Why is the differential of a little-o also a little-o?

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Let's consider a vector field ($C^{\infty}$) $$X : \mathcal{A} \rightarrow \mathcal{V}$$ where $\mathcal{A}$ is an affine space with underlying vector space $\mathcal{V}$.

Let's assume that, for all $t \in \mathbb{R}$, we have a diffeomorphism $$\phi_t : \mathcal{A} \rightarrow \mathcal{A}$$ s.t. $\phi_t(A) = \gamma_A(t)$ for all $t \in \mathbb{R}$, $A \in \mathcal{A}$, where $$\gamma_A : \mathbb{R} \rightarrow \mathcal{A}$$ is the solution of the Cauchy problem \begin{cases} (d\gamma_A/dt)(t) = X(\gamma_A(t)) \\ \gamma_A(0) = A \end{cases}

With this assumptions, we have that, for fixed $A \in \mathcal{A}$, the difference $$\phi_t(A) - A - tX(A) = \gamma_A(t) - A - tX(A)$$ is a little-o of $t$, for $t \rightarrow 0$.

Why can we conclude that also its differential (recall that $\phi_t$ is a diffeomorhpism) $$\nabla_A\phi_t - \text{id}_{\mathcal{V}} - t\nabla_AX$$ is a little-o of $t$, for $t \rightarrow 0$?