How to differentiate $\int_{\tau}^t \Phi(t, s)b(s)ds$

144 Views Asked by At

The solution $x(t, \tau, \xi)$ of $\dot{x} = A(t)x + b(t)$ , which passes the point $(\tau, \xi)$, is $$x(t, \tau, \xi)=\Phi(t, \tau)\xi + \int_{\tau}^t \Phi(t, s)b(s)ds, $$ where $\Phi(t, s) = (x(t, s, e_1), \ldots, x(t, s, e_n))$.

I want to confirm it, but how do I differentiate $\displaystyle\int_{\tau}^t \Phi(t, s)b(s)ds$ ?