For any infinite cardinal $\kappa$, $\sf DC_\kappa$ is the statement that whenever is a non-empty set, and is a relation on $S^{<\kappa}\times S$ with domain $S^{<\kappa}$, then there is a function $f\colon\kappa \rightarrow $ such that $f\restriction \alpha\mathrel{R}f(\alpha)$ for all $\alpha < \kappa$.
DC is the axiom of dependent choice.
Let $\kappa > \omega$. Does $\sf DC_\kappa$ imply $\sf DC$ over $\sf ZF$?
Yes, more generally it implies DC$_{\lambda}$ for any $\lambda \le \kappa.$ Let $R$ satisfy the premises of DC$_\lambda.$ Extend $R$ to $S^{<\kappa}$ by choosing some $x\in S$ and letting $sRx$ hold for all $s\in S^{<\kappa}\setminus S^{<\lambda}.$ Then $R$ satisfies the premise of DC$_\kappa,$ so let $f:\kappa\to S$ satisfy the conclusion of DC$_\kappa.$ Then, $f\upharpoonright\lambda$ satisfies the conclusion of DC$_\lambda.$
If it's unclear that DC$_{\aleph_0}$ means the same thing as DC (as it's usually formulated in terms of a total $R\subseteq S\times S$), observe that if $R$ is a total relation on a nonempty $S,$ we can let $R'\subseteq S^{<\omega}\times S$ have $sR'x$ if and only if $yRx$ where $y$ is the last element of $s$ (and let $\emptyset R'x_0$ for some arbitrary $x_0\in S$). Then an $f:\omega\to S$ satisfying the conclusion of DC$_{\aleph_0}$ for $R'$ will be a DC sequence in the usual sense.
(Actually, the preceding argument only shows DC$_{\aleph_0}$ implies DC, which is all that is needed... the other direction of the equivalence makes for a nice exercise.)
The easiest way to prove this is by using an equivalent formulation of $\sf DC_\kappa$. This is much like how it's easier to prove that $\Bbb R$ can be well-ordered by appealing to the well-ordering theorem, rather than the proof using Zorn's lemma or transfinite recursion and a choice function.
Theorem. $\sf DC_\kappa$ holds if and only if whenever $T$ is a $\kappa$-closed tree, then $T$ has a chain of length $\kappa$.
Here $T$ is a tree if it is a partial order with a minimum element, such that for every $t\in T$, the cone $\{s\in T\mid s<t\}$ is well-ordered (in the tree order). We say that $T$ is $\kappa$-closed if whenever $\gamma<\kappa$, and $\{t_\alpha\mid\alpha<\gamma\}$ is a chain in $T$, then there is some $t$ such that $t_\alpha<t$ for all $\alpha<\gamma$.
Proof. Assume $\sf DC_\kappa$ and suppose that $T$ is a $\kappa$-closed tree. Define the relation $R$ on increasing sequences in $T$ of length $\gamma<\kappa$: $s=\{t_\alpha\mid\alpha<\gamma\}\mathrel{R}t$ if $t$ is an upper bound of $s$. If $s$ is not an increasing sequence, we simply define $s\mathrel{R}0_T$, where $0_T$ is the minimum element.
Now let $f$ be a function with domain $\kappa$ given by $\sf DC_\kappa$. By induction, $f(\gamma)$ is an increasing sequence in $T$ for any $\gamma<\kappa$, and therefore $f$ is giving up a sequence of length $\kappa$ in the tree, as wanted.
In the other direction, if $R$ is a relation on $S^{<\kappa}\times S$, we can associate the natural tree $T$, given by increasing sequence in the relation. Since it satisfies the conditions of $\sf DC_\kappa$, this tree is $\kappa$-closed, and so it has a chain of type $\kappa$. Then this chain is $f$ as wanted.
Now it's easy. If $\sf DC_\lambda$ holds, and $\kappa\leq\lambda$, note that every $\kappa$-closed tree without a sequence of length $\kappa$ is also $\lambda$-closed; but if it had a sequence of length $\lambda$, it would have an initial segment of length $\kappa$. That's impossible.