Answer
$$\eqalign{
& \nabla h\left( {x,y} \right) = \left\langle {\frac{x}{{\sqrt {2 + {x^2} + 2{y^2}} }},\frac{{2y}}{{\sqrt {2 + {x^2} + 2{y^2}} }}} \right\rangle \cr
& \nabla h\left( {2,1} \right) = \left\langle {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right\rangle \cr
& {D_{\bf{u}}}h\left( {2,1} \right) = \frac{{7\sqrt 2 }}{{10}} \cr} $$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = \sqrt {2 + {x^2} + 2{y^2}} ;{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} P\left( {2,1} \right);{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\bf{u}} = \left\langle {\frac{3}{5},\frac{4}{5}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{3}{5}} \right)}^2} + {{\left( {\frac{4}{5}} \right)}^2}} = 1,{\text{ }}{\bf{u}}{\text{ is a unit vector}} \cr
& {\text{The gradient of }}h\left( {x,y} \right){\text{ is}} \cr
& {h_x}\left( {x,y} \right) = \frac{{2x}}{{2\sqrt {2 + {x^2} + 2{y^2}} }} = \frac{x}{{\sqrt {2 + {x^2} + 2{y^2}} }} \cr
& {h_y}\left( {x,y} \right) = \frac{{4y}}{{2\sqrt {2 + {x^2} + 2{y^2}} }} = \frac{{2y}}{{\sqrt {2 + {x^2} + 2{y^2}} }} \cr
& \nabla h\left( {x,y} \right) = \frac{x}{{\sqrt {2 + {x^2} + 2{y^2}} }}{\bf{i}} + \frac{{2y}}{{\sqrt {2 + {x^2} + 2{y^2}} }}{\bf{j}} \cr
& \nabla h\left( {x,y} \right) = \left\langle {\frac{x}{{\sqrt {2 + {x^2} + 2{y^2}} }},\frac{{2y}}{{\sqrt {2 + {x^2} + 2{y^2}} }}} \right\rangle \cr
& {\text{Calculate the gradient at the point }}P\left( {2,1} \right) \cr
& \nabla h\left( {2,1} \right) = \left\langle {\frac{2}{{\sqrt {2 + {{\left( 2 \right)}^2} + 2{{\left( 1 \right)}^2}} }},\frac{{2\left( 1 \right)}}{{\sqrt {2 + {{\left( 2 \right)}^2} + 2{{\left( 1 \right)}^2}} }}} \right\rangle \cr
& \nabla h\left( {2,1} \right) = \left\langle {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right\rangle \cr
& {\text{Computing the directional derivatives of }}h{\text{ at }}\left( {2,1} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{3}{5},\frac{4}{5}} \right\rangle \cr
& {D_{\bf{u}}}h\left( {a,b} \right) = \nabla h\left( {a,b} \right)\cdot{\bf{u}} \cr
& {D_{\bf{u}}}h\left( {2,1} \right) = \left\langle {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right\rangle \cdot\left\langle {\frac{3}{5},\frac{4}{5}} \right\rangle \cr
& {D_{\bf{u}}}h\left( {2,1} \right) = \frac{{3\sqrt 2 }}{{10}} + \frac{{4\sqrt 2 }}{{10}} \cr
& {D_{\bf{u}}}h\left( {2,1} \right) = \frac{{7\sqrt 2 }}{{10}} \cr} $$
