Answer
$$ - \frac{2}{{\sqrt 5 }}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {4 - {x^2} - 2y} ;\,\,\,\,\,\,P\left( {2, - 2} \right);\,\,\,\,\,\left\langle {\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \cr
& {\text{Let }}{\bf{u}} = \,\left\langle {\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{1}{{\sqrt 5 }}} \right)}^2} + {{\left( {\frac{2}{{\sqrt 5 }}} \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) = \frac{{ - 2x}}{{2\sqrt {4 - {x^2} - 2y} }} = - \frac{x}{{\sqrt {4 - {x^2} - 2y} }} \cr
& {f_y}\left( {x,y} \right) = \frac{{ - 2}}{{2\sqrt {4 - {x^2} - 2y} }} = - \frac{1}{{\sqrt {4 - {x^2} - 2y} }} \cr
& \nabla f\left( {x,y} \right) = - \frac{x}{{\sqrt {4 - {x^2} - 2y} }}{\bf{i}} - \frac{1}{{\sqrt {4 - {x^2} - 2y} }}{\bf{j}} \cr
& \nabla f\left( {x,y} \right) = - \frac{1}{{\sqrt {4 - {x^2} - 2y} }}\left( {x{\bf{i}} + {\bf{j}}} \right) \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( {2, - 2} \right) \cr
& \nabla f\left( {2, - 2} \right) = - \frac{1}{{\sqrt {4 - {{\left( 2 \right)}^2} - 2\left( { - 2} \right)} }}\left( {2{\bf{i}} + {\bf{j}}} \right) \cr
& \nabla f\left( {2, - 2} \right) = - \frac{1}{2}\left( {2{\bf{i}} + {\bf{j}}} \right) \cr
& \nabla f\left( {2, - 2} \right) = \left\langle { - 1, - \frac{1}{2}} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {2, - 2} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle { - 1, - \frac{1}{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( {2, - 2} \right) = \left\langle { - 1, - \frac{1}{2}} \right\rangle \cdot \left\langle {\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {2, - 2} \right) = - \frac{1}{{\sqrt 5 }} - \frac{1}{{\sqrt 5 }} \cr
& {D_{\bf{u}}}f\left( {2, - 2} \right) = - \frac{2}{{\sqrt 5 }} \cr} $$
