If

$$F(s)=\mathcal L{f(t)},$$

then differentiating in time becomes multiplication by $s$ together with terms containing the initial values.

For the first derivative,

$$\mathcal L{f'(t)}=sF(s)-f(0).$$

For the second derivative,

$$\mathcal L{f''(t)}=s^2F(s)-sf(0)-f'(0).$$

More generally, higher derivatives produce higher powers of $s$ plus the required initial derivative values.

Under zero initial conditions, these expressions simplify to

$$\mathcal L{f^{(n)}(t)}=s^nF(s).$$

This property is the reason linear constant-coefficient differential equations become algebraic in the Laplace domain. Nonzero initial conditions appear explicitly as additional terms rather than disappearing.