= Solution
The unilateral <Laplace transform> is $\bar f(p)=\int_0^\infty e^{-pt}f(t)dt$ in its <region of absolute convergence of a Laplace transform>. Under the usual inversion hypotheses, the <Bromwich inversion formula> is
$$
f(t)=\frac1{2\pi i}\lim_{R\to\infty}\int_{c-iR}^{c+iR}e^{pt}\bar f(p)dp.
$$
The vertical <Bromwich contour> is directed upward, with $c$ inside the transform's convergence half-plane and to the right of all its singularities. At discontinuities the standard symmetric inversion gives the mean of the one-sided values.
For a piecewise continuously <differentiable> $f$ with an integrable transformed <derivative>, <integration by parts> gives the <Laplace transform of a derivative>:
$$
\int_0^\infty e^{-pt}f'(t)dt=\left[e^{-pt}f(t)\right]_0^\infty+p\int_0^\infty e^{-pt}f(t)dt=\boxed{p\bar f(p)-f(0)}.
$$
Here the upper endpoint vanishes by the given decay condition, and right continuity at zero supplies the lower endpoint.
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