In my research work, I differentiated the following equation with respect to t
$$\frac{\partial^2 u}{\partial t^2}=x(1-x)\frac{\partial^2 u}{\partial x^2}-(\omega^2-2)u$$
to get
$$\frac{\partial^3 u}{\partial t^3}=x(1-x)\frac{\partial^3 u}{\partial t\,\partial x^2}-(\omega^2-2)\frac{\partial u}{\partial t}.$$
Differentiating again, with respect to t gives
$$\frac{\partial^4 u}{\partial t^4}=x(1-x)\frac{\partial^4 u}{\partial t^2\,\partial x^2}-(\omega^2-2)\frac{\partial^2 u}{\partial t^2}.$$
However, I have a problem representing $\frac{\partial^3 u}{\partial t\,\partial x^2}$ and $\frac{\partial^4 u}{\partial t^2\,\partial x^2}$ as central difference mixed derivative approximations. I understand that, for central difference
$$\left(\frac{\partial u}{\partial x}\right)_{i,j}=\frac{u_{i+1,j}-u_{i-1,j}}{2\Delta x}$$
$$\left(\frac{\partial^2 u}{\partial x^2}\right)_{i,j}=\frac{u_{i+1,j}-2u_{i,j}+u_{i-1,j}}{(\Delta x)^2}$$
$$\left(\frac{\partial u}{\partial t}\right)_{i,j}=\frac{u_{i,j+1}-u_{i,j}}{\Delta t}$$
My question is: how do I derive $\frac{\partial^4 u}{\partial t^2\,\partial x^2},\;\;\frac{\partial^3 u}{\partial t\,\partial x^2}, \;\; \frac{\partial^2 u}{\partial t^2}\;\; \text{and} \;\; \frac{\partial^3 u}{\partial t^3}$?