Evaluate derivative of reachability Gramian: is this correct? So I have the reachability Gramian matrix for a linear time-invariant system:
\begin{align}
W(t_{0},t) = \int_{t_{0}}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}.
\end{align}
In this case I have $t_{0}=0$. Let us differentiate this w.r.t. to $t$:
\begin{align}
\dot{W}(0,t) 
=& \frac{d}{dt}\left[\int_{0}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}ds\right] \\
=& \frac{d}{dt}\left[e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}\right] \\
=& Ae^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t} \\
&+e^{At}\left[e^{-As}BB^{\intercal}e^{-A^{\intercal}s}\bigg\vert_{0}^{t}\right]e^{A^{\intercal}t} + e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}A^{\intercal} \\[3mm]
=& AW(0,t) + BB^{\intercal} - e^{At}BB^{\intercal}e^{A^{\intercal}t}+W(0,t)A^{\intercal}.
\end{align}
So this is the result I obtain. However, in the solution to this problem that I was trying to solve the term $- e^{At}BB^{\intercal}e^{A^{\intercal}t}$ did not appear, or was zero. So I am wondering if I made some error, or if there is some control theory result that make this term vanish?

When you applied the product rule, you incorrectly computed the derivative of the integral. By the fundamental theorem of calculus, we have
$$
\frac d{dt} \int_{t_0}^t e^{-As}BB^\top e^{-A^\top s}\,ds = e^{-At}BB^\top e^{-A^\top t}.
$$
There is no $BB^\top$ term that should appear here.
