If $t \to \lVert f(\cdot,t) \rVert_{L^2_x}^2$ is absolutely continuous, can we interchange the spatial integtal and time derivative?

Question

Let $f(x,t) : S^1 \times [0,\infty) \to \mathbb{R}$ be a function such that

1. $f(x,t) \in L^q_t\bigl([0,\infty), L^p_x(S^1)\bigr)$
2. $\partial_t f(x,t) \in L^{q'}_t\bigl([0,\infty), L^{p'}_x(S^1)\bigr)$. Here the time derivative is weak one in the context of Bochner spaces.

where $p,p' \in (1,\infty)$ satisfy $1/p+1/p'=1$ but $q, q' \in (1,\infty)$ are just $1/q + 1/q' >1$.

However, suppose additionally that $t \to \lVert f(\cdot,t) \rVert_{L^2_x}^2$ is in fact "absolutely continuous".

Then, is it true that \begin{equation} \frac{d}{dt}\lVert f(\cdot,t) \rVert_{L^2_x}^2= 2\int_{S^1} f(x,t) \partial_t f(x,t) dx \end{equation} for a.e. $t \in [0,\infty)$?

I tried to prove this myself using Aubin-Lions lemma somehow, but things do not work for me since $q,q'$ are not conjugate exponents for now.

Could anyone please help me?