Given the PDE $$a_{11} u_{x x}+2 a_{12} u_{x y}+a_{22} u_{y y}+a_{1} u_{x}+a_{2} u_{y}+a_{0} u=0$$ where $$\begin{bmatrix} x \\ y\\ \end{bmatrix} =\left(\begin{array}{rr} \cos\left(t\right) & -\sin\left(t\right) \\ \sin\left(t\right) & \cos\left(t\right) \end{array}\right) \begin{bmatrix} x' \\ y' \end{bmatrix}$$
$$\begin{bmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} x'\\ y' \end{bmatrix}$$
How does one deduce that
$\begin{aligned} \frac{\partial^{2} u}{\partial x^{2}} &=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right) \\ &=\left(\frac{\partial x^{\prime}}{\partial x} \frac{\partial}{\partial x^{\prime}}+\frac{\partial y^{\prime}}{\partial x} \frac{\partial}{\partial y^{\prime}}\right)\left(\frac{\partial u}{\partial x}\right) \end{aligned}$?
We have $x'=x\cos(t)+y\sin(t)$ and $y'=-x\sin(t)+y\cos(t)$ then by the chain rule
$$\frac{\partial}{\partial x}=\frac{\partial x'}{\partial x}\frac{\partial}{\partial x'}+\frac{\partial y'}{\partial x}\frac{\partial }{\partial y'}=\cos(t)\frac{\partial}{\partial x'}-\sin(t)\frac{\partial}{\partial y'}$$