Answer
$$\nabla {\text{ }}f\left( {x,y} \right) = \left\langle {6x, - 10y} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {2, - 1} \right) = \left\langle {12,10} \right\rangle $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 2 + 3{x^2} - 5{y^2};\,\,\,\,\,\,P\left( {2, - 1} \right) \cr
& {\text{compute the gradient of }}f\left( {x,y} \right),{\text{ }} \cr
& {\text{first find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right).{\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2 + 3{x^2} - 5{y^2}} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = 6x \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2 + 3{x^2} - 5{y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = - 10y \cr
& {\text{using }}\nabla {\text{ }}f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr
& {\text{derivatives}} \cr
& \nabla {\text{ }}f\left( {x,y} \right) = 6x{\bf{i}} - 10y{\bf{j}} \cr
& or \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {6x, - 10y} \right\rangle \cr
& {\text{evaluate }}\nabla {\text{ }}f\left( {x,y} \right){\text{ at the point }}P \cr
& \nabla {\text{ }}f\left( {2, - 1} \right) = 6\left( 2 \right){\bf{i}} - 10\left( { - 1} \right){\bf{j}} \cr
& \nabla {\text{ }}f\left( {2, - 1} \right) = 12{\bf{i}} + 10{\bf{j}} \cr
& or \cr
& \nabla {\text{ }}f\left( {2, - 1} \right) = \left\langle {12,10} \right\rangle \cr
& \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {6x, - 10y} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {2, - 1} \right) = \left\langle {12,10} \right\rangle \cr} $$
