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