Laplace transform of derivative

Theorem.  If the real function  $t\mapsto f(t)$  and its derivative are Laplace-transformable and $f$ is continuous for  $t>0$,  then

$ℒ\{f^{\prime }(t)\}=sF(s)-\underset{t\to 0+}{lim}f(t).$(1)

Proof.  By the definition of Laplace transform and using integration by parts, the left hand side of (1) may be written

$${\int }_0^{\mathrm{\infty }}{e}^{-st}{f}^{\prime }(t)𝑑t=\underset{t=0}{\overset{\mathrm{\infty }}{/}}{e}^{-st}f(t)+s{\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)𝑑t=\underset{t\to \mathrm{\infty }}{lim}{e}^{-st}f(t)-\underset{t\to 0}{lim}{e}^{-st}f(t)+sF(s).$$

The Laplace-transformability of $f$ implies that $e^{-st}f(t)$ tends to zero as $t$ increases boundlessly.  Thus the last expression leads to the right hand side of (1).