Why does the Laplace Transform of a derivative have an initial value term, but the Z-transform of a delay not? I believe (?) they are similar concepts; a delay in a difference equation is considered synonymous to a derivative in a differential equation. But
$$ f'(t) \rightarrow sF(s) - f(0) $$
$$f[k-1] \rightarrow zF(z)$$
This continues for higher derivatives/greater delays. Is there a reason for the non-existent initial value term in the z-transform?
 A: It comes from the fact that Laplace transforms are usually unilateral (one-sided), while Z-transforms are bilateral (two-sided). Indeed, one has
$$
\mathscr{L}[f'(t)](s) = \int_0^\infty f'(t)e^{-st}\,\mathrm{d}t = \left.f(t)e^{-st}\right|_0^\infty + s\int_0^\infty f(t)e^{-st}\,\mathrm{d}t = sF(s) - f(0),
$$
by integration by parts, but
$$
\mathcal{Z}[f(k\color{red}{+}1)](z) = \sum_{k\in\mathbb{Z}}f(k+1)z^{-k} = \sum_{k\in\mathbb{Z}}f(k)z^{-(k-1)} = zF(z).
$$
If, on the contrary, we consider the two-sided Laplace transform and the one-sided Z-transform, the same expressions become
$$
\mathscr{L}[f'(t)](s) = \int_{-\infty}^\infty f'(t)e^{-st}\,\mathrm{d}t = \left.f(t)e^{-st}\right|_{-\infty}^\infty + s\int_{-\infty}^\infty f(t)e^{-st}\,\mathrm{d}t = sF(s)
$$
$-$ with appropriate conditions on the asymptotic behaviour of $f$ $-$ and
$$
\mathcal{Z}[f(k+1)](z) = \sum_{n\ge0}f(k+1)z^{-k} = f(1) + \sum_{n\ge0}f(k)z^{-(k-1)} = zF(z) + f(1).
$$
In conlusion, unilateral transforms produce boundary terms.

Addendum.
In order to avoid this problem but keep the one-sided Laplace transform at the same time (because its two-sided version often has convergence issues), some people work implicitly with $f(t)u(t)$ instead of $f(t)$, where $u(t)$ is the Heaviside function, and modify a little bit the definition of the Laplace transform, as follows :
$$
F(s) := \int_{\color{red}{0^\mathbf{-}}}^\infty f(t)e^{-st}\,\mathrm{d}t \quad\Rightarrow\quad \mathscr{L}[f'(t)](s) = sF(s),
$$
since boundary terms are killed by the fact that $u(0^-) = 0$.
