I am trying to derive Eq. (7.25) (p. 117) of Polyakov's book:

$$ \delta \Psi (C) = \int_{0}^{2\pi} {\rm P} \left(F_{\mu\nu}(x(s)) \exp \oint_C A_\mu dx^\mu \right)\dot{x}_\nu \delta x_\mu(x) \, {\rm d} s $$

where the non-abelian phase factor around a closed loop $C$ is defined as

$$ \Psi(C) = {\rm P}\exp \left(\oint A_\mu dx^\mu \right) = {\rm P}\exp \left(\int_{0}^{2\pi} A_\mu \dot{x}_\mu\, {\rm d}s \right)$$

It seems that he is using the relation given on p. 116:

$$ \delta \, {\rm P} \exp \int_{0}^{2\pi} M(\tau) {\rm d}\tau = \int_{0}^{2\pi}{\rm d}t\,{\rm P} \left(\delta M(t) \exp \int_{0}^{2\pi}M(\tau){\rm d}\tau\right)$$

Matching with (7.25) I find $\delta A_\nu = F_{\mu\nu} \delta x_\mu$. This relation seems to be saying that if I change the position of the loop at the parameter $s$ by $\delta x_\mu(x)$ then the vector potential changes by $\delta A_\nu(x(s)) = F_{\mu\nu}(x(s)) \delta x_\mu(s)$.

I don't know how to derive this relation. Is it legitimate?