Distributional derivative as linear approximation

Is it possible to characterize the distributional derivative as some sort of "best linear approximation" of a distribution (a la the Fréchet/Gâteaux derivatives), if viewed in the appropriate spaces?

For example, let $f \in D'$, $\phi \in D$, and let $\psi_x \in D$ be a mollified ramp function so that $\psi_x(x,y,z,...)=x$ on the support of $\phi$. What can we say about the quantity:

$$\frac{f(\phi)-f(\phi+s\psi_x)}{s}-\frac{\partial f}{\partial x}(\phi)$$

where $\frac{\partial f}{\partial x}$ is the usual distributional derivative?

Going further, if we relax possibilities for "derivative direction", so that simply $\psi \in D$, does it even make sense to talk about the quantity "$Df:D'\rightarrow D'$":

$$Df(\phi)(\psi):=\lim_{s\rightarrow 0} \frac{f(\phi)-f(\phi+s\psi)}{s}$$ ?

In the space of distributions we don't have a norm, but we still have a vector space structure and a notion of convergence, so it seems that something like this is at least plausible.

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(I use below $f$ for the distribution and $\phi$ for the test function...)
Let $\phi\in D'$ and $f\in D$. Let $\tau_x:D\to D$ be translation, so that $\tau_x(f)(y)=f(y+x)$, and let $\tau^t_x:D'\to D'$ be the transpose of $\tau_x$. Then $\frac{\tau_{h}f-f}h\to f'$ in $D$ when $h\to0$, so $$\langle\frac{\tau^t_h\phi-\phi}h,f\rangle=\langle\phi,\frac{\tau_{h}f-f}h\rangle\to\langle\phi,f'\rangle=\langle \phi',f\rangle$$ as $h\to 0$. Since the topology in $D'$ is that of pointwise convergence on $D$, it follows that $$\lim_{h\to0}\frac{\tau^t_h\phi-\phi}h=\phi',$$ which is something along the lines of what you wanted.
NB: Differentialting the actual map $\phi:D\to\mathbb R$ is not going to produce anything interesting, because the map is linear itself.
Thanks, this is the sort of thing I was looking for. That $\phi:D\rightarrow \Re$ is already linear explains why I was going around and around in circles getting nowhere with my original approach. (: –  Nick Alger Dec 11 '10 at 15:19