# Derivative of a convolution integral

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.

• 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. Commented May 26 at 1:35
• The function $g$ is the same Kernel defined one pages above your equation ? Commented May 26 at 1:53
• @Conreu, thanks. I actually did that setting x(t)=g(t)=t. It seems wrong indeed. Commented May 26 at 2:10
• @Hamdiken, I don't think so. g(t) is unspecified. Commented May 26 at 2:34