| written 4.8 years ago by |
$\dfrac {\partial z}{\partial u} = \dfrac {\partial z}{\partial x} \dfrac {\partial x}{\partial u}+ \dfrac {\partial z}{\partial y}\dfrac {\partial y}{\partial u} = \dfrac {\partial z}{\partial x} e^{u} - \dfrac {\partial z}{\partial y}e^{-u}$ $\dfrac {\partial z}{\partial v} = \dfrac {\partial z}{\partial x} \dfrac {\partial x}{\partial v} + \dfrac {\partial z}{\partial y}\dfrac {\partial y}{\partial v} = - \dfrac {\partial z}{\partial x} e^{-v} - \dfrac {\partial z}{\partial y} e^v$ By subtraction, $\dfrac {\partial z}{\partial u} - \dfrac {\partial z}{\partial v} = (e^u + e^{-v}) \dfrac {\partial z}{\partial x } (e^{-u} -e^v) \dfrac {\partial z}{\partial y} = x \dfrac {\partial z}{\partial x}- y \dfrac {\partial z}{\partial y}$

and 2 others joined a min ago.