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When we map from $\mathbb{R} \to \mathbb{R}$, the derivative is given by $\lim_{x \to a} \dfrac {f(x) - f(a)}{x-a}.$ When we map from $\mathbb{C} \to \mathbb{C}$ (basically $\mathbb{R}^2 \to \mathbb{R}^2$) we also have a definition for the derivative, $\lim_{z \to w} \dfrac {f(z) - f(w)}{z-w}.$

However, when we map $\mathbb{R}^2 \to \mathbb{R}$, we have only partial derivatives, $\dfrac {\partial f}{\partial x}$, $\dfrac {\partial f}{\partial y}$, and directional derivatives. Why is it that we can't define the derivative just as we do in the other $2$ cases? Is it because there is no nice way to define division between a member of $\mathbb{R}$ and a member if $\mathbb{R}^2$?

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    Think about it this way: The derivative that you are talking about is the differential map in $\mathbb{R}^n$, and in the special case $n=1$ that differential coincides with the only partial. – 2017-02-20
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    @Smurf What do you mean by "the differential map in $\mathbb{R}^n$? – 2017-02-20
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    *"Is it because there is no nice way to define division between a member of R and a member if R²?"* Yes. – 2017-02-20
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    @JackM Patrick Stevens' answer seems to imply that we indeed can define the derivative $f: \mathbb{R}^n \to \mathbb{R}^m$ for any $m, n$. – 2017-02-20
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    @Ovi You might find the general definition of the derivative to be quite unfamiliar, though. JackM is saying you're right about why the one-dimensional derivative you already know and love is hard to generalise to higher dimensions in a recognisable way. – 2017-02-20
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    @PatrickStevens Ah okay thanks. – 2017-02-20
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    You say $\mathbb{C}\mapsto\mathbb{C}$ is 'basically $\mathbb{R}^2\mapsto\mathbb{R}^2$', but that's not at all true. There's a _lot_ of structure on $\mathbb{C}$ that isn't in $\mathbb{R}^2$ and that structure is essential to making sense of the notion of 'the' derivative. – 2017-02-20

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We can define the derivative: a map $f: \mathbb{R}^n \to \mathbb{R}^m$ is said to be differentiable at $x \in \mathbb{R}^n$ if there is a linear map $\alpha: \mathbb{R}^n\to \mathbb{R}^m$ such that for every $h \in \mathbb{R}^n$, we have $f(x+h) = f(x) + \alpha(h) + \epsilon(h) |h|$ where $\epsilon(h) \to 0$ as $h \to 0$, and where $|\cdot|$ is the standard Euclidean norm. (It's pretty easy to prove that such an $\alpha$ is unique if it exists; we call it the derivative.)

The idea is to express the map as the sum of something linear and something small. Then the linear bit is the "derivative".

In the one-dimensional case (e.g. $f(x) = x^2$), we have $\alpha(h) = f'(x) h$ (in this example, $\alpha(h) = 2xh$) at each point $x$.

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One way to look at is that, as you say, there's no way to "divide" vectors in $\mathbf{R}^2$ that really makes sense.

But you can think of it this way. When you differentiate a function $y = f(x)$ at a point $x_0$, what you're doing is saying that that function can be approximated by a linear function for values of $x$ close to $x_0$, namely $$y \approx f(x_0) + f'(x_0)(x-x_0).$$

When you differentiate a function $z = f(x,y)$ of two variables, you likewise want to write that the function can be approximated by a linear function near $(x_0,y_0)$: $$z \approx k + a(x-x_0) + b(y-y_0),$$ whose graph is a plane. In this approximation, the constant $k$ is $f(x_0,y_0)$, and the coefficients $a$ and $b$ are $\frac{\partial f}{\partial x} (x_0,y_0)$ and $\frac{\partial f}{\partial y} (x_0,y_0)$, respectively. In any case you need two coefficients.

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Because $\mathbb{C} \to \mathbb{C}$ is not ''basically the same'' as $\mathbb{R}^2 \to \mathbb{R^2}$.

The possibility of define a derivative for a function $ f:\mathbb{C} \to \mathbb{C}$ use , as you noted, the fact that $ \mathbb{C}$ is a field, and this implies that the derivative, if it exists, is much more than the linear approximation of the function near a point.

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    Hmm in most textbooks I have seem, $\mathbb{C}$ is defined as $\mathbb{C}:= \{ (x, y): x, y \in \mathbb{R} \}$. I'm not sure I see how this is different from $\mathbb{R}^2$. – 2017-02-20
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    $\mathbb{C}$ is a field. This means that it is the set of couple $\{(x,y)\in \mathbb{R}^2\}$ with two operations defined in such away that satisfies some properties. So it is an algebraic structure very different from a simple cartesian product of sets. And it is this structure that gives a lot of special properties to holomorphic functions. – 2017-02-20
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I just want to elaborate a bit more on a point that the existing answers has made: The key to there being a "single" derivative on $\mathbb R$ and $\mathbb C$ is indeed because we have division in those sets (they are fields), but it's worth the time to think a little about what that changes.

Take a function $f$ on $\mathbb R$ (to $\mathbb R$, say). If we want to differentiate it at a point $x$, we look at the limit $$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. $$ One way of looking at this is to say that we start at $f(x)$ and deform the function a little in the direction $h$ to get $f(x+h)$, and look at how the deformations behave when we make the direction $h$ smaller and smaller.

Consider now a holomorphic function $f$ on $\mathbb C$ (again, let's say to $\mathbb C$). We can play the same game again to differentiate it at a point $z$: Deform it a little in a complex direction $h$, and look at how the deformations behave as the direction $h$ gets smaller and smaller.

Now let's take a function $f : \mathbb R^2 \to \mathbb R$. Again we fix a point $(x,y)$ and want to differentiate it at that point, so we try to play the deformation game again. But now we have a whole circle of directions we could deform the function in, and there is no single one that's better than the others, so we have to consider each one of them separately. This gives rise to the differential operator $D$ on $\mathbb R^2$ (or $\mathbb R^n$), where we both need to say at what point we want to differentiate our function, and in what direction, which corresponds to asking how the deformations of the function in that direction behave.

When we can divide things, there is only one direction to go into, so we get a privileged derivative kind of by accident.