Answer
$$12$$
Work Step by Step
$$\eqalign{ & f\left( {x,y} \right) = 13{e^{xy}};\,\,\,\,\,\,P\left( {1,0} \right);\,\,\,\,\,\left\langle {5,12} \right\rangle \cr & {\bf{u}} = \frac{{\left\langle {5,12} \right\rangle }}{{\sqrt {{5^2} + {{12}^2}} }} = \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr & {\bf{u}}{\text{ is a unit vector in the direction of }}\left\langle {5,12} \right\rangle \cr & {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr & {f_x}\left( {x,y} \right) = 13y{e^{xy}} \cr & {f_y}\left( {x,y} \right) = 13x{e^{xy}} \cr & \nabla f\left( {x,y} \right) = 13y{e^{xy}}{\bf{i}} + 13x{e^{xy}}{\bf{j}} \cr & \nabla f\left( {x,y} \right) = 13{e^{xy}}\left( {{y\bf{i}} + {x\bf{j}}} \right) \cr & \cr & {\text{Calculate the gradient at the point }}P\left( {1,0} \right) \cr & \nabla f\left( {1,0} \right) = 13{e^0}\left( {{0\bf{i}} + {\bf{j}}} \right) \cr & \nabla f\left( {1,0} \right) = \left\langle {0,13} \right\rangle \cr & \cr & {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {1,0} \right) \cr & \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr & {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {1,0} \right) = \left\langle {0,13} \right\rangle \cdot \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr & {D_{\bf{u}}}f\left( {1,0} \right) =12 \cr} $$
