Let $f: \mathbf{C} \to \mathbf{D}$ be a functor between small categories and $\mathbf{M}$ a cocomplete category. Then we have the coend formula $$ (f_! X)(d) = \int^{c \in \mathbf{C}} \mathbf{D} (f(c),d) \otimes X(c) \quad (*) $$ for the left Kan extension of a functor $X : \mathbf{C} \to \mathbf{M}$ along $f$.

Now assume that all categories and functors involved here are enriched over simplicial sets (for example). Then the same formula holds for the enriched left Kan extension when interpreted correctly (in terms of tensoring over simplicial sets).

In the next step assume that $\mathbf{C}, \mathbf{D}$ and $\mathbf{M}$ are actually a simplicial model categories. This is the setup discussed in:

ncatlab.org/nlab/show/homotopy+Kan+extension

I would like to see a formula like $(*)$ for the $derived\ enriched$ left Kan extension, preferrably using the derived hom and maybe a replacement of $X$?

For the case where $\mathbf{C}$ and $\mathbf{D}$ are just enriched over simplicial sets, but not treated as model categories, I found a model for the derived enriched left Kan extension in Riehl's $Categorical\ Homotopy\ Theory$ using the bar construction, but can we maybe express it directly using the appropriate homotopical analogon of $(*)$?

Thank you for any hints.

isthe "appropriate homotopical analogon of (*)". $\endgroup$ – Mike Shulman Mar 14 '18 at 18:48modelcategories. I don't really even know what the concept means in this case. For one thing, an arbitrary functor between model categories doesn't even induce a functor between their induced $(\infty,1)$-categories. There are two canonical ways that it might, its left and right derived functors. So when you talk about $f:C\to D$ and $X:C\to M$ and the "derived enriched" Kan extension, do you mean a representative for the $(\infty,1)$ Kan extension of their left derived functors? Or right? $\endgroup$ – Mike Shulman Mar 16 '18 at 11:215more comments