Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 18

$$\eqalign{ & \frac{{\partial z}}{{\partial u}} = 2u/{v^2} - {u^2}v{\sec ^2}\left( {u/v} \right) - 2u{v^2}\tan \left( {u/v} \right) \cr & \frac{{\partial z}}{{\partial v}} = - 2{u^2}/{v^3} - u{\sec ^2}\left( {u/v} \right) - 2{u^2}v\tan \left( {u/v} \right) \cr} $$

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