Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 21

$$\frac{{\partial z}}{{\partial u}} = {e^u}{\text{ and }}\frac{{\partial z}}{{\partial v}} = 0$$

$$\eqalign{ & z = {e^{{x^2}y}},\,\,\,\,\,\,x = \sqrt {uv} ,\,\,\,\,y = 1/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[ {{e^{{x^2}y}}} \right] \cr & {\text{Treat }}y{\text{ as a constant}} \cr & \frac{{\partial z}}{{\partial x}} = 2xy{e^{{x^2}y}} \cr & \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{e^{{x^2}y}}} \right] \cr & {\text{Treat }}x{\text{ as a constant}} \cr & \frac{{\partial z}}{{\partial y}} = {x^2}{e^{{x^2}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[ {\sqrt {uv} } \right] = \frac{{\sqrt v }}{{2\sqrt u }} = \frac{1}{2}\sqrt {\frac{v}{u}} \cr & \frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {\sqrt {uv} } \right] = \frac{1}{2}\sqrt {\frac{u}{v}} \cr & \frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {1/v} \right] = 0 \cr & \frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {1/v} \right] = - 1/{v^2} \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( {2xy{e^{{x^2}y}}} \right)\left( {\frac{1}{2}\sqrt {\frac{v}{u}} } \right) + \left( {{x^2}{e^{{x^2}y}}} \right)\left( 0 \right) \cr & {\text{simplifying}} \cr & \frac{{\partial z}}{{\partial u}} = xy{e^{{x^2}y}}\sqrt {\frac{v}{u}} \cr & {\text{where }}x = \sqrt {uv} ,\,\,\,\,y = 1/v \cr & \frac{{\partial z}}{{\partial u}} = \left( {\sqrt {uv} } \right)\left( {1/v} \right){e^{{{\left( {\sqrt {uv} } \right)}^2}\left( {1/v} \right)}}\sqrt {\frac{v}{u}} \cr & \frac{{\partial z}}{{\partial u}} = \sqrt {uv} {e^{{u^2}}}/v\sqrt {\frac{v}{u}} \cr & \frac{{\partial z}}{{\partial u}} = {e^u} \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( {2xy{e^{{x^2}y}}} \right)\left( {\frac{1}{2}\sqrt {\frac{u}{v}} } \right) + \left( {{x^2}{e^{{x^2}y}}} \right)\left( { - \frac{1}{{{v^2}}}} \right) \cr & \frac{{\partial z}}{{\partial v}} = xy{e^{{x^2}y}}\sqrt {\frac{u}{v}} - \frac{{{x^2}{e^{{x^2}y}}}}{{{v^2}}} \cr & {\text{where }}x = \sqrt {uv} ,\,\,\,\,y = 1/v \cr & \frac{{\partial z}}{{\partial v}} = \sqrt {uv} \left( {\frac{1}{v}} \right){e^{{{\left( {\sqrt {uv} } \right)}^2}\left( {1/v} \right)}}\sqrt {\frac{u}{v}} - \frac{{{{\left( {\sqrt {uv} } \right)}^2}{e^{{{\left( {\sqrt {uv} } \right)}^2}\left( {1/v} \right)}}}}{{{v^2}}} \cr & {\text{simplifying}} \cr & \frac{{\partial z}}{{\partial v}} = \frac{u}{v}{e^u} - \frac{{u{e^{{u^2}}}}}{v} \cr & \frac{{\partial z}}{{\partial v}} = 0 \cr} $$