How to prove the one-variable calculus definition of derivative extends to $\Bbb C$ *only* because $\Bbb C$ is a field? I have been told the one-variable calculus definition of derivative extends to $\Bbb C$ only because $\Bbb C$ is a field.
See : Higher dimensional analogues of the argument principle?
$$ \frac{df}{dx}=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $$
But I wonder why this definition does not hold for commutative rings such as $\Bbb C^n $ or $\Bbb R \Bbb \times \Bbb C^n $?
How to prove the one-variable calculus definition of derivative extends to $\Bbb C$ only because $\Bbb C$ is a field ?
I think I understand why $\Bbb C$ is the only possible field, but I am clueless why we need a field.
 A: The main problem is that the usual definition of derivative needs to calculate
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
in which you are assuming that in some neighbourhood of zero all elements distinct from zero are inversible in order to make the expression $\frac{f(x+h)-f(x)}{h}$ meaningful near zero. Furthermore, one needs also a topology, as said in other answer, in order to make the limit in the expression meaningful.
For making this more concrete, let's consider $\mathbb{C}^n$. For this object we have a natural topology, so there is no problem a priori with the limit but when we copy that definition directly, we obtain the following
$$\lim_{(h_1,\ldots,h_n)\to 0} \frac{f((x_1,\ldots,x_n)+(h_1,\ldots,h_n)-f(x_1,\ldots,x_n)}{(h_1,\ldots,h_n)} \text{.}$$
And what does dividing by a vector mean? But more important, even if it is meaningful, is it always possible? And except over division algebras (which containing $\mathbb{R}$ and being topologically "nice" -connected and locally compact with a non trivial topology-, you only have $\mathbb{R}$ itself, $\mathbb{C}$ and $\mathbb{H}$ -cuaternions-), in the rest of the cases this is senseless and gives us a limit with infinitely man non-defined values.
However, these are not the only ways to approach the problem of derivatives. We have two other approaches, equivalent over the reals, but different over other objects. These are:


*

*The Fréchet derivative, or more colloquially the differentiability approach, says that $f$ has at a point $x$ "derivative" $a$ if you can write
$$f(x+h)=f(x)+a\cdot x+o(h)$$
with $o(h)$ being a function of $h$ that approach to zero faster that $h$ itself.

*The Gâteaux derivative, or more colloquially the directional derivative, in which we consider for each vector $v$ the derivative of $f$ at $x$ in the direction of $v$ to be
$$\lim_{h\to 0} \frac{f(x+hv)-f(x)}{h}\text{,}$$
i.e. we consider the derivative in some domain that is a vector space over some field and recycle the previous definition by restricting ourselves to a straight line that essentially is the field we were working in.



In order to make this more concrete, I will expose some parts in order to make these ideas more concrete and detailed. In some ways these exposition will be non-standard.
Topological fields and the usual derivative definition
A topological field $\mathbb{F}$ is a field $\mathbb{F}$ together with some non-trivial topology $\tau$ that is compatible with the internal operations of the field, i.e., such that the following functions are continuous


*

*the addition $+:\mathbb{F}\times\mathbb{F}\rightarrow \mathbb{F}$, given by $+(x,y)=x+y$;

*the multiplication $\cdot:\mathbb{F}\times\mathbb{F}\rightarrow \mathbb{F}$, given by $\cdot(x,y)=x\cdot y$; and

*the inversion $i:\mathbb{F}\backslash \{0\}\rightarrow \mathbb{F}\backslash\{0\}$, given by $i(x)=x^{-1}=1/x$.


(Here, $\mathbb{F}\times\mathbb{F}$ has the product topology and $\mathbb{F}\backslash \{0\}$ the subspace topology.) Here, the non-trivial topology requirement is only to guarantee that we have more open sets that the total set and the empty set. However, one easy exercise in topology shows that a topological field must be Hausdorff. 
Being technical, it is usual to ask for other conditions such as completeness -although we don't have a metric, it can be define since topological fields have an uniform structure- or a non-discrete topology in order to have nice properties. Nevertheless, this is not necessary in order to define derivatives although this is makes at one's risk of them not behaving well.
So the usual definition of derivative translate exactly as follows:

Let $D\subseteq \mathbb{F}$ be an open set, a function $f:D \rightarrow \mathbb{F}$ is derivable at a point $x\in D$ if
  $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
  exists. In this case, we call the value of that derivative of $f$ at $x$ and write $f'(x)$ for the value of the limit.

Note that here there is no problem with the definition since in the limit $h$ never takes the value zero, although for having sense $lim_{h\to 0}$ we need a non-discrete topology -otherwise is senseless approaching to zero without being zero-.
Here, one can deduce the typical laws for derivatives such as the usual formula for the derivative of the sum, product and composition of two functions. However, depending on the field properties of analysis can change drastically anlysis over $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{Q}_p$ are radically different due to their topological propierties.
Relationship with the other definitions
When considering the concept of the Fréchet derivative, the given here is equivalent to this since given the previous $f$ derivable at $x$ we have taking
$$o(h)=f(x+h)-f(x)-f'(x)h\text{,}$$
that
$$f(x+h)=f(x)+f'(x)h+o(h)$$
with $o(h)$ going to zero faster that $h$ since
$$\lim_{h\to 0}\frac{o(h)}{h}=\lim_{h\to 0}\frac{f(x+h)-f(x)-f'(x)h}{h}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}-f'(x)=0$$
due to the continuity of the difference -which is the deduce from the continuity of the sum and the product-.
And when considering the Gâteaux derivative, the concept is equivalent if we consider the derivative with a vector over the field as a vector space over itself. In that case,
$$\lim_{h\to 0}\frac{f(x+hv)-f(x)}{h}=f'(x)v\text{.}$$
However, some fields can be considered as vector spaces over different fields. More concretely, $\mathbb{C}$ as a vector space over $\mathbb{R}$. In that cases the difference in the product would be in the meaning of $f'(x)v$, in the complez case, usual complex multiplication; in the real case, the application of the linear map $z\mapsto f'(x)z$ to $v$.
Metrical rings and the Fréchet definition
Topological rings are defined in a similar fashion to topological fields, by asking the sum and multiplication to be continuous -not the inversion, since it is not defined in all rings-. However, when reading the definition of Fréchet, we have the problem that in a topological ring we cannot compare the notion of $o(h)$ approach to zero faster than $h$ does. Therefore, we are force to put inside metric concepts in order to obtain means to do that comparison.
From here, a metrical ring $R$ will be a ring $R$ together with a metric $d$ such that addition and multiplication are continuous with respect to the topology induced by this metric in the same sense we considered in the case of topological field. In many cases, this metric can be seen also as a norm of the ring as a topological space over $\mathbb{R}$, $\mathbb{C}$ or other field such as the cases of $\mathbb{R}^n$ and $\mathbb{C}^n$ -furthermore in those cases, since all norms are equivalent, we can rely in the fact that the meaning of $o(h)$ going to zero faster than $h$ doesn't rely in the norm taken-.
Once the background is fixed and assuming whatever one needs to assume in order to not have strange behaviors with the derivatives (completeness, local compactnes, etc.) if you have to, the Fréchet definition is perfect in this case and gives the following definition:

Given a metrical ring $(R,d)$, an open set $D\subseteq R$ and a function $f:D\mapsto R$, we say that $f$ has Fréchet derivative $a$ if we can write
  $$f(x+h)=f(x)+ah+o(h)$$
  for some function $o(h)$ that approaches zero faster than $h$ in the sense that
  $$\lim_{h\to 0}\frac{d(0,o(h))}{d(0,h)}=0\text{.}$$

As thing to point out, the commutative case doesn't give us many problems, but in the non-commutative case we have some problems as one can note from the fact that it is not the same multiplying at the left and at the right. In those cases, other approaches may be necessary and maybe we have to abandon the concept of derivative as an element of the ring itself. More colloquially, in the non-commutative case this definition is a little bit leftist.
Nevertheless, the previous definition is perfect for the case you are looking for of $\mathbb{C}^n$ and $\mathbb{R}\times\mathbb{C}^n$ as rings. If one knows a little about several variables differentiation, it is easy to characterize the function differentiable in this sense. 
Normed vector spaces and the Fréchet and Gâteaux derivative
In the previous cases we had an internal multiplication, in this case the internal multiplication is substituted by an external one and we obtain the concept of vector space. As before, we must consider some sort of metrical/topological structure for our considered for our topological vector space.
First, we say that a topological field $\mathbb{F}$ is normed if there is a function
$$|\cdot|:\mathbb{F}\rightarrow \mathbb{R}$$
such that:


*

*It is non-negative, i.e., for all $x\in \mathbb{F}$, $|x|\geq 0$.

*For all $x\in\mathbb{F}$, $|x|=0$ iff $x=0$.

*It satisfies the triangle inequality, i.e., for all $x,y\in\mathbb{F}$, $|x+y|\leq |x|+|y|$.

*It is multiplicative, i.e., for all $x,y\in\mathbb{F}$, $|xy|=|x||y|$.


Second, given a normed field $(\mathbb{F},|\cdot|)$, a normed $\mathbb{F}$-vector space $(V,||\cdot ||)$ is vector space over $\mathbb{F}$ $V$ togethe with a function called norm
$$||\cdot||:V\rightarrow \mathbb{R}$$
such that:


*

*It is non-negative, i.e., for all $v\in V$, $||v||\geq 0$.

*For all $v\in V$, $||v||=0$ iff $v=0$.

*It satisfies the triangle inequality, i.e., for all $v,w\in V$, $||v+w||\leq ||v||+||w||$.

*It is compatible with $|\cdot|$, i.e., for all $x\in\mathbb{F}$ and $v\in V$, $||xv||=|x|\,||v||$.


And here the definitions take other form, i.e., we are not more interested in vectors for acting as derivatives, but more in linear applications acting as derivatives. In this sense the multiplication
$$ah$
of the Fréchet definition passes to be the application of a linear application. Similar reinterpretations can be made given the case, but the more popular one is the following:

Let $(\mathbb{F},|\cdot|)$ be a normed field, $V$ and $W$ normed vector spaces over $(\mathbb{F},|\cdot|)$ and $D\subset V$ an open set of $V$. A function $f:D\rightarrow W$ is "derivable" at $x\in D$ in the Fréchet sense if there is a linear map $A:V\rightarrow W$ (satisfying some nice properties depending on the case) such that we can write
  $$f(x+h)=f(x)+Ah+o(h)$$
  for some function $o(h)$ in a neighborhood of zero in $V$ $N$ that appraches to zero faster than $h$ in the sense that
  $$\lim_{h\to 0}\frac{||o(h)||_W}{||h||_V}\text{.}$$
  The linear map $A$ is called the "derivative" of $f$ at $x$.

However, sometimes this is a very strong concept and we are interested in a weaker version. That is, we are no interested in the case when we approaches uniformly, but in the case when we approach in one direction or another. In this case, the Gâteaux definition is the appropriate one. This is the following one:

Let $(\mathbb{F},|\cdot|)$ be a normed field, $V$ and $W$ normed vector spaces over $(\mathbb{F},|\cdot|)$ and $D\subset V$ an open set of $V$. The Gâteaux derivative of the function $f:D\rightarrow W$ at $x\in D$ in the direction of $v\in V$ is
  $$D_x(f)(v)=\lim_{t\to 0} \frac{f(x+vt)-f(x)}{t}$$
  where $t\in \mathbb{F}$.

An easy modification shows this can be written as
$$f(x+tv)=f(x)+D_x(f)(v)+o(v,t)$$
with
$$\lim_{t\to 0}\frac{o(v,t)}{t}=0$$
in $(W,||\cdot||_W)$ for all $v\in V$. So the big difference between the Fréchet and Gâteaux derivative is that the first one is direction-independent and the second one direction-dependent. Furthermore, the Gâteaux derivative has the advantage that you don't need a norm in order to make sense of it.
Even more
In all the previous cases, I have give a very broad introduction. One may wish to consider further cases such as modules (the equivalent of vector spaces for rings) or the non-commutative case. Furthermore, the exposition I have given is highly superficial since it does not deal how the topology you are working in influence the properties if the derivable functions you are working with.
I hope this helps to understand the different ways in which a derivative can be actually seem.
NOTE: For any inquiries, please tell me.
A: Whoever said this is wrong. The field also needs to have a topology for the limit to make sense. For finite fields, the only real topologies don't really work here, so you also want a field that is a manifold or at least a topological vector space for this to make sense.
Edit: I need to mention that it must be a field as well, still.
A: the purely algebraic notion of derivative is the following
Let $f$ be the function on the ring $R$, $R[d]=R[x]/(x^2)$ is the ring of dual numbers. Then $f$ is differentiable if there exists an extension of $f$ to $R[d]$ such that the ratio $f'(x)=\frac{f(x+dy)-f(x)}{dy}$ does not depend on $y$. 
This definition does not need any topology.
Of course this is heuristic there are a lot of different ways to extend $f$ but if you want the extension to extend some structure that was given on function $f$ it will work (you can define derivatives of polynomials, exponents or trigonometric functions in such way)
