Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 20

$$\frac{{\partial z}}{{\partial u}} = 3 + \frac{{3v}}{u} - 4u{\text{ and }}\frac{{\partial z}}{{\partial v}} = 3\ln u + 2 + 2\ln v$$

$$\eqalign{ & z = 3x - 2y,\,\,\,\,\,\,x = u + v\ln u,\,\,\,\,y = {u^2} - v\ln v \cr & {\text{Calculate the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {3x - 2y} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial z}}{{\partial x}} = 3 \cr & \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {3x - 2y} \right] \cr & {\text{Treat }}x{\text{ as a constant}} \cr & \frac{{\partial z}}{{\partial y}} = - 2 \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial x}}{{\partial u}}{\text{,}}\,\,\frac{{\partial x}}{{\partial v}},\,\,\,\,\frac{{\partial y}}{{\partial u}}{\text{ and }}\frac{{\partial y}}{{\partial v}} \cr & \frac{{\partial x}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {u + v\ln u} \right] = 1 + v/u \cr & \frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {u + v\ln u} \right] = \ln u \cr & \frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {{u^2} - v\ln v} \right] = 2u \cr & \frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {{u^2} - v\ln v} \right] = - 1 - \ln v \cr & \cr & {\text{Use the theorem 13}}{\text{.5}}{\text{.2 }}\left( {{\text{see page 952}}} \right){\text{ to find }}\frac{{\partial z}}{{\partial u}}{\text{ and }}\frac{{\partial z}}{{\partial v}} \cr & \frac{{\partial z}}{{\partial u}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial u}} \cr & {\text{substitute the derivatives}} \cr & \frac{{\partial z}}{{\partial u}} = \left( 3 \right)\left( {1 + v/u} \right) + \left( { - 2} \right)\left( {2u} \right) \cr & {\text{simplifying}} \cr & \frac{{\partial z}}{{\partial u}} = 3 + 3v/u - 4u \cr & \frac{{\partial z}}{{\partial u}} = 3 + \frac{{3v}}{u} - 4u \cr & \cr & and \cr & \cr & \frac{{\partial z}}{{\partial v}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial v}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial v}} \cr & {\text{substitute the derivatives}} \cr & \frac{{\partial z}}{{\partial v}} = \left( 3 \right)\left( {\ln u} \right) + \left( { - 2} \right)\left( { - 1 - \ln v} \right) \cr & \frac{{\partial z}}{{\partial v}} = 3\ln u + 2 + 2\ln v \cr} $$