Let $G$ be a Lie group. Let $i : G \longrightarrow G$ be the inversion map given by $g \mapsto g^{-1}.$ For $x,y \in G$ let $\rho_{y} : G \longrightarrow G$ and $\lambda_{x} : G \longrightarrow G$ respectively denote the right multiplication by $y$ and the left multiplication by $x.$ Then for any $x \in G$ $$d_{x} i = - (d_{e} \lambda_{x^{-1}}) \circ (d_{x} \rho_{x^{-1}}).$$
By chain rule $$(d_{e} \lambda_{x^{-1}}) \circ (d_{x} \rho_{x^{-1}}) = d_{x} (\lambda_{x^{-1}} \circ \rho_{x^{-1}}).$$ So in order to show the required equality we need to show that for every $f \in C^{\infty} (G)$ $$X_{x} (f \circ i) = - X_{x} (f \circ \lambda_{x^{-1}} \circ \rho_{x^{-1}})$$ for every $X_{x} \in T_{x} G.$ But I am not getting it. Any help in this regard would be greatly appreciated.
Thanks for your time.
Source $:$ Pavel Etingof lecture notes on quantum groups.