Answer
$$\nabla {\text{ }}g\left( {x,y} \right) = \left\langle {2x - 8xy - 8{y^2}, - 4{x^2} - 16xy} \right\rangle {\text{ and }}\nabla g\left( { - 1,2} \right) = \left\langle { - 18,28} \right\rangle $$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = {x^2} - 4{x^2}y - 8x{y^2};\,\,\,\,\,\,P\left( { - 1,2} \right) \cr
& {\text{compute the gradient of }}g\left( {x,y} \right),{\text{ }} \cr
& {\text{first find the partial derivatives }}{g_x}\left( {x,y} \right){\text{ and }}{g_y}\left( {x,y} \right).{\text{ then}} \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - 4{x^2}y - 8x{y^2}} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& {g_x}\left( {x,y} \right) = 2x - 8xy - 8{y^2} \cr
& and \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - 4{x^2}y - 8x{y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& {g_y}\left( {x,y} \right) = - 4{x^2} - 16xy \cr
& {\text{using }}\nabla {\text{ }}g\left( {x,y} \right) = {g_x}\left( {x,y} \right){\bf{i}} + {g_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr
& {\text{derivatives}} \cr
& \nabla {\text{ }}g\left( {x,y} \right) = \left( {2x - 8xy - 8{y^2}} \right){\bf{i}} + \left( { - 4{x^2} - 16xy} \right){\bf{j}} \cr
& or \cr
& \nabla {\text{ }}g\left( {x,y} \right) = \left\langle {2x - 8xy - 8{y^2}, - 4{x^2} - 16xy} \right\rangle \cr
& {\text{evaluate }}\nabla {\text{ }}g\left( {x,y} \right){\text{ at the point }}P \cr
& \nabla g\left( { - 1,2} \right) = \left( {2\left( { - 1} \right) - 8\left( { - 1} \right)\left( 2 \right) - 8{{\left( 2 \right)}^2}} \right){\bf{i}} + \left( { - 4{{\left( { - 1} \right)}^2} - 16\left( { - 1} \right)\left( 2 \right)} \right){\bf{j}} \cr
& \nabla g\left( { - 1,2} \right) = \left( { - 2 + 16 - 32} \right){\bf{i}} + \left( { - 4 + 32} \right){\bf{j}} \cr
& \nabla g\left( { - 1,2} \right) = - 18{\bf{i}} + 28{\bf{j}} \cr
& or \cr
& \nabla g\left( { - 1,2} \right) = \left\langle { - 18,28} \right\rangle \cr
& \cr
& \nabla {\text{ }}g\left( {x,y} \right) = \left\langle {2x - 8xy - 8{y^2}, - 4{x^2} - 16xy} \right\rangle {\text{ and }}\nabla g\left( { - 1,2} \right) = \left\langle { - 18,28} \right\rangle \cr} $$
