Given the continuous version of the $3x+1$ (Collatz) function by Chamberland: $$f(x)=x+\frac{1}{4}-\frac{2x+1}{4}\cos(\pi x)$$
Define $g:\mathbb{R} \times \mathbb{N} \rightarrow \mathbb{R}$ by iteration: $g(x,y)=f^y(x)$. Now extend this to $y \in \mathbb{R}_{\geq 1}$ by interpolation, i.e., for $0 < \epsilon < 1$ define:
$$g(x,y+\epsilon)=f\big((1-\epsilon) \cdot g(x,y-1) + \epsilon \cdot f(g(x,y-1) \big)$$
$g$ is continuous in $y$ because for $\epsilon=0$, we would get $g(x,y)=f(g(x,y-1))$ and for $\epsilon=1$: $g(x,y+1)=f(f(g(x,y-1))$.
Is $g$ also differentiable/smooth in $y$? If yes, does it follow from $f$ being differentiable/smooth?