Laplace Transform of Derivatives

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 nth order derivative:

$\mathcal{L} \left\{ f^n(t) \right\} = s^n \mathcal{L} \left\{ f(t) \right\} - s^{n - 1} f(0) - s^{n - 2} \, f'(0) - \dots - f^{n - 1}(0)$

 

Proof of Laplace Transform of Derivatives
$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \int_0^\infty e^{-st} f'(t) \, dt$
 

Using integration by parts,
$u = e^{-st}$
$du = -se^{-st} \, dt$

$dv = f'(t) \, dt$
$v = f(t)$
 

Thus,
$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \Big[ e^{-st} f(t) \Big]_0^\infty - \int_0^\infty f(t) \, (-se^{-st} \, dt)$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \left[ \dfrac{f(t)}{e^{st}} \right]_0^\infty + s\int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \left[ \dfrac{f(t)}{e^{st}} \right]_0^\infty + s \, \mathcal{L} \left\{ f(t) \right\}$
 

Apply the limits from 0 to :
$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \left[ \dfrac{f(\infty)}{e^\infty} - \dfrac{f(0)}{e^0} \right] + s \, \mathcal{L} \left\{ f(t) \right\}$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = -f(0) + s \, \mathcal{L} \left\{ f(t) \right\}$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$           okay
 

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