I'm reading this material Topics in inverse problems and I'm having difficulty in understanding how the derivative of a convolution integral was obtained.

In equation 4.5 (page 91), the function $y(t)$ is defined as

$y(t) = \int_0^T g(t-s)x(s)ds$, with $t \in [0,T]$ and functions $x$ and $g : [0,T]\longrightarrow \mathbb{R}$.

Then, in equation 4.7, the derivative of $y(t)$ with respect to $t$ is

$y'(t) = g(t)x(t) +\int_0^T g'(t-s)x(s)ds$, $t \in [0,T]$.

I don't get why the term $g(t)x(t)$ appears in this derivative.

  • 1
    $\begingroup$ It doesn't seem to make sense it should just be $$y'(t)=\int_0^T\partial_t\,g(t-s)x(s)\,\mathrm ds.$$ It can be checked manually using some example function that it isn't the same. $\endgroup$
    – Conreu
    Commented May 26 at 1:35
  • $\begingroup$ The function $g$ is the same Kernel defined one pages above your equation ? $\endgroup$
    – Hamdiken
    Commented May 26 at 1:53
  • $\begingroup$ @Conreu, thanks. I actually did that setting x(t)=g(t)=t. It seems wrong indeed. $\endgroup$
    – zooond
    Commented May 26 at 2:10
  • $\begingroup$ @Hamdiken, I don't think so. g(t) is unspecified. $\endgroup$
    – zooond
    Commented May 26 at 2:34


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