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How does this hold? $$ \frac{d}{dt} \int_0^t g(t-r) f(r) dr = g(0)f(t) + \int_0^t \frac{dg}{dt} (t-r) f(r) dr $$ Assume that all the function are sufficiently integrable and differentiable.

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Let $\Phi(t) := (t,t)$ and $h(x,y) := \int_0^x g(y-r) \cdot f(r) \, dr$. Then

$$\frac{d}{dt} \int_0^t g(t-r) \cdot f(r) \, dr = \frac{d}{dt}(h \circ \Phi)(t) = h'(\Phi(t))^T \cdot \Phi'(t) \\ = \begin{pmatrix} g(y-x) \cdot f(x)\bigg|_{(x,y)=\Phi(t)} & \int_0^x \frac{d}{dy} g(y-r) \cdot f(r) \, dr \bigg|_{(x,y)=\Phi(t)} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

where we used the fundamental theorem of calculus, i.e.

$$\frac{d}{dt} \int_0^t h(r) \, dr = h(t)$$

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Perhaps treat $\int_0^t g(t-r) f(r) dr$ as a bivariate function $g(u,v)=\int_0^u g(v-r) f(r) dr$ with $u(t)=v(t)=t$ and compute the total derivative of $g$?

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