 }}{{37}} \cr & {D_u}f\left( {2, - 3} \right) = \frac{{5\sqrt 2 }}{{74}} \cr} $$