Why is the definition of $\lim^1$ via a cokernel the first derived functor of $\lim$? Let $A_*=\ldots\to A_n\to A_{n-1}\to\ldots\to A_0$ be a linear system of abelian groups. The limit of this system may be defined as the kernel of the map
$$
\prod A_n\xrightarrow{g-1}\prod A_n
$$
where $g$ comes from the structure maps of the system. The limit functor $\lim\colon\mathbf{Ab}^{\mathbb{N^{op}}}\to\mathbf{Ab}$ is not exact since it does not send short exact sequences in $\mathbf{Ab}^{\mathbb{N^{op}}}$ (exactness is tested in each degree) to exact sequences in $\mathbf{Ab}$ and one can therefore with the usual method construct its right derived functors $R^q\lim\colon\mathbf{Ab}^{\mathbb{N^{op}}}\to\mathbf{Ab}$ and we set $\lim^1=R^1\lim$.
On the other hand, $\lim^1$ is often defined in an other way (let me denote this object temporarily as $\lim^1'$):
$$\textstyle
\lim^1'A_*=coker(\prod A_n\xrightarrow{g-1}\prod A_n).
$$

Why is $\lim^1 A_*=\lim^1'A_*$? I would prefer an abstract argument with derived categories, if it exists.

Since countable products 'are exact' in abelian groups, the product over $\mathbb{N}$ is already the derived product and if all the structure maps from the system would be surjective, there would be an exact triangle
$$
Rlim A_*\to \prod A_n\xrightarrow{g-1}\prod A_n
$$
in the derived category (of chain complexes of abelian groups). Then, the shifted $Rlim$ would equal the mapping cone of $g-1$ (which is also the homotopy cofiber of $g-1$). But this mapping cone is usually not concentrated in degree zero and in particular it is not the same as $coker(\prod A_n\xrightarrow{g-1}\prod A_n)$ which defines $\lim^1'A_*$. Hence, I don't see a connection between $\lim^1A_*$ and $\lim^1'A_*$.

Can't notate because I'm on a tablet. Given a short exact of directed systems, the snake lemma tells us we get (functorially), an exact sequence
0 to lim A to lim B to lim C to lim 1'A to lim 1'B to lim 1'C to 0
That's enough to guarantee that lim1 and lim1' agree.
