Time Differentiation Property of The Laplace Transform

Given that $ F(s) $ is the Laplace transform of $ f(t) $, the Laplace transform of its derivative is
$$\mathcal{L}\left[\frac{d f}{d t}\right]=\int_{0^{-}}^{\infty} \frac{d f}{d t} e^{-s t} d t$$
To integrate this by parts, we let $ u=e^{-s t}, d u=-s e^{-s t} d t $, and $ d v= $ $ (d f / d t) d t=d f(t), v=f(t) $. Then
$$\begin{aligned}\mathcal{L}\left[\frac{d f}{d t}\right] &=\left.f(t) e^{-s t}\right|_{0^{-}} ^{\infty}-\int_{0^{-}}^{\infty} f(t)\left[-s e^{-s t}\right] d t \\&=0-f\left(0^{-}\right)+s \int_{0^{-}}^{\infty} f(t) e^{-s t} d t=s F(s)-f\left(0^{-}\right)\end{aligned}$$
$$\bbox[10px,border:1px solid grey]{\mathcal{L}\left[f^{\prime}(t)\right]=s F(s)-f\left(0^{-}\right)} \tag{1}$$
The Laplace transform of the second derivative of $ f(t) $ is a repeated application of Eq. (1) as
$$\begin{aligned}\mathcal{L}\left[\frac{d^{2} f}{d t^{2}}\right] &=s \mathcal{L}\left[f^{\prime}(t)\right]-f^{\prime}\left(0^{-}\right)=s\left[s F(s)-f\left(0^{-}\right)\right]-f^{\prime}\left(0^{-}\right) \\&=s^{2} F(s)-s f\left(0^{-}\right)-f^{\prime}\left(0^{-}\right)\end{aligned}$$
$$\mathcal{L}\left[f^{\prime \prime}(t)\right]=s^{2} F(s)-s f\left(0^{-}\right)-f^{\prime}\left(0^{-}\right)$$
Continuing in this manner, we can obtain the Laplace transform of the $ n $th derivative of $ f(t) $ as
$$\begin{aligned}\mathcal{L}\left[\frac{d^{n} f}{d t^{n}}\right]=& s^{n} F(s)-s^{n-1} f\left(0^{-}\right) \\& \quad-s^{n-2} f^{\prime}\left(0^{-}\right)-\cdots-s^{0} f^{(n-1)}\left(0^{-}\right)\end{aligned}$$
As an example, we can use Eq. (1) to obtain the Laplace transform of the sine from that of the cosine. If we let $ f(t)=\cos \omega t $, then $ f(0)=1 $ and $ f^{\prime}(t)=-\omega \sin \omega t $. Using Eq. (1) and the scaling property,
$$\begin{aligned}\mathcal{L}[\sin \omega t] &=-\frac{1}{\omega} \mathcal{L}\left[f^{\prime}(t)\right]=-\frac{1}{\omega}\left[s F(s)-f\left(0^{-}\right)\right] \\&=-\frac{1}{\omega}\left(s \frac{s}{s^{2}+\omega^{2}}-1\right)=\frac{\omega}{s^{2}+\omega^{2}}\end{aligned}$$
as expected.