Partial differentation and Lie group action

Question

Let $G$ be a Lie group, for each $g \in G$ let us denote by $L_g$ the left multiplication by $g$. Let $(U,\varphi)$ be a coordinate chart around $e$ and let $f \in C^{\infty}(G)$ be a smooth function. How to show that $\frac{\partial(f \circ L_g \circ \varphi^{-1})}{\partial x_i}(\varphi(h))=\frac{\partial(f \circ L_{g'} \circ \varphi^{-1})}{\partial x_i}(\varphi(h'))$ if $h,h' \in U$ and $g,g'$ are such that $gh=g'h'$? I'm pretty sure that it should be true: I came across this question considering the triviality of the tangent bundle of the Lie group.