= Solution
Inside the half-plane of convergence, the differentiation theorem for the <Laplace transform> permits differentiation with respect to the transform variable. Since $\partial_p e^{-pt}=-te^{-pt}$,
$$
\boxed{\overline x'(p)=-\int_0^\infty t x(t)e^{-pt}\,dt=\mathcal L[-tx(t)](p).}
$$
Repeating the argument, or applying induction using $\partial_p^n e^{-pt}=(-t)^n e^{-pt}$, gives
$$
\boxed{\overline x^{(n)}(p)=\int_0^\infty(-t)^n x(t)e^{-pt}\,dt
=(-1)^n\mathcal L[t^n x(t)](p),\qquad n\geq1.}
$$
The identities hold for $\operatorname{Re}p>0$, within the stated convergence half-plane. The usual transform differentiation theorem uses an interior margin in the real part of $p$ to control the polynomial factors in $t$; it makes no assertion on the boundary $\operatorname{Re}p=0$.
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