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Let $\varphi(t)$ be a smooth monoparametric family in $\operatorname{Aut}\mathfrak g$, $\mathfrak g$ being a Lie algebra, and $\varphi(0)=\operatorname{id}$. Prove that $\varphi'(0)\in\operatorname{Der}\mathfrak g$.($\operatorname{Der}\mathfrak g$ is the derivations on $\mathfrak g$.) Could someone help me prove the statement?

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For each $t\in\mathbb R$,$$(\forall Y,Z\in\mathfrak{g}):\varphi(t)\bigl([Y,Z]\bigr)=\bigl[\varphi(t)(Y),\varphi(t)(Z)\bigr].$$This means that you have two identical maps from $\mathbb R$ into $\mathfrak g$. Differentiating both of them at the origin and using the fact that the Lie bracket is bilinear, you get that$$\varphi'(0)\bigl([Y,Z]\bigr)=\bigl[\varphi'(0)(Y),Z\bigr]+\bigl[Y,\varphi'(0)(Z)\bigr].$$