# Differentiating under the integral sign with variable limit

The formula

$\frac{d}{dt}\int_0^t \phi(t,s) u(s)\, ds = \phi(t,t)u(t) + \int_0^t \phi_t(t,s) u(s)\, ds$

is not new but it doesn’t seem so easy to track down a simple derivation. A simple (formal) derivation of this runs as follows. First define,

$F(t,\xi) := \int_0^t f(\xi, s)\, ds = \digamma(\xi, t) - \digamma(\xi,0)$

where $\digamma$ is a primitive such that $\digamma_s(\cdot,s) = f(\cdot,s)$. Then

$\frac{dF}{dt}= \frac{\partial F}{\partial t} + \frac{\partial F}{\partial\xi}\frac{d\xi}{d t}$

But on substituting

$\frac{\partial F}{\partial t}(t,\xi) = \frac{\partial\digamma}{\partial t}(\xi,t) = f(\xi,t)$

and

$\frac{\partial F}{\partial\xi}(t,\xi) = \int_0^t f_\xi(\xi,s)\, ds$

this becomes

$\frac{dF}{dt}= f(\xi,t) + \frac{d\xi}{d t} \int_0^t f_\xi(\xi,s)\, ds.$

That’s it really. As a corollary take $\xi = t$ and $f(\xi,s) = \phi(t,s)u(s)$ and then we see that

$\frac{d}{dt}\int_0^t \phi(t,s)u(s)\, ds = \phi(t,t)u(t) + \int_0^t(\phi(t,s)u(s))_t\, ds$

which is what we set out to demonstrate.