Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 27

$$\frac{{\partial z}}{{\partial x}} = \frac{{{x^3}}}{{{y^{3/5}} + x}} + 3{x^2}\ln \left( {1 + x{y^{ - 3/5}}} \right){\text{ and }}\frac{{\partial z}}{{\partial y}} = \frac{{ - \frac{3}{5}{x^4}}}{{{y^{8/5}} + xy}}$$

$$\eqalign{ & z = {x^3}\ln \left( {1 + x{y^{ - 3/5}}} \right) \cr & {\text{Calculate the partial derivative }}\frac{{\partial z}}{{\partial x}}{\text{ }} \cr & {\text{Fifferentiating both sides with respect to }}x \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{x^3}\ln \left( {1 + x{y^{ - 3/5}}} \right)} \right] \cr & {\text{Applying the product rule for derivatives}} \cr & \frac{{\partial z}}{{\partial x}} = {x^3}\frac{\partial }{{\partial x}}\left[ {\ln \left( {1 + x{y^{ - 3/5}}} \right)} \right] + \ln \left( {1 + x{y^{ - 3/5}}} \right)\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] \cr & {\text{Solving the derivatives}} \cr & \frac{{\partial z}}{{\partial x}} = {x^3}\left( {\frac{{{y^{ - 3/5}}}}{{1 + x{y^{ - 3/5}}}}} \right) + \ln \left( {1 + x{y^{ - 3/5}}} \right)\left( {3{x^2}} \right) \cr & {\text{Simplifying the resultant derivative}} \cr & \frac{{\partial z}}{{\partial x}} = {x^3}\left( {\frac{{{y^{ - 3/5}}}}{{1 + x{y^{ - 3/5}}}}} \right) + 3{x^2}\ln \left( {1 + x{y^{ - 3/5}}} \right) \cr & \frac{{\partial z}}{{\partial x}} = \frac{{{x^3}{y^{ - 3/5}}}}{{1 + x{y^{ - 3/5}}}} + 3{x^2}\ln \left( {1 + x{y^{ - 3/5}}} \right) \cr & \frac{{\partial z}}{{\partial x}} = \frac{{{y^{3/5}}}}{{{y^{3/5}}}}\left( {\frac{{{x^3}{y^{ - 3/5}}}}{{1 + x{y^{ - 3/5}}}}} \right) + 3{x^2}\ln \left( {1 + x{y^{ - 3/5}}} \right) \cr & \frac{{\partial z}}{{\partial x}} = \frac{{{x^3}}}{{{y^{3/5}} + x}} + 3{x^2}\ln \left( {1 + x{y^{ - 3/5}}} \right) \cr & \cr & and \cr & {\text{Calculate the partial derivative }}\frac{{\partial z}}{{\partial y}}{\text{ }} \cr & {\text{differentiating both sides with respect to }}y \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^3}\ln \left( {1 + x{y^{ - 3/5}}} \right)} \right] \cr & {\text{Applying the constant multiple rule}} \cr & \frac{{\partial z}}{{\partial y}} = {x^3}\frac{\partial }{{\partial y}}\left[ {\ln \left( {1 + x{y^{ - 3/5}}} \right)} \right] \cr & {\text{Solving the derivative}} \cr & \frac{{\partial z}}{{\partial y}} = {x^3}\left( {\frac{{ - \frac{3}{5}x{y^{ - 8/5}}}}{{1 + x{y^{ - 3/5}}}}} \right) \cr & {\text{Simplifying the resultant derivative}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{ - \frac{3}{5}{x^4}{y^{ - 8/5}}}}{{1 + x{y^{ - 3/5}}}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{{y^{8/5}}\left( { - \frac{3}{5}{x^4}{y^{ - 8/5}}} \right)}}{{{y^{8/5}}\left( {1 + x{y^{ - 3/5}}} \right)}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{ - \frac{3}{5}{x^4}}}{{{y^{8/5}} + xy}} \cr & \cr & {\text{The partial derivatives are:}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{{x^3}}}{{{y^{3/5}} + x}} + 3{x^2}\ln \left( {1 + x{y^{ - 3/5}}} \right){\text{ and }}\frac{{\partial z}}{{\partial y}} = \frac{{ - \frac{3}{5}{x^4}}}{{{y^{8/5}} + xy}} \cr} $$