I forgot the theorems or the proposition that allow the operator D (derivative) to enter inside the integral and the one allow the inner product to enter inside too
$D L(x) \cdot v=\int_{0}^{\infty} D\left(\left\|e^{s A} x\right\|^{2}\right) \cdot v d s$
with $L(x) =\int_{0}^{\infty} \left\|e^{s A} x\right\|^{2}d s$