Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 32

$$\eqalign{ & {z_{xx}}\left( {x,y} \right) = - \frac{{3\left( {y + 1} \right)}}{{{x^2}}},\,\,\,\,{z_{yy}}\left( {x,y} \right) = \frac{{y - 1}}{{{y^2}}} \cr & {z_{xy}}\left( {x,y} \right) = {z_{yx}}\left( {x,y} \right) = \frac{3}{x} \cr} $$

$$\eqalign{ & z = \left( {y + 1} \right)\ln \left| {{x^3}y} \right| \cr & {\text{Find }}{z_x}\left( {x,y} \right){\text{ and }}{z_y}\left( {x,y} \right) \cr & {z_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\left( {y + 1} \right)\ln \left| {{x^3}y} \right|} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}} \cr & {z_x}\left( {x,y} \right) = \left( {y + 1} \right)\frac{\partial }{{\partial x}}\left[ {\ln \left| {{x^3}y} \right|} \right] \cr & {z_x}\left( {x,y} \right) = \left( {y + 1} \right)\left( {\frac{1}{{{x^3}y}}} \right)\frac{\partial }{{\partial x}}\left[ {{x^3}y} \right] \cr & {z_x}\left( {x,y} \right) = \left( {y + 1} \right)\left( {\frac{1}{{{x^3}y}}} \right)\left( {3{x^2}y} \right) \cr & {z_x}\left( {x,y} \right) = \frac{{3\left( {y + 1} \right)}}{x} \cr & \cr & {z_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\left( {y + 1} \right)\ln \left| {{x^3}y} \right|} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then by the power rule}} \cr & {z_y}\left( {x,y} \right) = \left( {y + 1} \right)\frac{\partial }{{\partial y}}\left[ {\ln \left| {{x^3}y} \right|} \right] + \ln \left| {{x^3}y} \right|\frac{\partial }{{\partial y}}\left[ {\left( {y + 1} \right)} \right] \cr & {z_y}\left( {x,y} \right) = \left( {y + 1} \right)\left( {\frac{{{x^3}}}{{{x^3}y}}} \right) + \ln \left| {{x^3}y} \right|\left( 1 \right) \cr & {z_y}\left( {x,y} \right) = \frac{{y + 1}}{y} + \ln \left| {{x^3}y} \right| \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {z_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{3\left( {y + 1} \right)}}{x}} \right] \cr & {z_{xy}}\left( {x,y} \right) = \frac{3}{x}\frac{\partial }{{\partial y}}\left[ {y + 1} \right] \cr & {z_{xy}}\left( {x,y} \right) = \frac{3}{x} \cr & \cr & {z_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{y + 1}}{y} + \ln \left| {{x^3}y} \right|} \right] \cr & {z_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{y + 1}}{y}} \right] + \frac{\partial }{{\partial x}}\left[ {\ln \left| {{x^3}y} \right|} \right] \cr & {z_{yx}}\left( {x,y} \right) = \frac{{3{x^2}y}}{{{x^3}y}} \cr & {z_{yx}}\left( {x,y} \right) = \frac{3}{x} \cr & \cr & {z_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{3\left( {y + 1} \right)}}{x}} \right] \cr & {z_{xx}}\left( {x,y} \right) = 3\left( {y + 1} \right)\frac{\partial }{{\partial x}}\left[ {\frac{1}{x}} \right] \cr & {z_{xx}}\left( {x,y} \right) = 3\left( {y + 1} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr & {z_{xx}}\left( {x,y} \right) = - \frac{{3\left( {y + 1} \right)}}{{{x^2}}} \cr & {\text{and}} \cr & {z_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{y + 1}}{y} + \ln \left| {{x^3}y} \right|} \right] \cr & {z_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {1 + \frac{1}{y} + \ln \left| {{x^3}y} \right|} \right] \cr & {z_{yy}}\left( {x,y} \right) = - \frac{1}{{{y^2}}} + \frac{{{x^3}}}{{{x^3}y}} \cr & {z_{yy}}\left( {x,y} \right) = - \frac{1}{{{y^2}}} + \frac{1}{y} \cr & {z_{yy}}\left( {x,y} \right) = \frac{{y - 1}}{{{y^2}}} \cr} $$