Answer
$$\nabla h\left( {x,y} \right) = \tan y{\bf{i}} + x{\sec ^2}y{\bf{j}},{\text{ }}\sqrt {17} $$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = x\tan y,{\text{ point }}\left( {2,\frac{\pi }{4}} \right) \cr
& {\text{Find the partial derivatives }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right) \cr
& {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x\tan y} \right] \cr
& {h_x}\left( {x,y} \right) = \tan y \cr
& and \cr
& {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x\tan y} \right] \cr
& {h_y}\left( {x,y} \right) = x{\sec ^2}y \cr
& {\text{Calculate }}\nabla h\left( {x,y} \right) \cr
& \nabla h\left( {x,y} \right) = {h_x}\left( {x,y} \right){\bf{i}} + {h_y}\left( {x,y} \right){\bf{j}} \cr
& \nabla h\left( {x,y} \right) = \tan y{\bf{i}} + x{\sec ^2}y{\bf{j}} \cr
& {\text{At the given point the gradient is}} \cr
& \nabla h\left( {2,\frac{\pi }{4}} \right) = \tan \left( {\frac{\pi }{4}} \right){\bf{i}} + 2{\sec ^2}\left( {\frac{\pi }{4}} \right){\bf{j}} \cr
& \nabla h\left( {2,\frac{\pi }{4}} \right) = {\bf{i}} + 4{\bf{j}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}h\left( {x,y} \right){\text{ is }}\left\| {\nabla h\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla h\left( {2,\frac{\pi }{4}} \right)} \right\| = \left\| {{\bf{i}} + 4{\bf{j}}} \right\| \cr
& \left\| {\nabla h\left( {2,\frac{\pi }{4}} \right)} \right\| = \sqrt {1 + 16} \cr
& \left\| {\nabla h\left( {2,\frac{\pi }{4}} \right)} \right\| = \sqrt {17} \cr} $$
