Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 28

$$\frac{{\partial u}}{{\partial x}} = {\left( {4y + 1} \right)^2}\left( {x + 2x\ln x{y^3}} \right){\text{ and }}\frac{{\partial u}}{{\partial y}} = {\left( {4y + 1} \right)^3}\left( {\frac{3}{y}\left( {4y + 1} \right) + 8\ln \left( {x{y^3}} \right)} \right)$$

$$\eqalign{ & u = r{s^2}\ln t,\,\,\,\,\,r = {x^2},\,\,\,\,s = 4y + 1,\,\,\,\,t = x{y^3} \cr & \cr & {\text{Substitute }}r = {x^2},\,\,\,s = 4y + 1,\,\,\,t = x{y^3}{\text{ into the given function }}u = r{s^2}\ln t \cr & u = \left( {{x^2}} \right){\left( {4y + 1} \right)^2}\ln \left( {x{y^3}} \right) \cr & {\text{Simplify}} \cr & u = {x^2}{\left( {4y + 1} \right)^2}\ln \left( {x{y^3}} \right) \cr & \cr & {\text{Find the partial derivatives }}\frac{{\partial u}}{{\partial x}}{\text{ and }}\frac{{\partial u}}{{\partial y}} \cr & \frac{{\partial u}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{x^2}{{\left( {4y + 1} \right)}^2}\ln \left( {x{y^3}} \right)} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial u}}{{\partial x}} = {\left( {4y + 1} \right)^2}\frac{\partial }{{\partial x}}\left[ {{x^2}\ln \left( {x{y^3}} \right)} \right] \cr & {\text{use the product rule}} \cr & \frac{{\partial u}}{{\partial x}} = {\left( {4y + 1} \right)^2}\left( {{x^2}\frac{\partial }{{\partial x}}\left[ {\ln \left( {x{y^3}} \right)} \right] + \ln \left( {x{y^3}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2}} \right]} \right) \cr & \frac{{\partial u}}{{\partial x}} = {\left( {4y + 1} \right)^2}\left( {{x^2}\left( {\frac{1}{x}} \right) + \ln \left( {x{y^3}} \right)\left( {2x} \right)} \right) \cr & \frac{{\partial u}}{{\partial x}} = {\left( {4y + 1} \right)^2}\left( {x + 2x\ln x{y^3}} \right) \cr & \cr & and \cr & \cr & \frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^2}{{\left( {4y + 1} \right)}^2}\ln \left( {x{y^3}} \right)} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial u}}{{\partial y}} = {x^2}\frac{\partial }{{\partial y}}\left[ {{{\left( {4y + 1} \right)}^2}\ln \left( {x{y^3}} \right)} \right] \cr & {\text{use the product rule}} \cr & \frac{{\partial u}}{{\partial y}} = {\left( {4y + 1} \right)^2}\left( {{{\left( {4y + 1} \right)}^2}\frac{\partial }{{\partial y}}\left[ {\ln \left( {x{y^3}} \right)} \right] + \ln \left( {x{y^3}} \right)\frac{\partial }{{\partial y}}\left[ {{{\left( {4y + 1} \right)}^2}} \right]} \right) \cr & \frac{{\partial u}}{{\partial y}} = {\left( {4y + 1} \right)^2}\left( {{{\left( {4y + 1} \right)}^2}\left( {\frac{{3x{y^2}}}{{x{y^3}}}} \right) + 8\ln \left( {x{y^3}} \right)\left( {4y + 1} \right)} \right) \cr & \frac{{\partial u}}{{\partial y}} = {\left( {4y + 1} \right)^3}\left( {\frac{3}{y}\left( {4y + 1} \right) + 8\ln \left( {x{y^3}} \right)} \right) \cr} $$