The scheme of “large dependent choice” (LDC) consists of statements of the form
Suppose $\forall x \exists y P(x,y)$. Then for all $z$, there is some infinite sequence $\{s_i\}_{i \in \mathbb{N}}$ such that $s_0 = z$ and $\forall i \in \mathbb{N} P(s_i, s_{i + 1})$.
The ordinary version of dependent choice (DC) can be phrased as a single axiom, which is (the universal closure of) the following:
Suppose $P \subseteq A^2$. And further suppose $\forall x \in A \exists y \in A ((x, y) \in P)$. Then for all $z \in A$, there exists a sequence $\{s_i\}_{i \in \mathbb{N}}$ such that $s_0 = z$ and $\forall i \in \mathbb{N} (s_i, s_{i + 1}) \in P$
In ZF, every instance of LDC follows from DC. This can be proved using Scott’s trick.
Is the same true in IZF (with collection)?
Your large dependence choice is known as Relativized Dependent Choice ($\mathsf{RDC}$). It is known by Aczel that $\mathsf{CZF}$ proves $\mathsf{RDC}$ is equivalent to the combination of $\mathsf{DC}$ and Relation Reflection Scheme ($\mathsf{RRS}$). You can find Aczel's proof from
As far as I know, it is open whether $\mathsf{IZF+DC}$ proves $\mathsf{RRS}$ or not. However, it is known by Swan and Blechschmidt (on their unpublished note) that $\mathsf{RRS_2}$, a strengthening of $\mathsf{RRS}$, is equivalent to the reflection principle over $\mathsf{IZF}$. (Note that, it is also open whether $\mathsf{IZF}$ proves the reflection principle.)