Deduce from $\displaystyle \alpha ' (t) [f] = \frac {d(f(\alpha))}{dt} (t)$ that in the definition of directional derivative, $\displaystyle {\bf v_p}[f] = \frac {d}{dt} (f({\bf p} + t{\bf v}) \mid _{t = 0}$, the straight line $\displaystyle t \to {\bf p} + t{\bf v}$ can be replaced by any curve $\displaystyle \alpha$ with initial velocity $\displaystyle {\bf v_p}$, that is, such that $\displaystyle \alpha (0) = {\bf p}$ and $\displaystyle \alpha ' (0) = {\bf v_p}$

Not quite sure how to proceed. Any help would be appreciated.