Let $G$ be a Poisson-Lie group with Poisson bivector field $\pi.$ Let $\pi^{R}$ be the trivialization of $\pi$ with respect to the right translations i.e. $$\pi^{R} (g) = (d_{g} R_{g^{-1}} \otimes d_{g} R_{g^{-1}}) \pi (g), \ \ g \in G.$$ Since $\{f_{1}, f_{2} \} (x) = (d_x f_1 \otimes d_x f_2) \pi (x)$ it follows that $$\{f_1, f_2 \} (x) = (d_x f_1 \otimes d_x f_2) (d_{e} R_x \otimes d_e R_x) \pi^R (x).$$ Let $d_e f_1 = \xi_1$ and $d_e f_2 = \xi_2.$ If $\delta = d_e \pi^R$ then by differentiating the above equation at $x = e$ in the direction of $X \in \mathfrak {g}$ and noting that $\pi^{R} (e) = e$ it follows that $$d_{e} \{f_1, f_2\} (X) = (\xi_1 \otimes \xi_2) \delta (X).$$
I have understood almost everything except the differentiation part. Could anyone please shed some light on it? I am trying to use chain rule but I failed as I am not familiar with differential of a tensor product at a point. Any help regarding this would be warmly appreciated.
Thanks for your time.
Source $:$ Chari and Pressley.
