Expansion of $f(\vec v +F(\vec v)\varepsilon)$? Naturally I would think that the Taylor expansion of $f(\vec v +\vec F(\vec v)\varepsilon)$ would be:
$$f(\vec v+F(\vec v)\varepsilon)=f(\vec v)+\frac{df(\vec v)}{d\vec v} \cdot\vec F(\vec v) \varepsilon+...$$
But is this still correct even though $\vec F(\vec v)$ depends on $\vec v$? If so why and if not what is correct?
 A: Let $\vec{\omega}(\epsilon)=\vec{v}+\epsilon{\vec{F}(\vec{v})}$. The taylor expansion in epsilon is
$$f(\vec{\omega}(\epsilon))=f(\vec{\omega}(0)))+\frac{df(\vec{\omega}(\epsilon))}{d\epsilon}\Big]_{\epsilon=0}\epsilon+\mathcal{O}(\epsilon^{2})$$
Then
$$\frac{df(\vec{\omega}(\epsilon))}{d\epsilon}\Big]_{\epsilon=0}=\frac{d\vec{\omega}(\epsilon)}{d\epsilon}\cdot\frac{df(\vec{\omega}(\epsilon))}{d\vec{\omega}(\epsilon)}\Big]_{\epsilon=0}=\vec{F}(\vec{v})\cdot\frac{df(\vec{v})}{d\vec{v}}$$
Hence
$$f(\vec{\omega}(\epsilon))=f(\vec{v})+\vec{F}(\vec{v})\cdot\frac{df(\vec{v})}{d\vec{v}}\epsilon+\mathcal{O}(\epsilon^{2})$$
In the second line, to be more specific about the chain rule. You have $$df(\vec{\omega}(\epsilon))=\sum_{k}\frac{\partial{f}(\vec{\omega}(\epsilon))}{\partial{\omega_{k}(\epsilon)}}d\omega_{k}(\epsilon)$$
Then divide by $d\epsilon$ to give 
$$\frac{df(\vec{\omega}(\epsilon))}{d\epsilon}=\sum_{k}\frac{\partial{f}(\vec{\omega}(\epsilon))}{\partial{\omega_{k}(\epsilon)}}\frac{d\omega_{k}(\epsilon)}{d\epsilon}=\frac{df(\vec{\omega}(\epsilon))}{d\vec{\omega}(\epsilon)}\cdot\frac{d\vec{\omega}(\epsilon)}{d\epsilon}$$
Where $\omega_{k}(\epsilon)$ is the $k^{th}$ component of $\vec{\omega}(\epsilon)$.
