Laplace transform
Transform of a function's derivative

It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows: L { f (t) } = ∫ 0 − ∞ e − s t f (t) d t = 0 − ∞ − ∫ 0 − ∞ e − s t − s f ′ (t) d t (by parts) = + 1 s L { f ′ (t) }, {\displaystyle {\begin{aligned}{\mathcal {L}}\left\{f(t)\right\}&=\int _{0^{-}}^{\infty }e^{-st}f(t)\,dt\\&=\left_{0^{-}}^{\infty }-\int _{0^{-}}^{\infty }{\frac {e^{-st}}{-s}}f'(t)\,dt\quad {\text{(by parts)}}\\&=\left+{\frac {1}{s}}{\mathcal {L}}\left\{f'(t)\right\},\end{aligned}}} yielding L { f ′ (t) } = s ⋅ L { f (t) } − f (0 −), {\displaystyle {\mathcal {L}}\{f'(t)\}=s\cdot {\mathcal {L}}\{f(t)\}-f(0^{-}),} and in the bilateral case, L { f ′ (t) } = s ∫ − ∞ ∞ e − s t f (t) d t = s ⋅ L { f (t) }. {\displaystyle {\mathcal {L}}\{f'(t)\}=s\int _{-\infty }^{\infty }e^{-st}f(t)\,dt=s\cdot {\mathcal {L}}\{f(t)\}.}

The general result L { f (n) (t) } = s n ⋅ L { f (t) } − s n − 1 f (0 −) − ⋯ − f (n − 1) (0 −), {\displaystyle {\mathcal {L}}\left\{f^{(n)}(t)\right\}=s^{n}\cdot {\mathcal {L}}\{f(t)\}-s^{n-1}f(0^{-})-\cdots -f^{(n-1)}(0^{-}),} where f (n) {\displaystyle f^{(n)}} denotes the n th derivative of f, can then be established with an inductive argument.
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