\begin{equation}
\mathscr{L}[f(t)] =F(s)=\int\limits_{0}^{\infty}e^{-st}f(t)\,dt\tag{6.3.2}
\end{equation}
Differentiating equation (6.3.2), w. r. t. ’s’, we get -
\begin{equation*}
\frac{d}{\,ds}[F(s)] = \frac{d}{\,ds}\left[\int\limits_{0}^{\infty}e^{-st}f(t)\,dt\right] = \int\limits_{0}^{\infty}\frac{d}{\,ds}(e^{-st})f(t)\,dt
\end{equation*}
\begin{equation*}
= \int\limits_{0}^{\infty}(-t e^{-st})f(t)\,dt =\int\limits_{0}^{\infty}e^{-st}[-tf(t)]\,dt = L[-t f(t)]
\end{equation*}
or,
\begin{equation*}
L[tf(t)] = (-1)^{1}\frac{d}{\,ds}F(s)
\end{equation*}
Similarly,
\begin{equation*}
L[t^{2} f(t)] = (-1)^{2}\frac{d^{2}}{\,ds^{2}}F(s)
\end{equation*}
or,
\begin{equation*}
L[t^{3} f(t)] = (-1)^{3}\frac{d^{3}}{\,ds^{3}}[F(s)]
\end{equation*}
\begin{equation*}
\vdots
\end{equation*}
\begin{equation*}
\mathscr{L}[t^{n} f(t)] = (-1)^{n}\frac{d^{n}}{\,ds^{n}}[F(s)]
\end{equation*}
