# Partial derivative of convolution integral

1. Oct 17, 2011

### cdsi385

Does anyone know how to take the partial derivative of a convolution integral where the derivative is taken with respect to one of the functions of the convolution integral?

In the following example, the best I can come up with is:

$\frac{\partial}{\partial g(t)}\int L(t-\tau)g(t)\,d\tau=\int L(t-\tau)\,d\tau$

Is this correct, or does it even make sense???

To put this in context, what I usually do (successfully) is perform the convolution integral in a simulation (without the partial differentiation) where $L(t)$ is the impulse response function of a system and $g(t)$ is the velocity of my system which is calculated on the fly during the simulation.

What I'm trying to do now is make a new simulation which relies on this partial derivative which I'm trying to express analytically before simulating it. If what I've expressed above is correct then all I need to simulate is: $\int L(t-\tau)\,d\tau$

Thanks in advance...
cdsi385

2. Oct 17, 2011

### mathman

To be a convolution, it should have g(τ) inside the integral, not g(t).

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