## Calculus: Early Transcendentals 8th Edition

$(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$
$(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$ Here, we have $\dfrac{\partial u}{\partial s}=e^s \cos t \dfrac{\partial u}{\partial x}+ e^s \sin t \dfrac{\partial u}{\partial y}$ and $\dfrac{\partial u}{\partial t}=-e^s \sin t \dfrac{\partial u}{\partial x}+ e^s \cos t \dfrac{\partial u}{\partial y}$ Now, $(\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2=e^{2s}[(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2]$ Hence, it has been proved that $(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=e^{-2s}[ (\dfrac{\partial u}{\partial s})^2+(\dfrac{\partial u}{\partial t})^2]$