Given two finite dimensional connected quiver algebras $A$ and $B$. Is there an easy example such that $A$ and $B$ are derived equivalent and $A$ has finite representation type but $B$ has infinite representation type? Easy should mean, that it is not hard to see that the algebras are derived equivalent and not hard to see that one is representation finite and the other representation infinite.
addition: The easiest example I found yet can be found in http://link.springer.com/article/10.1007%2Fs10468-009-9169-y#page-1 where some Nakayama algebras are derived equivalent to tame heredtiary algebra of type $\overline{E_7}$ or $\overline{E_8}$.
Take the path algebra $B$ of the $\tilde{D}_4$ quiver
$$\require{AMScd}\begin{CD} @.3@.\\ @.@AAA@.\\ 1@>>>2@>>>4\\ @.@VVV@.\\ @.5@. \end{CD}$$
This has infinite representation type. The module $T=I_1\oplus P_1\oplus P_3\oplus P_4\oplus P_5$ is a tilting module whose endomorphism algebra $A$ has the same quiver, but with zero relations for all paths of length two. So $A$ and $B$ are derived equivalent.
If $V$ is a representation of $A$ with a nonzero vector space at vertex $1$, then the subrepresentation generated by a nonzero element of this vector space is either the injective at vertex $1$ or the injective at vertex $2$, and in particular is a direct summand. So the representation theory of $A$ reduces to that of the subquiver spanned by vertices $2,3,4,5$, which has finite representation type.