Chapter 12 - Functions of Several Veriables - 12.4 Partial Derivatives - 12.4 Exercises - Page 904: 34

$$\eqalign{ & {f_{xx}}\left( {x,y} \right) = - {y^2}\cos xy,\,\,\,\,\,\,\,\,{\text{ }}{f_{yy}}\left( {x,y} \right) = - {x^2}\cos xy \cr & {f_{xy}}\left( {x,y} \right) = - xy\cos xy - \sin xy{\text{ and }}{f_{yx}}\left( {x,y} \right) = - xy\cos xy - \sin xy \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = \cos xy \cr & {\text{Find the first 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[ {\cos xy} \right] \cr & {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr & {f_x}\left( {x,y} \right) = - \sin xy\frac{\partial }{{\partial x}}\left[ {xy} \right] \cr & {f_x}\left( {x,y} \right) = - \sin xy\left( y \right) \cr & {f_x}\left( {x,y} \right) = - y\sin xy \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\cos xy} \right] \cr & {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr & {f_y}\left( {x,y} \right) = - \sin xy\frac{\partial }{{\partial y}}\left[ {xy} \right] \cr & {f_y}\left( {x,y} \right) = - \sin xy\left( x \right) \cr & {f_y}\left( {x,y} \right) = - x\sin xy \cr & \cr & {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - y\sin xy} \right] \cr & {\text{use product rule}} \cr & {f_{xy}}\left( {x,y} \right) = - y\frac{\partial }{{\partial y}}\left[ {\sin xy} \right] + \sin xy\frac{\partial }{{\partial y}}\left[ { - y} \right] \cr & {f_{xy}}\left( {x,y} \right) = - y\left( {x\cos xy} \right) + \sin xy\left( { - 1} \right) \cr & {f_{xy}}\left( {x,y} \right) = - xy\cos xy - \sin xy \cr & and \cr & {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - x\sin xy} \right] \cr & {\text{use product rule}} \cr & {\text{ }}{f_{yx}}\left( {x,y} \right) = - x\frac{\partial }{{\partial x}}\left[ {\sin xy} \right] + \sin xy\frac{\partial }{{\partial x}}\left[ { - x} \right] \cr & {\text{ }}{f_{yx}}\left( {x,y} \right) = - x\left( {y\cos xy} \right) + \sin xy\left( { - 1} \right) \cr & {\text{ }}{f_{yx}}\left( {x,y} \right) = - xy\cos xy - \sin xy \cr & \cr & {\text{Find the second partial derivatives }}{f_{xx}}\left( {x,y} \right){\text{ and }}{f_{yy}}\left( {x,y} \right){\text{ then}} \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - y\sin xy} \right] \cr & {f_{xx}}\left( {x,y} \right) = - y\frac{\partial }{{\partial x}}\left[ {\sin xy} \right] \cr & {f_{xx}}\left( {x,y} \right) = - y\left( {y\cos xy} \right) \cr & {f_{xx}}\left( {x,y} \right) = - {y^2}\cos xy \cr & and \cr & {\text{ }}{f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - x\sin xy} \right] \cr & {\text{ }}{f_{yy}}\left( {x,y} \right) = - x\frac{\partial }{{\partial y}}\left[ {\sin xy} \right] \cr & {\text{ }}{f_{yy}}\left( {x,y} \right) = - x\left( {x\cos xy} \right) \cr & {\text{ }}{f_{yy}}\left( {x,y} \right) = - {x^2}\cos xy \cr} $$