What is the Laplace transform of tcos4t using the derivative of a transform?

[Jtechguy21]

Homework Statement

Evaluate the Laplace of {tcos4t} using the derivative of a transform

Ofcourse i know the shortcut way of doing this, but I need to do it the long way.

Homework Equations

shortcut way
t cos bt = $\frac{s^2-b^2}{(s^2+b^2)^2}$

long way transform of a derivative
(-1)^n $\frac{d^n}{ds^n}$ F(s)

F(s)=L{f(t)}

The Attempt at a Solution

n=1 f(t)=cos4t
f(s)= Laplace of cos4t

L{cos4t}= $\frac{s}{s^2+16}$

-$\frac{d}{ds}$ $\frac{s}{s^2+16}$

Quotient rule

-$\frac{s^2+16-2s^2}{(s^2+16)^2}$

This is as far as I get on my own.

In my notes for class she's goes a step further and I'm not quite sure what happens to the
-2s^2 in the next step

my notes continue as follows:
-$\frac{-s^2+16}{(s^2+16)^2}$

now we distribute the negative

and get
$\frac{s^2-16}{(s^2+16)^2}$

I checked the answer using a Laplace transform table(easy way) and did receive $\frac{s^2-16}{(s^2+16)^2}$

but when i do it the long way i don't know what happens to -2s^2

 

[Jtechguy21]

wow I am so stupid. i missed such an easy thing. combing the like terms s^2 and -2s^2. i get it now.