PDE standard form chain rule derivations I am currently reading a document on analytical solution of second-order PDE's which reads the following:

We could define new independent variables $\xi(x,y)$ and $\eta(x,y)$.... As before we compute the chain rule derivations:
$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}$$
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \xi}{\partial x}\right)^2+2 \frac {\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \eta}{\partial x}\right)^2+\frac{\partial u}{\partial \xi}\frac{\partial^2 \xi}{\partial x^2}+\frac{\partial u}{\partial \eta}\frac{\partial ^2 \eta}{\partial x^2}$$

With similar computations for $\frac {\partial u}{\partial y}$. Regarding the second derivative, I am not happy about the last two terms. I am not sure why they appear. According to me, the second derivative can be taken as follows, since $u$ depends on $x$ and $y$ which in turn depend each on $\eta $ and $\xi$.
$$\frac{\partial^2 u}{\partial x^2} = \frac {\partial}{\partial x }\left(\frac {\partial u} {\partial x}\right)$$
We can reuse the very first chain rule as follows:
$$\frac {\partial}{\partial x }\left(\frac {\partial u} {\partial x}\right)=\left(\frac{\partial }{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial }{\partial \eta}\frac{\partial \eta}{\partial x}\right)\left(\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u }{\partial \eta}\frac{\partial  \eta}{\partial x}\right)$$
which when multiplied through using multiplicative diistribution would yield:
$$\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \xi}{\partial x}\right)^2+2 \frac {\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \eta}{\partial x}\right)^2$$
Comparing my answer with the books answer I am missing two terms. Why and how do these appear?
 A: Picking up close to where you left,
$$
\begin{alignedat}{4}
\frac {\partial}{\partial x } \left(\frac {\partial u} {\partial x} \right) 
&= \frac {\partial}{\partial x } \left(\frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x}+\frac{\partial u }{\partial \eta}\frac{\partial  \eta}{\partial x}\right)
\\&= \frac {\partial}{\partial x}  \left(\xi_x u_\xi+\eta_x u_\eta\right)
\\&= \xi_{xx}u_\xi + \xi_x\frac{\partial u_\xi}{\partial x} + \eta_{xx}u_\eta + \eta_x\frac{\partial u_\eta}{\partial x} 
\\
&= \xi_{xx}u_\xi + \xi_x
\left(\frac{\partial u_\xi}{\partial \xi}\frac{\partial \xi}{\partial x} +   \frac{\partial u_\xi}{\partial \eta}\frac{\partial \eta}{\partial x}\right) + \eta_{xx}u_\eta + \eta_x 
\left(\frac{\partial u_\eta}{\partial \xi}\frac{\partial \xi}{\partial x} +   \frac{\partial u_\eta}{\partial \eta}\frac{\partial \eta}{\partial x}\right) 
\\&= 
\xi_{xx}u_\xi + \xi_x
\left(\xi_x u_{\xi\xi} +   \eta_x u_{\xi\eta}\right) + \eta_{xx}u_\eta + \eta_x 
\left(\xi_x u_{\xi\eta} +   \eta_x u_{\eta\eta}\right) 
\\&=  
\xi^2_x  u_{\xi\xi} + 2\xi_x\eta_x u_{\xi\xi} + \eta_x^2 u_{\eta\eta} + \xi_{xx} u_\xi + \eta_{xx} u_\eta.
\end{alignedat}
$$
Thus we got 
$$\bbox[1.5ex, border:solid 2pt #e10000]{
\frac{\partial^2 u}{\partial x^2} 
=\left(\frac{\partial\xi}{\partial x}\right)^2\frac{\partial^2u}{\partial\xi^2} 
+ 2\left(\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}\right) \frac {\partial^2 u}{\partial \xi \partial \eta} 
+ \left(\frac{\partial \eta}{\partial x}\right)^2 \frac{\partial^2 u}{\partial \eta^2} 
+ \frac{\partial u}{\partial \xi}\frac{\partial^2 \xi}{\partial x^2} 
+ \frac{\partial u}{\partial \eta}\frac{\partial ^2 \eta}{\partial x^2}}
$$
