This is a problem when I'm doing inversion using Tikhonov regularisation. It can be simplified to (only the regularization part)
$$ f(m) = \sum_{\textrm{all i, j}} (\bigtriangleup m(x_i, y_j))^2 $$
$\bigtriangleup$ is a discrete Laplacian operator using finite-difference method. Because I use the iterative inversion strategy, I need to calculate the derivative of $f(m)$ on $m$.
I have known $m(x_i, y_i)$, how to calculate $\frac{\partial f}{\partial m}(x_i, y_i)$
Nothing complicate here.
$$(\Delta m)_{i,j} = \sum_{l,k} c_{l,k} m_{i+l,j+k}$$ where $c$ are the coefficients of your discrete Laplacian matrix
Note that $$\frac{\partial (\Delta m)_{i',j'}}{\partial m_{i,j}} = \sum_{l,k} c_{l,k} \frac{\partial m_{i'+l,j'+k}}{\partial m_{i,j}} = c_{i-i',j-j'} $$
Then $f(m) = \sum_{i,j} (\Delta m)_{i,j}^2$ is a function of $m_{i,j}$ and what you want is $$\frac{\partial f}{\partial m_{i,j}} = \sum_{i',j'} \frac{\partial(\Delta m)^2_{i',j'}}{\partial m_{i,j}} = \sum_{i',j'} 2(\Delta m)_{i',j'}\frac{\partial(\Delta m)_{i',j'}}{\partial m_{i,j}}= \sum_{i',j'} 2(\Delta m)_{i',j'}c_{i-i',j-j'} $$