proof

Theorem
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then
 

For first-order derivative:
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$
 

For second-order derivative:
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$
 

For third-order derivative:
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$
 

For n^th order derivative:

Division by $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

provided   $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$   exists.
 

Multiplication by Power of $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

where   $n = 1, \, 2, \, 3, \, ...$
 

Change of Scale Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

Second Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   and   $g(t)
= \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$
 

then,

First Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   when   $s > a$   then,
 

In words, the substitution   $s - a$   for   $s$   in the transform corresponds to the multiplication of the original function by   $e^{at}$.
 

Linearity Property
If   $a$   and   $b$   are constants while   $f(t)$   and   $g(t)$   are functions of   $t$   whose Laplace transform exists, then
 

Proof of Linearity Property
$\displaystyle \mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = \int_0^\infty e^{-st}\left[ a \, f(t) + b \, g(t) \right] \, dt$