Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 22

$$\eqalign{ & \frac{{\partial z}}{{\partial u}} = 2u\cos \left( {u - v} \right)\cos \left( {{u^2} + {v^2}} \right) - \sin \left( {u - v} \right)\sin \left( {{u^2} + {v^2}} \right) \cr & \frac{{\partial z}}{{\partial v}} = \sin \left( {{u^2} + {v^2}} \right)\sin \left( {u - v} \right) + 2v\cos \left( {u - v} \right)\cos \left( {{u^2} + {v^2}} \right) \cr} $$

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