What is $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$? I am trying to follow through this proof involving solutions to $\dot{x}=A(t)x+g(t)$. There is a part in the proof where you have to write $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$ as a function. This is a bit tricky since I cannot exactly use the fundamental theorem of calculus. I think the proof states that $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds=f(t,t)g(t)+\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds$, but I am not sure, nor am I convinced since I do not have the proof. Any help is appreciated.
I tried getting dirty with the limit definition of the derivative. If I set $H(t):=\int_{t_0}^tf(t,s)g(s)ds$, then I can manipulate $\frac{dH}{dt}(t)=\lim_{\epsilon \rightarrow 0}\frac{H(t+\epsilon)-H(t)}{\epsilon}$ into $\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds+\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx$. I do not know where to go from here. I guess my intuition says $\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx=f(t,t)g(t)$, but how do I show this?
 A: Here is a rigorous proof. I assume that $f$ and $g$ are continuous. (I am not even sure if this is necessary since one can approximate integrable functions by $C^\infty$-functions)
Notice that
$$
f(t, t)g(t) = \frac{1}{\varepsilon}  \int^{t+\varepsilon}_t f(t, t) g(t)~\mathrm{d}s.
$$
Hence:
$$
\left \lvert \frac{1}{\varepsilon} \int^{t+\varepsilon}_t f(t + \varepsilon, s)g(s)~\mathrm{d}s - f(t, t)g(t)\right \rvert = \frac{1}{\varepsilon} \left \lvert \int^{t+\varepsilon}_t f(t + \varepsilon, s)g(s) -f(t, t)g(t)~\mathrm{d}s \right \rvert \leq \\
\frac{1}{\varepsilon} \int^{t+\varepsilon}_t \left \lvert f(t + \varepsilon, s)g(s) -f(t, t)g(t)~\mathrm{d}s \right \rvert \leq \frac{1}{\varepsilon} \int^{t+\varepsilon}_t \sup_{x \in [t, t + \varepsilon]} \left \lvert f(t + \varepsilon, x)g(x) -f(t, t)g(t) \right \rvert~\mathrm{d}s  = \\
\sup_{x \in [t, t + \varepsilon]} \left \lvert f(t + \varepsilon, x)g(x) -f(t, t)g(t) \right \rvert
$$
Since $[t, t+ \varepsilon]$ is compact and our functions are continuous, this $\sup$ is attained at some $x_\varepsilon \in [t, t + \varepsilon]$. Clearly,  $x_\varepsilon \rightarrow t$ for $\varepsilon \rightarrow 0$. So continuity gives us:
$$
\sup_{x \in [t, t + \varepsilon]} \left \lvert f(t + \varepsilon, x)g(x) -f(t, t)g(t) \right \rvert = \left \lvert f(t + \varepsilon, x_\varepsilon)g(x_\varepsilon) -f(t, t)g(t) \right \rvert \rightarrow 0
$$
as $\varepsilon \rightarrow 0$.
A: You are in the right track since
$$ \lim_{\epsilon\to}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx=f(t,t)g(t).$$
A: Here is a complete proof that can be easily extended for functions defined on Banach spaces. Assume that $(t,s)\mapsto(\partial _tf)(t,s)g(s)$ is integrable over $[T_1,T_2]_t\times[T_1,T_2]_s$ and $s\mapsto f(t,s)g(s)$ is continuous where $[T_1,T_2]$ contains $t_0$ and consider $t\in(T_1,T_2)$. Set
\begin{align*}
 F(t)&:=\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s.
 \end{align*}
Pick $h\in\mathbb{R}$ sufficiently small so that $t+h\in(T_1,T_2)$ then write:
\begin{align*}
 F(t+h)-F(t)&=\int_{t_0}^{t+h}f(t+h,s)g(s)\,\mathrm{d}s-\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s.\tag{$\star$}
 \end{align*}
By definition of the partial derivative, there exists a neighborhood $\mathcal{V}\subset\mathbb{R}$ of $0$ as well as a function $\psi:\mathcal{V}\times(T_1,T_2)\to\mathbb{R}$ such that $\psi(h,s)\to0$ as $h\to0$ and
\begin{align*}
f(t+h,s)&=f(t,s)+h\partial_tf(t,s)+|h|\psi(h,s)
\end{align*}
for any $s\in(T_1,T_2)$. This relation implies that $\psi(h,s)$ is continuous and integrable in $s$ as are all the other terms. In particular, $\max\{|\psi(h,s)|\,\vert\,s\in[T_1,T_2]\}\to0$ as $h\to0$ by uniform continuity over a compact set. Hence $(\star)$ becomes:
\begin{align*}
F(t+h)-F(t)&=\int_{t_0}^{t+h}\big(f(t,s)+h\partial_tf(t,s)+|h|\psi(h)\big)g(s)\,\mathrm{d}s-\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s\\
%
&=\int_{t_0}^{t+h}f(t,s)g(s)\,\mathrm{d}s+h\int_{t_0}^{t+h}\partial_tf(t,s)g(s)\,\mathrm{d}s+|h|\int_{t_0}^{t+h}\psi(h,s)g(s)\,\mathrm{d}s\nonumber\\&\quad-\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s\\
%
&=\int_{t}^{t+h}f(t,s)g(s)\,\mathrm{d}s+h\int_{t_0}^{t+h}\partial_tf(t,s)g(s)\,\mathrm{d}s+|h|\int_{t_0}^{t+h}\psi(h,s)g(s)\,\mathrm{d}s.\tag{$\star\star$}
\end{align*}
Now $fg$ is continuous in $s$ so that there exists a neighborhood $\mathcal{U}\subset\mathbb{R}$ of $0$ as well as a function $\xi:\mathcal{U}\to\mathbb{R}$ (modulus of continuity) such that $\xi(h)\to0$ as $h\to0$ and
\begin{align*}
 f(t,s)g(s)&=f(t,t)g(t)+\xi(t-s)
 \end{align*}
when $|t-s|$ is small enough to belong to $\mathcal{U}$. It follows:
\begin{align*}
 \int_{t}^{t+h}f(t,s)g(s)\,\mathrm{d}s&=\int_{t}^{t+h}\big(f(t,t)g(t)+\xi(t-s)\big)\,\mathrm{d}s\\
 %
 &=hf(t,t)g(t)+\int_{t}^{t+h}\xi(t-s)\,\mathrm{d}s.
 \end{align*}
Thus, for all $\varepsilon>0$, there exists $h_0>0$ such that, for all $h\in(-h_0,h_0)$, we have $|\xi(h)|\leq\varepsilon$ and then
\begin{align*}
 \left|\int_{t}^{t+h}f(t,s)g(s)\,\mathrm{d}s-hf(t,t)g(t)\right|&\leq\left|\int_{t}^{t+h}\xi(t-s)\,\mathrm{d}s\right|\leq \varepsilon|h|.
 \end{align*}
Back into $(\star\star)$, we finally find for $h_0$ sufficiently small so that $\sup\{|\psi(h,s)|\,\vert\,s\in[T_1,T_2]\}<\varepsilon$:
\begin{align*}
 \left|\frac{F(t+h)-F(t)}{h}-f(t,t)g(t)-\int_{t_0}^{t}\partial_tf(t,s)g(s)\,\mathrm{d}s\right|&\leq\varepsilon+\int_{t}^{t+h}|\partial_tf(t,s)||g(s)|\,\mathrm{d}s+\int_{t_0}^{t+h}|\psi(h,s)||g(s)|\,\mathrm{d}s\\
 %
 &\leq\varepsilon+\int_{t}^{t+h}|\partial_tf(t,s)||g(s)|\,\mathrm{d}s+\varepsilon\int_{t_0}^{t+h}|g(s)|\,\mathrm{d}s\\
 %
 &\leq\varepsilon+\int_{t}^{t+h}|\partial_tf(t,s)||g(s)|\,\mathrm{d}s+\varepsilon\int_{t_0}^{T_1}|g(s)|\,\mathrm{d}s.
 \end{align*}
The right-hand side above is easily shown to converges to $\varepsilon\left(1+\int_{t_0}^{T_1}|g(s)|\,\mathrm{d}s\right)$ as $h\to0$, proving the formula.
More generally, under continuity or boundedness assumptions on functions $a,b:[T_1,T_2]\to\mathbb{R}$, we can show the more general formula:
\begin{align*}
 \frac{\mathrm{d}}{\mathrm{d}t}\int_{a(t)}^{b(t)}f(t,s)g(s)\,\mathrm{d}s&=f(t,b(t))g(b(t))b'(t)-f(t,a(t))g(a(t))a'(t)+\int_{a(t)}^{b(t)}\partial_tf(t,s)g(s)\,\mathrm{d}s
 \end{align*}
