I don't understand how in the following paper https://users.cs.duke.edu/~tomasi/papers/shi/TR_93-1399_Cornell.pdf (Good features to track) the residual (Eq. 3.1) is differentiated so that Equations 3.2 and 3.3 are obtained. Can someone explain this to me?
$\epsilon = \int\int_W [J(A$x$)+ d]^2 \omega$(x)$d$x (3.1)
$\partial\epsilon/\partial D = 2\int\int_W [J(A$x$)+ d] (\partial J/\partial x, \partial J/\partial y)^T$ x$^T\omega$(x)$d$x (3.2)
$\partial\epsilon/\partial d = 2\int\int_W [J(A$x$)+ d] (\partial J/\partial x, \partial J/\partial y)\omega$(x)$d$x (3.3)
Consider the (scalar) integrand $\phi = \left[ J(\mathbf{u})-I(\mathbf{x}) \right]^2$ where $\mathbf{u}=\mathbf{Ax+d}$.
The derivative wrt vector $\mathbf{d}$ is easily computed \begin{eqnarray} d\phi &=& 2 \left[ J(\mathbf{u})-I(\mathbf{x}) \right] \left( \frac{\partial J}{\partial \mathbf{u}}:d\mathbf{u} \right) \tag{*}\\ &=& 2 \left[ J(\mathbf{u})-I(\mathbf{x}) \right] \left( \frac{\partial J}{\partial \mathbf{u}}:d\mathbf{d} \right) \end{eqnarray} so the integrand derivative reads $$ \frac{\partial \phi}{\partial \mathbf{d}} = 2 \left[ J(\mathbf{u})-I(\mathbf{x}) \right] \nabla J(\mathbf{u}) $$
Looking at the paper $\mathbf{A}=\mathbf{D+I}$ \begin{eqnarray} d\phi &=& (*) = 2 \left[ J(\mathbf{u})-I(\mathbf{x}) \right] \left( \frac{\partial J}{\partial \mathbf{u}}:(d\mathbf{A})\mathbf{x} \right) \\ &=& 2 \left[ J(\mathbf{u})-I(\mathbf{x}) \right] \left( \frac{\partial J}{\partial \mathbf{u}} \mathbf{x}^T:d\mathbf{D} \right) \end{eqnarray} so the integrand derivative reads $$ \frac{\partial \phi}{\partial \mathbf{D}} = 2 \left[ J(\mathbf{u})-I(\mathbf{x}) \right] \nabla J(\mathbf{u})\mathbf{x}^T $$
Note that the gradient of $J$ should be evaluated at $\mathbf{u}$.