Let $T$ be a functor with $p$ variables with some contravariant and some covariant. Let $s$ variable be active variables of $T$ and those $s$ variables are to be resolved by either projective or injective resoltuion according to either contravariance or covariance respectively. The rest variables are left untouched.
$\textbf{Q:}$ What does it mean to derive those derived functors? Say $T(X_1,\dots, X_s,Y_{s+1},\dots, Y_p)$ is the functor with $X_i$ to be resolved. Does it mean take "higher dimensional" complex formed by the corresponding resolution of $X_i$, produce a total complex as in 2 complex case via taking $\sum_{j\leq s}i_j=n$ for indices, and compute its homology or cohomology?
Ref. Cartan, Eilenberg, Homological Algebra, Chpt V, Sec 8, Partial Derived Fucnctors.(pg94)
For concreteness, let's take a right derived functor of $T$. Dualize mutatis mutandis for left derived functors.
In the context I'm aware of, you want to have the following additional hypothesis on $T$:
Under this hypothesis, one can show that to compute a right derived functor of $T$, all one must do is choose any variable slot and treat $T$ as a single-variable functor in that slot, with all other slots fixed. Then you just build a derived functor of the single-variable functor like you always do: take an acyclic resolution, compute (co)homology. (The canonical example is treating the bifunctor $\operatorname{Hom}(-,-)$ as a functor $\operatorname{Hom}(-,B)$ or as $\operatorname{Hom}(A,-)$; both give you $\operatorname{Ext}$.)
A priori, this would give you different derived functors if you choose different slots, but one can show that, with the right balanced hypothesis, all the right derived functors are naturally isomorphic. (In the case of $-\otimes-$ and $\operatorname{Hom}(-,-)$ this is called "balancing" $\operatorname{Tor}$ and $\operatorname{Ext}$ and involves a quaint application of spectral sequences via a result called the Acyclic Assembly Lemma.) Below is the general computation:
Let $T$ be a right balanced functor, without loss of generality with all covariant variables. Choose $i\in\{2,\ldots,p\}$; we show that $R^{*}T(\widehat{X_{1}},X_{2},\ldots,X_{p})(X_{1})\cong R^{*}T(X_{1},\ldots,\widehat{X_{i}},\ldots,X_{p})(X_{i})$, where this notation means the (single-variable) functor with hatted-variable free and all others fixed.
Choose injective resolutions $X_{1}\xrightarrow{\varepsilon}I^{\bullet}$ and $X_{i}\xrightarrow{\eta}J^{\bullet}$. Fix $X_{j}$ for all $j\not\in\{1,i\}$, and write $T(-,-)=T(-,X_{2},\ldots,X_{i-1},-,X_{i+1},\ldots,X_{p})$. In other words, we're slightly abusing notation and just forgetting about all the fixed variables; they are absorbed as part of the functor. Form the first quadrant double cochain complex:
$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\ua}[1]{\uparrow\raise.5ex\rlap{\scriptstyle#1}}$ \begin{array}{c} \vdots & & \vdots & & \vdots & & \vdots & & \vdots & & \\ \ua{} & & \vdots & & \ua{} & & \ua{} & & \ua{} & & \\ T(X_{1},J^{2}) & & \vdots & & T(I^{0},J^{2}) & \ras{d} & T(I^{1},J^{2}) & \ras{d} & T(I^{2},J^{2}) & \ras{} & \cdots \\ \ua{} & & \vdots & & \ua{d} & & \ua{-d} & & \ua{d} & & \\ T(X_{1},J^{1}) & & \vdots & & T(I^{0},J^{1}) & \ras{d} & T(I^{1},J^{1}) & \ras{d} & T(I^{2},J^{1}) & \ras{} & \cdots \\ \ua{} & & \vdots & & \ua{d} & & \ua{-d} & & \ua{d} & & \\ T(X_{1},J^{0}) & & \vdots & & T(I^{0},J^{0}) & \ras{d} & T(I^{1},J^{0}) & \ras{d} & T(I^{2},J^{0}) & \ras{} & \cdots \\ & & \cdot & & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ & & & & T(I^{0},X_{i}) & \ra{} & T(I^{1},X_{i}) & \ra{} & T(I^{2},X_{i}) & \ra{} & \cdots \\ \end{array} where the differentials in the horizontal direction are the images of $d:I^{p}\to I^{p+1}$ under the functor $T(-,B)$ where $B$ is fixed, and the differentials in the vertical direction use the sign trick; i.e., they are $(-1)^{p}$ times the images of $d:J^{q}\to J^{q+1}$ under the functor $T(A,-)$ where $A$ is fixed.
Call the double complexes formed from $I$ and $J$, from $I$ and $X_{i}$, and from $X_{1}$ and $J$:
respectively. This means you should think of the above double cochain complex as $\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\ua}[1]{\uparrow\raise.5ex\rlap{\scriptstyle#1}}$ \begin{array}{c} T(X_{1},J)&\vdots&T(I,J)\\ &\cdot&\cdots\\ &&T(I,X_{i}) \end{array} (Those dotted lines are there to help you compartmentalize.)
Now consider the double complex $C$ obtained from $T(I,J)$ by adding $T(X_{1},J)[-1]$ in the column $p=-1$, as we have in the picture. (It's just forgetting the $T(I,X_{i})$ piece.) The translate $\operatorname{Tot}(C)[+1]$ is the mapping cone of the map induced by $\varepsilon$ from $\operatorname{Tot}(T(X_{1},J))=T(X_{1},J)$ to $\operatorname{Tot}(T(I,J))$. This induced map is a quasi-isomorphism if and only if its mapping cone is exact; i.e., the complex $\operatorname{Tot}(C)[+1]$ is acyclic, which is the case if and only if the nonshifted $\operatorname{Tot}(C)$ is.
Since $T$ is a right balanced functor, covariant in all variables, and $J^{\bullet}$ are injective modules, $T(-,J)$ is an exact functor. Thus, every row of $C$ is exact, and as $C$ is a right half-plane with exact rows, the Acyclic Assembly Lemma implies that $\operatorname{Tot}(C)$ is acyclic, as desired. Thus, the map induced by $\varepsilon$ is a quasi-isomorphism; that is, \begin{align*} H^{*}(\operatorname{Tot}(T(I,J)))\cong H^{*}(T(X_{1},J))=R^{*}T(X_{1},-)(X_{i})=R^{*}T(X_{1},\ldots,\widehat{X_{i}},\ldots,X_{p})(X_{i}). \end{align*}
Similarly, the double complex $D$ obtained from $T(I,J)$ by adding $T(I,X_{i})[-1]$ in the row $q=-1$ (take the picture and forget $T(X_{1},J)$) yields a totalization $\operatorname{Tot}(D)$. The translate $\operatorname{Tot}(D)[+1]$ is the mapping cone of the map induced by $\eta$ from $\operatorname{Tot}(T(I,X_{i}))=T(I,X_{i})$ to $\operatorname{Tot}(T(I,J))$. We again show that $\operatorname{Tot}(D)$ is acyclic to get the desired isomorphism on cohomology.
Indeed, as $T$ is right balanced, covariant, and $I^{\bullet}$ are injective, $T(I,-)$ is exact, so every column of $D$ is exact, and as $D$ is an upper half-plane with exact columnds, $\operatorname{Tot}(D)$ is acyclic, thanks to the Acyclic Assembly Lemma. Therefore, we may conclude \begin{align*} H^{*}(\operatorname{Tot}(T(I,J)))\cong H^{*}(T(I,X_{i}))=R^{*}T(-,X_{i})(X_{1})=R^{*}T(\widehat{X_{i}},X_{2},\ldots,X_{p}). \end{align*} By transitivity, \begin{align*} R^{*}T(X_{1},\ldots,\widehat{X_{i}},\ldots,X_{p})(X_{i})\cong R^{*}T(\widehat{X_{1}},X_{2},\ldots,X_{p})(X_{1}), \end{align*} as we wished to show, and therefore the derived functor of $T$ is well-defined. You get the same result independent of which single-variable functor you derive.
Check out Weibel's An introduction to homological algebra, section 2.7, for the meat of this discussion.