# Laplace transforms of derivatives

where

As shown in the parent entry (http://planetmath.org/LaplaceTransformOfDerivative), the Laplace transform of the first derivative of a Laplace-transformable function $f(t)$ is got from the formula

 $\displaystyle\mathcal{L}\{f^{\prime}(t)\}\;=\;s\,F(s)-\lim_{t\to 0+}\!f(t).$ (1)

The rule can be applied also to the function $f^{\prime}(t)$:

 $\mathcal{L}\{f^{\prime\prime}(t)\}\;=\;s[sF(s)-\lim_{t\to 0+}\!f(t)]-\lim_{t% \to 0+}\!f^{\prime}(t)\;=\;s^{2}F(s)-sf(0\!^{+})-f^{\prime}(0\!^{+})$

Here the short notation $0\!^{+}$ has been used for the right limits.

Further, one can use the rule to $f^{\prime\prime}(t)$, getting

 $\mathcal{L}\{f^{\prime\prime\prime}(t)\}\;=\;s[s^{2}F(s)-sf(0\!^{+})-f^{\prime% }(0\!^{+})]-f^{\prime\prime}(0\!^{+})\;=\;s^{3}F(s)-s^{2}f(0\!^{+})-sf^{\prime% }(0\!^{+})-f^{\prime\prime}(0\!^{+}).$

Continuing similarly, one comes to the general formula

 $\displaystyle\mathcal{L}\{f^{(n)}(t)\}\;=\;s^{n}F(s)-s^{n-1}f(0\!^{+})-s^{n-2}% f^{\prime}(0\!^{+})-\ldots-f^{(n-1)}(0\!^{+}).$ (2)

Use of (2) requires that $f(t)$, $f^{\prime}(t)$, $f^{\prime\prime}(t)$, …, $f^{(n)}(t)$ are Laplace-transformable and that $f(t)$, $f^{\prime}(t)$, $f^{\prime\prime}(t)$, …, $f^{(n-1)}(t)$ are continuous when  $t>0$ (not only piecewise continuous).

Remark.  Suppose that $f(t)$ and $f^{\prime}(t)$ are Laplace-transformable and that $f(t)$ is continuous for  $t>0$  except the point  $t=a$  where the function has a finite jump discontinuity.  Then

 $\mathcal{L}\{f^{\prime}(t)\}\;=\;sF(s)-f(0\!^{+})-e^{-as}(\lim_{s\to a+}f(s)-% \lim_{s\to a-}f(s)).$

Application.  Derive the Laplace transform of $\sin{at}$ using the derivatives of sine (cf. Laplace transform of cosine and sine).

We have

 $f(t)\;:=\;\sin{at},\qquad f^{\prime}(t)\;=\;a\cos{at},\qquad f^{\prime\prime}(% t)\;=\;-a^{2}\sin{at}.$

Using (2) with  $n=2$  we obtain

 $\mathcal{L}\{-a^{2}\sin{at}\}\;=\;s^{2}\mathcal{L}\{\sin{at}\}-s\sin 0-a\cos 0,$

i.e.

 $-a^{2}\mathcal{L}\{\sin{at}\}\;=\;s^{2}\mathcal{L}\{\sin{at}\}-a,$

which implies

 $\mathcal{L}\{\sin{at}\}\;=\;\frac{a}{s^{2}\!+\!a^{2}}.$
Title Laplace transforms of derivatives LaplaceTransformsOfDerivatives 2014-04-06 8:24:33 2014-04-06 8:24:33 pahio (2872) pahio (2872) 6 pahio (2872) Topic msc 44A10