## Calculus: Early Transcendentals 8th Edition

$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$ $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$
Consider $P=f_x=\dfrac{\partial f}{\partial x};Q=f_y=\dfrac{\partial f}{\partial y}$ and $R=f_z=\dfrac{\partial f}{\partial z}$ a) $\dfrac{\partial P}{\partial y}=\dfrac{\partial}{\partial y}(\dfrac{\partial f}{\partial x})=\dfrac{\partial^2 f}{\partial x \partial y}=\dfrac{\partial Q}{\partial x}$ b) $\dfrac{\partial P}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial x \partial z}=\dfrac{\partial R}{\partial x}$ c) $\dfrac{\partial Q}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial z \partial y}=\dfrac{\partial R}{\partial y}$ Hence, $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$ $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$