Answer
$${D_u}h\left( {1,\frac{\pi }{2}} \right) = - e$$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = {e^x}\sin y,{\text{ }}P\left( {1,\frac{\pi }{2}} \right),{\text{ }}{\bf{v}} = - {\bf{i}} \cr
& {\text{Calculate }}\nabla h\left( {x,y} \right) \cr
& \nabla h\left( {x,y} \right) = {h_x}\left( {x,y} \right){\bf{i}} + {h_y}\left( {x,y} \right){\bf{j}} \cr
& {h_x}\left( {x,y} \right) = {e^x}\sin y \cr
& {h_y}\left( {x,y} \right) = {e^x}\cos y \cr
& \nabla h\left( {x,y} \right) = {e^x}\sin y{\bf{i}} + {e^x}\cos y{\bf{j}} \cr
& {\text{Evaluate }}\nabla h\left( {1,\frac{\pi }{2}} \right) \cr
& \nabla h\left( {1,\frac{\pi }{2}} \right) = {e^1}\sin \left( {\frac{\pi }{2}} \right){\bf{i}} + {e^1}\cos \left( {\frac{\pi }{2}} \right){\bf{j}} \cr
& \nabla h\left( {1,\frac{\pi }{2}} \right) = e{\bf{i}} + 0{\bf{j}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| { - {\bf{i}}} \right| = 1 \cr
& {\bf{v}}{\text{ is a unit vector}} \cr
& {\text{The directional derivative at }}\left( {0,0} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}h\left( {x,y} \right) = \nabla h\left( {x,y} \right) \cdot {\bf{v}} \cr
& {D_u}h\left( {1,\frac{\pi }{2}} \right) = \nabla h\left( {1,\frac{\pi }{2}} \right) \cdot \left( { - {\bf{i}} + 0{\bf{j}}} \right) \cr
& {D_u}h\left( {1,\frac{\pi }{2}} \right) = \left( {e{\bf{i}} + 0{\bf{j}}} \right) \cdot \left( { - {\bf{i}} + 0{\bf{j}}} \right) \cr
& {D_u}h\left( {1,\frac{\pi }{2}} \right) = - e \cr} $$
