Solution 1:

This is an important conceptual question, indeed!

In both cases, what you're written as the definition/characterization of derivative and/or integral would reasonably be called a "weak" derivative/integral or other linear operator $T$, with the word "weak" in this context meaning (exactly as you put it) $TF=G$ weakly if, for every (continuous) linear functional $\lambda$, $\lambda(TF)=\lambda(G)$.

Notably, this doesn't have to be just about distributions, but does apply to many topological vector spaces ... of various types of functions and generalized functions, and mappings between, and so on. (Google "quasi-complete".)

Gelfand and Pettis showed, early in the 20th century, that in many circumstances, "weak integrals" of distributions are again distributions... so there is indeed a distribution that creates the desired effect.

L. Schwartz and A. Grothendieck proved that the same is true for derivatives of distribution-valued functions (and others...)

So, although these proofs are not trivial, they are not crazy-hard, either. They are robust. The corresponding facts seem to be taken for granted, routinely, by many people.

A fairly careful treatment of many aspects of this is in chapters 13, 14, 15 of my recent book "Modern Analysis of Automorphic Forms, by Example", freely available at http://www.math.umn.edu/~garrett/m/v/current_version.pdf (by agreement with the publisher), as well as in hardcopy and other from Cambridge University Press.

EDIT: and, to be clear, those chapters are independent of the rest, so no knowledge of number theory or automorphic forms or ... is required for that bit of "modern analysis". :)

Solution 2:

That's how I would define the integral and derivative with respect to a parameter. In general the results might not be distributions, but under some conditions they are.

On the compact set $K,$ for each distribution $f(\lambda)$ there are $N_K(\lambda) \in \mathbb{N}_0$ and $C_K(\lambda)>0$ such that for every $\phi$ with support in $K$ the following inequality is valid: $$ |\langle f(\lambda), \phi \rangle| \leq C_K(\lambda) \sum_{|\alpha|\leq N_K(\lambda)} \|\partial^\alpha \phi\|. $$ If there exists $N_K$ such that $N_K(\lambda)\leq N_K$ for all $\lambda,$ and $\int C(\lambda) \, d\lambda < 0$ then $\int f(\lambda) \, d\lambda$ is a distribution: $$ \left| \int \langle f(\lambda), \phi \rangle \, d\lambda \right| \leq \int \left| \langle f(\lambda), \phi \rangle \right| \, d\lambda \leq \int C_K(\lambda) \sum_{|\alpha|\leq N_K(\lambda)} \| \partial^\alpha \phi \| \, d\lambda \\ \leq \int C_K(\lambda) \sum_{|\alpha|\leq N_K} \| \partial^\alpha \phi \| \, d\lambda = \int C_K(\lambda) \, d\lambda \sum_{|\alpha|\leq N_K} \| \partial^\alpha \phi \| . $$

The derivative case seems more difficult, or it's just my brain being slow today.