derivative of integral function Consider the function defined by
$$
w(t)=\int_{-\infty}^t k(t-s)f(x(s)) \, ds
$$
where both $k(\cdot)$ and $f(\cdot)$ are "nice" real-valued functions. I would like to find $w^{\prime}(t)$. I think that it is given by
$$
w^{\prime}(t)=k(0)[f(x(t))-w(t)].
$$
Is this correct? If not, what is wrong with it?
 A: In general, $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{a(t)}^{b(t)}F(t,s)\,\mathrm{d}s\right)=F(t,b(t))\times b'(t)-F(t,a(t))\times a'(t)+\int_{a(t)}^{b(t)}\frac{\partial F(t,s)}{\partial t}\mathrm{d} s$$ is valid under some regularity conditions, provided that every function involved is “nice” (differentiable, etc.). In your example, the result should be $$k(0)f(x(t))+\int_{-\infty}^t k'(t-s)f(x(s))\,\mathrm{d} s.$$
A: When in doubt, use https://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement
Assuming $k$, $f$ and $x$ are bounded,
\begin{align*}
w^{\prime}\left(t\right) & =\frac{\partial}{\partial t}\int_{-\infty}^{t}k\left(t-s\right)f\left(x\left(s\right)\right)ds\\
 & =\int_{-\infty}^{t}\frac{\partial}{\partial t}\left[k\left(t-s\right)f\left(x\left(s\right)\right)\right]ds+k\left(t-t\right)f\left(x\left(t\right)\right)-\lim_{s\rightarrow\infty}k\left(t+s\right)f\left(x\left(s\right)\right)\cdot0\\
 & =\int_{-\infty}^{t}k^{\prime}\left(t-s\right)f\left(x\left(s\right)\right)ds+k\left(0\right)f\left(x\left(t\right)\right).
\end{align*}
