differentiation of Laplace transform with respect to parameter

We use the curved from the Laplace-transformed functions to the corresponding initial functions.

If

$$f(t,x)\curvearrowleft F(s,x),$$

then one can differentiate both functions with respect to the parametre $x$:

$f_x^{\prime }(t,x)\curvearrowleft F_x^{\prime }(s,x)$(1)

(1) may be written also as

$ℒ\{{\displaystyle \frac{\partial }{\partial x}}f(t,x)\}={\displaystyle \frac{\partial }{\partial x}}ℒ\{f(t,x)\}.$(2)

Proof.  We differentiate partially both sides of the definingequation

$$F(s,x):={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t,x)𝑑t,$$

on the right hand sideunder the integration sign (http://planetmath.org/differentiationundertheintegralsign), getting

$F_x^{\prime }(s,x)={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-st}{f}_x^{\prime }(t,x)𝑑x,$(3)

which means same as (1).  Q.E.D.

Example.  If the rule

$$\frac{s}{s^2-a^2}\curvearrowright \cosh at$$

is differentiated with respect to $a$, the result is

$$\frac{2as}{{(s^2-a^2)}^2}\curvearrowright t\sinh at.$$