Unilateral Laplace Transform's Differentiation Property
You need to look at the derivation of that property. Integration by parts gives
$$\begin{align}\mathcal{L}\{f'(t)\}&amp;=\int_{0^-}^{\infty}f'(t)e^{-st}dt\\&amp;=f(t)e^{-st}\Big|_{0^-}^{\infty}+s\underbrace{\int_{0^-}^{\infty}f(t)e^{-st}dt}_{F(s)}\\&amp;=\lim_{t\to\infty}f(t)e^{-st}-f(0^-)+sF(s)\tag{1}\end{align}$$
The first term in $(1)$ is only guaranteed to vanish for $t\to\infty$ if $\text{Re}\{s\}&gt;\alpha$ for some value $\alpha$. So even if you define a unilateral Fourier transform, that term may not vanish in general.
However, if $\lim_{t\to\infty}f(t)=0$ holds, we could use the same property with a unilateral Fourier transform. It's just much more common to use the well-established (unilateral) Laplace transform in cases where non-zero initial conditions need to be taken into account.