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I was wondering what the definition of a partial derivative of such a map is?

Is it $\partial_if_j$ or is it anything else? The reason why I have doubts is that I found the definition that a partial derivative is the total derivative of a function $f(\,\cdot\,,x_2): x_1 \mapsto f(x_1,x_2)$, where $(x_1,x_2)$ is some decomposition of $x \in \mathbb{R}^n$.

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    $\begingroup$ A partial derivative of a function $(x_1,...,x_n) \mapsto f(x_1,...,x_n)$ would be the derivative of a restricted function $x_k \mapsto f(x_1,...,x_n)$ (for some $k$, all other parameters are fixed). $\endgroup$
    – copper.hat
    Dec 29 '13 at 22:26
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    $\begingroup$ @GitGud Really? That is news to me. Why can't you define a partial derivative of a vector valued function by the exact same formula that is used for scalar functions? $\endgroup$ Dec 29 '13 at 22:28
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    $\begingroup$ @GitGud: Wrong, buddy. :) $\endgroup$ Dec 29 '13 at 22:29
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    $\begingroup$ @GitGud: The 'partial' has to do with the parameterization of the domain. $\endgroup$
    – copper.hat
    Dec 29 '13 at 22:30
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    $\begingroup$ @GitGud: Fir a beginning calculus student, the context of a scalar function is standard. That doesn't make it the universal definition! $\endgroup$ Dec 29 '13 at 22:30
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The most natural definition would be this, I think: A map $f: \mathbb{R}^n \to \mathbb{R}^m$ can always be decomposed into $m$ maps $f_1, f_2, \ldots, f_m : \mathbb{R}^n \to \mathbb{R}$ that satisfy $$f(x) = (f_1(x), f_2(x), \ldots, f_m(x))$$ Then one can define the partial derivative of $f$ with respect to the $i$th argument simply by $$\frac{\partial f}{\partial x_i}(x) = \left( \frac{\partial f_1}{\partial x_i}(x), \frac{\partial f_2}{\partial x_i}(x), \ldots, \frac{\partial f_m}{\partial x_i}(x) \right)$$ The Jacobian matrix of $f$ is formed by letting these "partial derivatives" be column vectors in an $m \times n$ matrix.

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You already accepted an answer, but there is an even simpler and more general way to look at it, if you know what a topological vector space is (for example, any vector space with a norm is a topological vector space). If $S$ is a (real) topological vector space, and $f:\mathbb{R}^n \to S$, then define

$$ \frac{\partial f}{\partial x_i}(\mathbf{x}) = \lim_{h \to 0}\frac{f(\mathbf{x}+h\mathbf{e}_i)- f(\mathbf{x})}{h}, $$

if the limit exists. Here $\mathbf{e}_i$ is the unit vector in $\mathbf{R}^n$ with $1$ in the $i$th coordinate and $0$ in all the other coordinates. If $S = \mathbf{R}^m$, this is equivalent to Svinepels's answer.

I do not know how much, if at all, this can be generalized to a more general algebraic structure than "topological vector space".

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It helps a lot to think the derivative in matrix terms as $$Jf=\left[\begin{array}{cccc} {\partial f_1\over\partial x_1}&{\partial f_1\over\partial x_2}&\cdots&{\partial f_1\over\partial x_n}\\ {\partial f_2\over\partial x_1}&{\partial f_2\over\partial x_2}&\cdots&{\partial f_2\over\partial x_n}\\ \vdots&&\ddots&\vdots\\ {\partial f_m\over\partial x_1}&{\partial f_m\over\partial x_2}&\cdots&{\partial f_m\over\partial x_n} \end{array}\right],$$ where you can see in each row the gradients of each component $f_i$. The reason to use the "name" $Jf$, it is because it is also dubbed as the "Jacobian matrix" of $f$. With this device, it becomes more transparent how is the chain's rule and many other things.

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