Show that for every curve $\gamma$ in $\mathbb{R^m}$ we obtain the same velocity vector in $\mathbb{R^n}$ Let $f:\mathbb R^m \rightarrow \mathbb R^n$ differentiable at the point $p\in \mathbb R^m$. Prove that for every curve $\gamma: I \rightarrow \mathbb R^m$, such that $\gamma(0)=p$ and $\gamma'(0)=v$, we obtain the same velocity vector at $t=0$ for a curve in $\mathbb R^n$ given by $\phi(t)=(f \circ \gamma) (t)$.
Attempt:
Let $\phi(t)=(f \circ \gamma) (t).$ Differentiating, the chain rule gives us that $\phi'(t) = (\nabla f) (\gamma(t)) \cdot \gamma'(t)$. In particular, for $t=0$ we have that $\phi'(0) = (\nabla f)(\gamma(0)) \cdot \gamma'(0)= \nabla f(p) \cdot v$.
Edit:
I realized that I was using the chain rule for a scalar field, which doesn't make sense, since $f$ is a vector field. So the chain rule should give us for $t=0$: $\phi'(t)=Df(\gamma(0)) \gamma'(0) = Df(p) v$, which is a matrix of dimension $m \times 1$. Does the fact that this derivative is independent of any $\gamma$ curve is sufficient to conclude that we obtain the same velocity vector?
 A: To expand on Daniel's answer, a more precise statement is that if $\gamma_1$ and $\gamma_2$ are two curves in $\mathbb{R}^m$ with $\gamma_1(0)=\gamma_2(0)$ and $\gamma_1^\prime(0)=\gamma_2^\prime(0)$ and $f$ a map $\mathbb{R}^m \rightarrow \mathbb{R}^n$ then $(f \circ \gamma_1) (0)= (f \circ \gamma_2)(0)$ and $(f \circ \gamma_1)^\prime (0)= (f \circ \gamma_2)^\prime(0)$.
To see that $(f \circ \gamma_1) (0)= (f \circ \gamma_2)(0)$, use the definition of composition
\begin{align*}
(f \circ \gamma_1) (0) &= f(\gamma_1(0)) \\
&= f(\gamma_2(0) \\
&= (f \circ \gamma_2)(0).
\end{align*}
To see that $(f \circ \gamma_1)^\prime (0)= (f \circ \gamma_2)^\prime(0)$, we use the chain rule to calculate
\begin{align*}
(f \circ \gamma_1)^\prime(0) &= \sum_{\mu=1}^m \frac{\partial f }{\partial x^\mu} \frac{\gamma_1^\mu}{dt}\Bigr|_{t=0} \\
&= \sum_{\mu=1}^m \frac{\partial f }{\partial x^\mu} \frac{\gamma_2^\mu}{dt}\Bigr|_{t=0} \\
&= (f \circ \gamma_2)^\prime(0)
\end{align*}
where $x^\mu$ are co-ordinates on $\mathbb{R}^m$ and $\gamma_1^\mu,\gamma_2^\mu$ are the co-ordinates of $\gamma_1,\gamma_2$ respectively (I swept a bit of stuff about where everything is evaluated under the carpet when I used the chain rule here).
