Laplace Transform of Derivatives
For first-order derivative:
L{f′(t)}=sL{f(t)}−f(0)
For second-order derivative:
L{f″(t)}=s2L{f(t)}−sf(0)−f′(0)
For third-order derivative:
L{f‴(t)}=s3L{f(t)}−s2f(0)−sf′(0)−f″(0)
For nth order derivative:
L{fn(t)}=snL{f(t)}−sn−1f(0)−sn−2f′(0)−⋯−fn−1(0)
Proof of Laplace Transform of Derivatives
L{f′(t)}=∫∞0e−stf′(t)dt
Using integration by parts,
u=e−st
du=−se−stdt
dv=f′(t)dt
v=f(t)
Thus,
L{f′(t)}=[e−stf(t)]∞0−∫∞0f(t)(−se−stdt)
L{f′(t)}=[f(t)est]∞0+s∫∞0e−stf(t)dt
L{f′(t)}=[f(t)est]∞0+sL{f(t)}
Apply the limits from 0 to ∞:
L{f′(t)}=[f(∞)e∞−f(0)e0]+sL{f(t)}
L{f′(t)}=−f(0)+sL{f(t)}
L{f′(t)}=sL{f(t)}−f(0) okay