This is a sequel to this question.
Let $k$ be a field, let $A$ be the $k$-algebra $k[\varepsilon]$ with $\varepsilon^2=0$, and consider the following three abelian categories:
$\bullet\ \text M(A)$ is the category of $A$-modules,
$\bullet\ \text{FD}(A)$ is the category of finite dimensional $A$-modules,
$\bullet\ \text M(k)$ is the category of $k$-vector spaces.
Let $\text K(A), \text K(\text{FD}(A)), \text K(k), \text D(A), \text D(\text{FD}(A)), \text D(k)$ be the corresponding homotopy and derived categories.
Denote again by $k$ the $A$-module $A/A\varepsilon$ and define the functors $$ \begin{matrix} \text M(A)&\overset{\widetilde F}{\to}&\text M(k)\\ \iota\uparrow&&||\\ \text{FD}(A)&\underset{F}{\to}&\text M(k) \end{matrix} $$ as follows:
$\bullet\ \widetilde F:=\text{Hom}_A(k,\hskip1em)$,
$\bullet\ \iota$ is the inclusion,
$\bullet\ F$ is the restriction of $\widetilde F$ to $\text{FD}(A)$.
Define $F':\text D(\text{FD}(A))\to\text D(k)$ by $$ F'(X):=\operatorname*{colim}_{X\to Y}\ \text K(F)(Y), $$ where $X\to Y$ runs over all the quasi-isomorphisms out of $X$ in $\text K(\text{FD}(A))$. (The colimit exists because $\text D(\text{FD}(A))$ is essentially small.)
Question 1: Is $F'$ the right derived functor of $F$?
Note that $F'$ is the unique (up to unique isomorphism) candidate for being the right derived functor of $F$.
Let $X$ be in $\text D(\text{FD}(A))$.
Question 2: Is the natural morphism $F'(X)\to\text{RHom}_A(k,X)$ an isomorphism?
Of course, "yes to Question 2 for all $X$" implies "yes to Question 1".
We also have "yes to Question 2 for $X$ in $\text D^+(\text{FD}(A))$" by Corollary 13.3.3 p. 330 and Remark 13.3.6 (iii) p. 331 in the book Categories and Sheaves by Kashiwara and Schapira.
