a hyperbolic PDE $\mathbf{u_{xx}-u_{yy}=0}$ transition of $u_x$ into $u_{xx}$ I am analysing a hyperbolic PDE $\mathbf{u_{xx}-u_{yy}=0}$ and I don't quite get how the the $u_x$ was transitioned into the $u_{XX}$ ($u_y$ into $u_{yy}$) using the second set of coordinates $\xi$ and  $\eta$. 
Would anybody clarify the steps of the transitions 3a->4a->5a,  3b->4b->5b?
$(1a) \ \ \ \xi = x + y, 
\\ (1b) \ \ \  \eta = x-y$ 
$(2a) \ \ \ \frac{\partial }{\partial x} = \frac{\partial }{\partial \xi} +  \frac{\partial }{\partial \eta} \\
\\
(2b) \ \ \ \frac{\partial }{\partial y} =  \frac{\partial }{\partial \xi} -  \frac{\partial }{\partial \eta}\\$
$(3a) \ \ \ u_x = u_\xi + u_\eta  \\  
(3b) \ \ \  u_y = u_\xi - u_\eta $
$\mathbf{(4a) \ \ \ u_{xx}=(\frac{\partial }{\partial \xi} + \frac{\partial }{\partial \eta})(u_\xi + u_\eta ) \\ 
(4b) \ \ \ u_{yy}=(\frac{\partial }{\partial \xi} - \frac{\partial }{\partial \eta})(u_\xi - u_\eta )}$
$\mathbf{(5a) \ \ \ u_{xx} = u_{\xi\xi} + 2 u_{\xi\eta}  + u_{\xi\xi} \\
(5b) \ \ \ u_{yy} = u_{\xi\xi} - 2 u_{\xi\eta}  + u_{\xi\xi}}$
 A: $u_{xx}=\dfrac{\partial u_x}{\partial x} = (\dfrac{\partial}{\partial \epsilon}+\dfrac{\partial}{\partial \eta})(u_{\epsilon} + u_{\eta})= (\dfrac{\partial}{\partial \epsilon}+\dfrac{\partial}{\partial \eta})(u_{\epsilon})   +(\dfrac{\partial}{\partial \epsilon}+\dfrac{\partial}{\partial \eta})(u_{\eta})$= $\dfrac{\partial u_{\epsilon}}{\partial \epsilon}+\dfrac{\partial u_{\epsilon}}{\partial \eta}+\dfrac{\partial u_{\eta}}{\partial \epsilon}+\dfrac{\partial u_{\eta}}{\partial \eta}=u_{\epsilon\epsilon}+2u_{\epsilon\eta}+u_{\eta\eta}$
(By Young's theorem we can change the order of taking a partial derivative.)
A: Since $du = u_\xi d\xi + u_\eta d\eta$, we get $u_x = u_\xi \xi_x + u_\eta \eta_x = u_\xi + u_\eta$. Now both $u_\xi$ and $u_\eta$ are functions of $\xi$ and $\eta$. Therefore, $du_\xi = u_{\xi\xi} d\xi + u_{\xi\eta} d\eta$ and hence
\begin{equation}
\frac{\partial u_\xi}{\partial x} = u_{\xi\xi} + u_{\xi\eta}
\end{equation}
Similarly,
\begin{equation}
\frac{\partial u_\eta}{\partial x} = u_{\eta\xi} + u_{\eta\eta}
\end{equation}
and hence,
\begin{equation}
u_{xx} = \frac{\partial u_\xi}{\partial x} + \frac{\partial u_\eta}{\partial x} = u_{\xi\xi} + 2u_{\xi\eta} + u_{\eta\eta}
\end{equation}
The other derivatives follow similarly.
