How to define derivative in Minkowski space My understanding of derivative is like this: it is the unique linear mapping that sends the difference in $x$ to the difference in $f(x)$ when the difference in $x$ is small. To put it more rigorously, $g$ is the derivative of $f$ if and only if for every positive real number $\epsilon$, there exists a positive real number $\delta$ such that
$$\lVert f(x+\Delta x)-f(x)-g(x)(\Delta x)\rVert \le \epsilon \lVert\Delta x\rVert$$
for every $\lVert\Delta x\rVert < \delta$. Is it the same in Minkowski space where $\lVert \Delta x \rVert$ may be equal to $0$?
 A: I've only seen this definition of derivative (The Fréchet Derivative) used for normed spaces - in the case of a semi-norm such as your example, things get very nasty. For example in order for $f$ to be differentiable by this definition we must have $f(x + \Delta x) = f(x) + g(x) \Delta x$ whenever $\Vert \Delta x \Vert \le 0$, i.e. $f$ must be affine outside the light-cone of any particular point. I'm pretty sure that by pasting together this fact at multiple base points you can conclude that $f$ must be affine.
The better abstract definition of derivative to use in this setting is the Gâteaux derivative, which is defined independent of the norm. It thus treats Minkowski space identically to Euclidean space, producing the familiar differential/gradient whenever it exists. If you require the Gâteaux derivative to be linear to call the function differentiable then you recover exactly the classically differentiable functions.
A: Minkowski space is $\mathbf{R}^4$, considered with the Lorentz form $g : \mathbf{R}^4 \times \mathbf{R}^4 \to \mathbf{R}$ defined by 
$$g(X,Y) = x^0y^0 - \mathbf{x}.\mathbf{y},$$
where we use the notation $X = (x^0, \mathbf{x}) \in \mathbf{R} \times \mathbf{R}^3$. 
In physics, differential calculus is done on Minkowski space using the standard Euclidean structure for $\mathbf{R}^4$. However, physicists use operators like the four-gradient rather than the Fréchet derivative. 
The four-gradient is defined for a function $f : \mathbf{R}^4 \to \mathbf{R}$ by 
$$\text{Grad } f = \bigg(\frac{1}{c}\frac{\partial f}{\partial t}, -\nabla f\bigg)$$
where $\nabla$ is the gradient on $\mathbf{R}^3$ and the partial derivatives are calculated using the usual definitions for $\mathbf{R}^4$.
Similarly, the Fréchet derivative of a function on Minkowski space could be considered using the Euclidean structure for $\mathbf{R}^4$: for example, to show that the transformation law for the Christoffel symbols implies that Poincaré transformations are affine.
A: In a Minkowski space such a derivative does not exists in general. To see this focus on direction of $\Delta x$ ,if its direction changes we have a different value for $g$ so we can not define $g$ as function of $x$.
A: Restating with bold vectors for clarity:

$\nabla {f}({\bf x})$ is the gradient of $f$ at point $\bf x$, iff for
  all $\epsilon$ there exists $\delta$ such that if $|{\bf \Delta x}|<\delta$: $$\left| f({\bf x}+ {\bf \Delta x})- f({\bf x})-\nabla{f}({\bf x})\cdot({\bf \Delta x}) \right| < \epsilon |{\bf \Delta x}|$$

In theory, when $|{\bf \Delta x}|=0$, the gradient vector thus defined can take on any value. In practice however, if $|{\bf \Delta x}|=0$, this means we are taking the derivative along the edge of the light cone. Since all physical phenomena occur only in the interior of the light cone, this isn't really a problem.
