How to compute the second derivatives? Motivation:
In isogeometric analysis, state variables(e.g. displacement) are defined in the parametric domain, which can be mapped to the physical domain by $\boldsymbol{\xi}\mapsto \boldsymbol{x}$ as shown beneath. However the quantity related to displacment such as stress, strain are spacial derivatives of displacement. The following procedure is commonly used for solving those derivatives. $\blacksquare$
Know
\begin{equation}
u(\xi,\eta) = \sum_{i} c_iN^i(\xi,\eta) 
\end{equation}
with 
$$x(\xi,\eta) = \sum_{i} x_i N^i(\xi,\eta), \quad y(\xi,\eta) = \sum_{i} y_i N^i(\xi,\eta),$$
where $c_i,x_i,y_i$ are constants, with assumption that $(\xi,\eta)\mapsto(x,y)$ is bijective, i.e. inverse exists, $$J :=[\frac{\partial x_i}{\partial \xi_j}],\: |J| \neq 0\quad (\text{where }x_2 = y,\,\xi_2 = \eta).$$
By chain rule, 
$$\frac{\partial u}{\partial \xi_j} = \frac{\partial u}{\partial x_i}\frac{\partial x_i}{\partial \xi_j}$$
or
$$
\begin{bmatrix}
\frac{\partial u}{\partial \xi}\\
\frac{\partial u}{\partial \eta}
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial x}{\partial \xi} &
\frac{\partial y}{\partial \xi}\\
\frac{\partial x}{\partial \eta} &
\frac{\partial y}{\partial \eta}
\end{bmatrix}
\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}
 = J^T
\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}.
$$
Hence,
$$
\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}
=
(J^T)^{-1}
\begin{bmatrix}
\frac{\partial u}{\partial \xi}\\
\frac{\partial u}{\partial \eta}
\end{bmatrix}
$$
However, now I need to compute the second derivatives $$\frac{\partial^2 u}{\partial x_i\partial x_j},\quad i \text{ and }j \in \{1,2\}.$$
One may think displacement $\boldsymbol{u} = [u, v]^T$, then the goal is to compute
$$\nabla(\nabla\boldsymbol{u})$$ with information as above.
 A: I think total differential concept is very useful notation for your derivation. Let
$$u\big(x(\xi,\eta),x(\xi,\eta)\big) $$
Then it follows that
$$du=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}d\xi+\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}d\eta+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}d\xi+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}d\eta$$
$$\Rightarrow du=\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}\bigg)d\xi+\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}\bigg)d\eta$$
and you can use it for partial derivatives to construct your Jacobian
$$d\eta=0\Rightarrow\frac{\partial u}{\partial \xi}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}$$
$$d\xi=0\Rightarrow\frac{\partial u}{\partial \eta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}$$
For second order
$$d^2u=d\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}\bigg) d\xi+d\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}\bigg)d\eta+d\bigg(\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}\bigg)d\xi+d\bigg(\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}\bigg)d\eta$$
where
$$d\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}\bigg)=d\bigg(\frac{\partial u}{\partial x}\bigg)\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}d\bigg(\frac{\partial x}{\partial \xi}\bigg)$$
$$=\frac{\partial^2 u}{\partial x^2}\bigg(\frac{\partial x}{\partial \xi}d\xi+\frac{\partial x}{\partial \eta}d\eta\bigg)\frac{\partial x}{\partial \xi}+\frac{\partial^2 u}{\partial x \partial y}\bigg(\frac{\partial y}{\partial \xi}d\xi+\frac{\partial y}{\partial \eta}d\eta\bigg)\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}\bigg(\frac{\partial^2 x}{\partial \xi^2}d\xi+\frac{\partial^2 x}{\partial \xi\partial\eta}d\eta\bigg)$$
$$=\bigg(\frac{\partial^2 u}{\partial x^2}\bigg(\frac{\partial x}{\partial \xi}\bigg)^2+\frac{\partial^2 u}{\partial x \partial y}\frac{\partial y}{\partial \xi}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial \xi^2}\bigg)d\xi+\bigg(\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial \xi}\frac{\partial x}{\partial \eta}+\frac{\partial^2 u}{\partial x \partial y}\frac{\partial y}{\partial \eta}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial \xi\partial \eta}\bigg)d\eta$$
You can calculate the other three differentials and find the total second order differential. Again by setting $d\eta$ and $d\xi$ to zero you can calculate second order partial differentials. Remember it's not an easy task...
