I was reading page 38 of this article.
How come be that answer at the top page 38.
Isn't
$$U_{xx} ={\partial U \over \partial \xi }.{\partial ^2 \xi \over \partial x^2 } +{\partial \xi \over \partial x }.{\partial ^2U \over \partial x \partial\xi }+.{\partial U \over \partial \eta } .{\partial ^2 \eta \over \partial x^2}+{\partial ^2U \over \partial x \partial \eta }.{\partial \eta\over \partial x } $$
What I did was I computed $U_x$ using chain rule and differentiated it again with respect to x
$\newcommand\d{\partial}$You want to use the chain rule so that $u$ itself is only partially differentiated by $\xi$ and $\eta$, not directly by $x$. For example $$ \frac{\d^2 u}{\d x\d\xi} = \frac{\d}{\d x} \frac{\d u}{\d\xi} = \frac{\d^2 u}{\d\xi^2}\frac{\d\xi}{\d x} + \frac{\d^2 u}{\d\xi\d\eta}\frac{\d\eta}{\d x}. $$
From $$ \frac{\d u}{\d x} = \frac{\d u}{\d\xi}\frac{\d\xi}{\d x} + \frac{\d u}{\d \eta}\frac{\d \eta}{\d x} $$ you obtain \begin{align} \frac{\d^2 u}{\d x^2} &= \frac{\d}{\d x} \left( \frac{\d u}{\d\xi}\frac{\d\xi}{\d x} + \frac{\d u}{\d \eta}\frac{\d \eta}{\d x} \right) \\ &= \frac{\d}{\d x} \left( \frac{\d u}{\d\xi}\frac{\d\xi}{\d x} \right) + \frac{\d}{\d x} \left( \frac{\d u}{\d \eta}\frac{\d \eta}{\d x} \right) \\ &= \left( \frac{\d}{\d x} \frac{\d u}{\d\xi} \right) \frac{\d\xi}{\d x} + \frac{\d u}{\d\xi} \left( \frac{\d}{\d x} \frac{\d\xi}{\d x} \right) + \left( \frac{\d}{\d x} \frac{\d u}{\d\eta} \right) \frac{\d\eta}{\d x} + \frac{\d u}{\d\eta} \left( \frac{\d}{\d x} \frac{\d\eta}{\d x} \right) \\ &= \left( \frac{\d^2 u}{\d\xi^2} \frac{\d\xi}{\d x} + \frac{\d^2 u}{\d\xi\d\eta} \frac{\d\eta}{\d x} \right) \frac{\d\xi}{\d x} + \frac{\d u}{\d\xi} \frac{\d^2\xi}{\d x^2} + \\&\qquad \left( \frac{\d^2 u}{\d\eta\d\xi}\frac{\d\xi}{\d x} + \frac{\d^2 u}{\d\eta}\frac{\d\eta}{\d x} \right) \frac{\d\eta}{\d x} + \frac{\d u}{\d\eta} \frac{\d^2\eta}{\d x^2} \\ &= \frac{\d^2 u}{\d\xi^2} \left( \frac{\d\xi}{\d x}\right)^2 + \frac{\d^2 u}{\d\xi\d\eta} \frac{\d\eta}{\d x} \frac{\d\xi}{\d x} + \frac{\d u}{\d\xi} \frac{\d^2\xi}{\d x^2} + \\&\qquad \frac{\d^2 u}{\d\eta\d\xi}\frac{\d\xi}{\d x} \frac{\d\eta}{\d x} + \frac{\d^2 u}{\d\eta}\left( \frac{\d\eta}{\d x}\right)^2 + \frac{\d u}{\d\eta} \frac{\d^2\eta}{\d x^2} \\ &= \frac{\d^2 u}{\d\xi^2} \left( \frac{\d\xi}{\d x}\right)^2 + 2 \frac{\d^2 u}{\d\xi\d\eta} \frac{\d\eta}{\d x} \frac{\d\xi}{\d x} + \frac{\d^2 u}{\d\eta}\left( \frac{\d\eta}{\d x}\right)^2 + \frac{\d u}{\d\xi} \frac{\d^2\xi}{\d x^2} + \frac{\d u}{\d\eta} \frac{\d^2\eta}{\d x^2}. \end{align}