Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 17

$$\frac{{\partial z}}{{\partial u}} = 24{u^2}{v^2} - 16u{v^3} - 2v + 3{\text{ and }}\frac{{\partial z}}{{\partial v}} = 16{u^3}v - 24{u^2}{v^2} - 2u - 3$$

$$\eqalign{ & z = 8{x^2}y - 2x + 3y,\,\,\,\,\,\,x = uv,\,\,\,\,y = u - v \cr & \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[ {8{x^2}y - 2x + 3y} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial z}}{{\partial x}} = 16xy - 2 \cr & \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {8{x^2}y - 2x + 3y} \right] \cr & {\text{Treat }}x{\text{ as a constant}} \cr & \frac{{\partial z}}{{\partial y}} = 8{x^2} + 3 \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[ {uv} \right] = v \cr & \frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {uv} \right] = u \cr & \frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {u - v} \right] = 1 \cr & \frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {u - v} \right] = - 1 \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( {16xy - 2} \right)\left( v \right) + \left( {8{x^2} + 3} \right)\left( 1 \right) \cr & {\text{where }}x = uv,\,\,\,\,y = u - v \cr & \frac{{\partial z}}{{\partial u}} = \left( {16\left( {uv} \right)\left( {u - v} \right) - 2} \right)\left( v \right) + \left( {8{{\left( {uv} \right)}^2} + 3} \right) \cr & {\text{simplifying}} \cr & \frac{{\partial z}}{{\partial u}} = \left( {16\left( {{u^2}v - u{v^2}} \right) - 2} \right)\left( v \right) + \left( {8{u^2}{v^2} + 3} \right) \cr & \frac{{\partial z}}{{\partial u}} = 16{u^2}{v^2} - 16u{v^3} - 2v + 8{u^2}{v^2} + 3 \cr & \frac{{\partial z}}{{\partial u}} = 24{u^2}{v^2} - 16u{v^3} - 2v + 3 \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( {16xy - 2} \right)\left( u \right) + \left( {8{x^2} + 3} \right)\left( { - 1} \right) \cr & {\text{where }}x = uv,\,\,\,\,y = u - v \cr & \frac{{\partial z}}{{\partial v}} = \left( {16\left( {uv} \right)\left( {u - v} \right) - 2} \right)\left( u \right) - \left( {8{{\left( {uv} \right)}^2} + 3} \right) \cr & {\text{simplifying}} \cr & \frac{{\partial z}}{{\partial v}} = \left( {16\left( {{u^2}v - u{v^2}} \right) - 2} \right)\left( u \right) - \left( {8{u^2}{v^2} + 3} \right) \cr & \frac{{\partial z}}{{\partial v}} = 16{u^3}v - 16{u^2}{v^2} - 2u - 8{u^2}{v^2} - 3 \cr & \frac{{\partial z}}{{\partial v}} = 16{u^3}v - 24{u^2}{v^2} - 2u - 3 \cr} $$