Chapter 12 - Functions of Several Veriables - 12.6 Directional Derivatives and the Gradient - 12.6 Exercises - Page 925: 16

$$\nabla h\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\left\langle {2x,4y} \right\rangle {\text{ and }}\nabla h\left( {2, - 3} \right) = \frac{4}{{23}}\left\langle {1, - 3} \right\rangle $$

$$\eqalign{ & h\left( {x,y} \right) = \ln \left( {1 + {x^2} + 2{y^2}} \right);\,\,\,\,\,\,P\left( {2, - 3} \right) \cr & {\text{compute the gradient of }}h\left( {x,y} \right),{\text{ }} \cr & {\text{first find the partial derivatives }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right).{\text{ then}} \cr & {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left( {1 + {x^2} + 2{y^2}} \right)} \right] \cr & {\text{treat }}y{\text{ as a constant}} \cr & {h_x}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\frac{\partial }{{\partial x}}\left[ {1 + {x^2} + 2{y^2}} \right] \cr & {h_x}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\left( {2x} \right) \cr & {h_x}\left( {x,y} \right) = \frac{{2x}}{{1 + {x^2} + 2{y^2}}} \cr & and \cr & {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left( {1 + {x^2} + 2{y^2}} \right)} \right] \cr & {\text{treat }}x{\text{ as a constant}} \cr & {h_y}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\frac{\partial }{{\partial y}}\left[ {1 + {x^2} + 2{y^2}} \right] \cr & {h_y}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\left( {4y} \right) \cr & {h_y}\left( {x,y} \right) = \frac{{4y}}{{1 + {x^2} + 2{y^2}}} \cr & {\text{using }}\nabla {\text{ }}h\left( {x,y} \right) = {h_x}\left( {x,y} \right){\bf{i}} + {h_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr & {\text{derivatives}} \cr & \nabla {\text{ }}h\left( {x,y} \right) = \frac{{2x}}{{1 + {x^2} + 2{y^2}}}{\bf{i}} + \frac{{4y}}{{1 + {x^2} + 2{y^2}}}{\bf{j}} \cr & or \cr & \nabla h\left( {x,y} \right) = \left\langle {\frac{{2x}}{{1 + {x^2} + 2{y^2}}},\frac{{4y}}{{1 + {x^2} + 2{y^2}}}} \right\rangle \cr & \nabla h\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\left\langle {2x,4y} \right\rangle \cr & {\text{evaluate }}\nabla {\text{ }}h\left( {x,y} \right){\text{ at the point }}P \cr & \nabla h\left( {2, - 3} \right) = \frac{1}{{1 + {{\left( 2 \right)}^2} + 2{{\left( { - 3} \right)}^2}}}\left\langle {2\left( 2 \right),4\left( { - 3} \right)} \right\rangle \cr & \nabla h\left( {2, - 3} \right) = \frac{1}{{23}}\left\langle {4, - 12} \right\rangle \cr & \nabla h\left( {2, - 3} \right) = \frac{4}{{23}}\left\langle {1, - 3} \right\rangle \cr & \cr & \nabla h\left( {x,y} \right) = \frac{1}{{1 + {x^2} + 2{y^2}}}\left\langle {2x,4y} \right\rangle {\text{ and }}\nabla h\left( {2, - 3} \right) = \frac{4}{{23}}\left\langle {1, - 3} \right\rangle \cr} $$