Derivative of a definite integral Could you help me to find the derivative of this definite integral?

$\frac{d}{dt}\int_0^t{e^{A(t-s)}f(s)ds}$,

where $A$ and $f$ are matrices.

Basically I want to find the maximum/minimum points for the general solution of a system of linear differential equations:

$\frac{d}{dt}x(t)=0=\frac{d}{dt}[e^{At}x_o+\int_0^t{e^{A(t-s)}f(s)ds}]$,

where $x_o$ is the initial conditions vector.

Regards,

Marcos.
Rewrite the integral as $\;\displaystyle\mathrm e^{At} \int_0^t\mathrm e^{-As}f(s)\,\mathrm d s $ and apply the First fundamental theorem of integral calculus:
\begin{align}\frac{\mathrm d}{\mathrm d t}\biggl(\mathrm e^{At} \int_0^t\mathrm e^{-As}f(s)\,\mathrm d s\biggr)&amp;=A\,\mathrm e^{At} \int_0^t\mathrm e^{-As}f(s)\,\mathrm d s+\mathrm e^{At}\frac{\mathrm d}{\mathrm d t}\biggl(\int_0^t\mathrm e^{-As}f(s)\,\mathrm d s\biggr)\\
&amp;=A\,\mathrm e^{At} \int_0^t\mathrm e^{-As}f(s)\,\mathrm d s+\mathrm e^{At}\,\mathrm e^{-At}f(t)\\
&amp;=A\,\mathrm e^{At}\int_0^t\mathrm e^{-As}f(s)\,\mathrm d s+f(t).
\end{align}
Your question makes sense but the motivation does not: these are vectors!

Aynways,
$$
\frac{d}{dt}\left[e^{At}x_0+\int_0^t{e^{A(t-s)}f(s)\,ds}\right]=\frac{d}{dt}x(t)=Ax(t)+f(t).
$$