Laplace Transform of $\cosh^2(3t)$

Question

Could someone help me on laplace transfrom ?

Using Laplace transform of derivative of $f(t)$,

Find the Laplace transform for

A) $\cosh^2(3t)$

How to derive it using Laplace transform of derivative ??

Then answer given is $\displaystyle\frac{S^2-18}{S(S-6)(S+6)} $

Answer 1

You can do it as follows:

1- Take @Ilya's suggestion to write $\cosh(3t)$ as:

$$\left(\frac{1}{2}\exp(3t)+\frac{1}2\exp(-3t)\right)^2=\frac{1}4\exp(6t)+\frac{1}2+\frac{1}4\exp(-6t)$$

Or

2- Use the fact that: $$\cosh^2(3t)=\frac{1+\cosh(6t)}{2}$$