Laplace Transformation of derivative

Question

I want to rules behind the step(with arrow). I cant get how that transformation happened.

Answer 1

$$ \mathcal{L}\left[y'(t)\right]=\int_{0}^{\infty}y'(t)e^{-st}\mathrm{d}t $$

$$ \mathrm{d}v=y'(t)\mathrm{d}t\longrightarrow v= y $$

$$ u=e^{-st}\longrightarrow \mathrm{d}u=-se^{-st}\mathrm{d}t $$ integrating by parts

$$ \mathcal{L}\left[y'(t)\right]=(e^{-st}y)|_{0}^{+\infty}+s\int_{0}^{\infty}y(t)e^{-st}\mathrm{d}t $$

$$ (e^{-st}y)|_{0}^{+\infty}=-y(0) $$

$$ \mathcal{L}\left[y(t)\right]=\int_{0}^{\infty}y(t)e^{-st}\mathrm{d}t $$

so one can then see

$$ \mathcal{L}\left[y'(t)\right]=s\mathcal{L}\left[y(t)\right]-y(0) $$