# Differentiation with convolution operators

1. Aug 19, 2013

### anja.ende

Hello,

I have really been banging my head the whole day and trying to figure this derivative out. I have a function of the following form:

F = W * (I.J(t)) - (W * I).(W*J(t))

where I and J are two images. J depends on some transformation parameters t and W is a gaussian kernel with some fixed standard deviation and zero mean. * represents the convolution operator. Now, I want to compute the derivative of F wrt to the transformation parameters 't'.

So, I try the following:

$\frac{dF}{dt} = \frac{d}{dt} [(I . W*J(t)) - (W*I)(W*J(t))]$

I can talk 'I' out as it can be treated as a constant. This gives (I think):

$\frac{dF}{dt} = (I. W*J'(t)) - (W*J'(t)) . (W*I)$

Can I treat the convolution operators this way or is this wrong? The convolution kernels are fixed width Gaussians and do not depend on the parameters 't'.

Anja

Last edited: Aug 19, 2013
2. Aug 20, 2013

### CompuChip

The definition is
$$(f \ast g)(t) = \int_\mathbb{R} f(\tau) g(t - \tau)$$
so with the appropriate smoothness conditions and all that, you can easily verify
$$(f \ast g)'(t) = \int_\mathbb{R} f(\tau) g'(t - \tau) = (f \ast g')(t)$$