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It seems to me that two very different definitions of derivative permeate the literature: in introductory calculus, the derivative $f'(x)$ of a (differentiable) function $f:U\subseteq\mathbb{R}\rightarrow\mathbb{R}$ at a point $x\in\mathbb{R}$ is a real number. When one moves on to multivariable calculus, the derivative $Df(x)$ of a function $f:U\subseteq\mathbb{R}^m\rightarrow\mathbb{R}^n$ at a point $x\in\mathbb{R}^m$ is a linear mapping $Df(x):\mathbb{R}^m\rightarrow\mathbb{R}^n$.
However, when one talk about curves $\gamma:\mathbb{R}\rightarrow\mathbb{R}^n$, we are already in the domain of multivariable calculus, so the derivative $\gamma'(x)=D\gamma(x)$ of $\gamma$ at a point $x\in\mathbb{R}$ should be a linear mapping $D\gamma(x):\mathbb{R}\rightarrow\mathbb{R}^n$. But usually I see people using $\gamma'(x)$ as an element of $\mathbb{R}^n$. How to reconcile those?

My guess is that $\mathcal{L}(\mathbb{R},\mathbb{R}^n)\simeq M_{1\times n}(\mathbb{R})\simeq\mathbb{R}^n$, so people just identify $\gamma'(x)$ with its tuple in $\mathbb{R}^n$, just like sometimes one speaks of a matrix as a linear map or a linear map as being a matrix.

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    Your guess is exactly right. – 2017-02-26
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    Think about it this way; the derivate tells you how much a function increases if you take another step. For a scalar function there is only one direction (onwards) but if the function isnt scalar there are more than one possible directions in which you could want to know the increase of the original function. Therefor the derivate is a linear mapping(bec it needs to know in which direction you would want to know the increase/decrease of the function) – 2017-02-26
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    @FeLix Yes, I know that: but that wasn't really the question. – 2017-02-26
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    So if you have a function from $\matbb{R}$ to $\mathbb{R}$ the increase/decrease is a vector, but there are only two possible directions + and -. So its a vector, multiplied by a scalar, so you just can identify it as a vector(like you did, this is perfectly fine, i just wanted to provide some context) – 2017-02-26
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    @Simones: yes i know, i just wanted to split up the comment, otherwise it would have been very long. (Using my mobile, so it took me some time, sorry) – 2017-02-26

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The relation between the velocity vector

$$ \gamma'(x) := \lim_{t \to 0} \frac{\gamma(x + t) - \gamma(x)}{t} $$

and the full differential $D\gamma|_{x} \colon \mathbb{R} \rightarrow \mathbb{R}^n$ is given by $\gamma'(x) = (D\gamma)|_{x}(1)$. Note that $D\gamma|_{x}(v)$ is the directional derivative of $\gamma$ at the point $x$ in the direction $v$, so it is defined by

$$ (D\gamma)|_{x}(v) := \lim_{t \to 0} \frac{\gamma(x + tv) - \gamma(x)}{t}. $$

By taking $v = 1 \in \mathbb{R}$, we get back the usual derivative. Since $D\gamma|_{x}$ is a linear map and $\mathbb{R}$ is one-dimensional, once you know the value of $D\gamma|_{x}(1)$, you know the value of $D\gamma|_{x}(c)$ for all $c \in \mathbb{R}$ using $$D\gamma|_{x}(c) = D\gamma|_{x}(c \cdot 1) = c D\gamma|_{x}(1).$$