Derivative of a little-o remainder

Question

If we have a $\phi: \mathbb{R} \times \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$, $\phi = \phi(t, \mathbf{q},\alpha)$ one-parameter group of infinitesimal transformation which is $\mathcal{C}^2$ with respect to $t$ and $\mathbf{q}$, is it sufficient to $\phi$ to be differentiable with respect to $\alpha$ to write $$\phi = \phi(t, \mathbf{q}, 0) + \alpha \frac{\partial \phi}{\partial \alpha}(t, \mathbf{q}, 0) + o(\alpha)$$ where we want to be $\frac{\mathrm{d}}{\mathrm{d}\alpha} o(\alpha) = o(1)$ and we want $\eta (t, \mathbf{q}) = \frac{\partial \phi}{\partial \alpha}(t, \mathbf{q}, 0)$ to be $\mathcal{C}^1$ ? In particular, the fact that $\phi$ is $\mathcal{C}^2$ with respect to $t$ and $\mathbf{q}$, guarantees $\eta$ to be $\mathcal{C}^1$? I suspect that $\eta(t, \mathbf{q})$ is $\mathcal{C}^2$ as $\phi$ is $\mathcal{C}^2$ with respect to $t$ and $\mathbf{q}$.

Answer 1

That the function $\phi$ is differentiable with respect to $\alpha$ at $(t,q,0)$ for every $(t,q)$ means that there exists some functions $\eta$ and $\epsilon$ such that $\phi(t,q,\alpha)=\phi(t,q,0)+\alpha\eta(t,q)+\epsilon(t,q,\alpha)$ and $\lim\limits_{\alpha\to0}\alpha^{-1}\epsilon(t,q,\alpha)=0$. This is less than your hypotheses ${}^{(1)}$ hence your hypotheses guarantee that $\phi$ is differentiable with respect to $\alpha$ at $(t,q,0)$ for every $(t,q)$ (but these guarantee nothing about the differentiability at $(t,q,\alpha)$ with $\alpha\ne0$).

${}^{(1)}$ The only point to check is that if $\epsilon(t,q,\ )$ is differentiable in a neighborhood of $0$ with derivative $\varrho(t,q,\ )$ such that $\lim\limits_{\alpha\to0}\varrho(t,q,\alpha)=0$, then $\lim\limits_{\alpha\to0}\alpha^{-1}\epsilon(t,q,\alpha)=0$. This is direct from the mean value theorem.