Answer
$$\nabla {\text{ }}f\left( {x,y} \right) = \left\langle {8x - 2y, - 2x + 2y} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( { - 1, - 5} \right) = \left\langle {2, - 8} \right\rangle $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 4{x^2} - 2xy + {y^2};\,\,\,\,\,\,P\left( { - 1, - 5} \right) \cr
& {\text{compute the gradient of }}f\left( {x,y} \right),{\text{ }} \cr
& {\text{first find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right).{\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2} - 2xy + {y^2}} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = 8x - 2y \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{x^2} - 2xy + {y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = - 2x + 2y \cr
& {\text{using }}\nabla {\text{ }}f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr
& {\text{derivatives}} \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left( {8x - 2y} \right){\bf{i}} + \left( { - 2x + 2y} \right){\bf{j}} \cr
& or \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {8x - 2y, - 2x + 2y} \right\rangle \cr
& {\text{evaluate }}\nabla {\text{ }}f\left( {x,y} \right){\text{ at the point }}P \cr
& \nabla {\text{ }}f\left( { - 1, - 5} \right) = \left( {8\left( { - 1} \right) - 2\left( { - 5} \right)} \right){\bf{i}} - \left( { - 2\left( { - 1} \right) + 2\left( { - 5} \right)} \right){\bf{j}} \cr
& \nabla {\text{ }}f\left( { - 1, - 5} \right) = \left( { - 8 + 10} \right){\bf{i}} - \left( {2 - 10} \right){\bf{j}} \cr
& \nabla {\text{ }}f\left( { - 1, - 5} \right) = 2{\bf{i}} - 8{\bf{j}} \cr
& or \cr
& \nabla {\text{ }}f\left( { - 1, - 5} \right) = \left\langle {2, - 8} \right\rangle \cr
& \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {8x - 2y, - 2x + 2y} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( { - 1, - 5} \right) = \left\langle {2, - 8} \right\rangle \cr} $$
