differentiation of e^(-3t/2)[Acos(2t) + Bsin(2t)]

[charlesmci]

Hi, If anyone could help me with this differentiation?
I need to differentiate with respect to t.
x(t) = e[sup:36rmiuyn](-3t/2)[/sup:36rmiuyn][Acos(2t) + Bsin(2t)]
Where A and B are arbitrary constants.
Thank you!

 

[Deleted member 4993]

Hi, If anyone could help me with this differentiation?
I need to differentiate with respect to t.
x(t) = e[sup:2kmw1qis](-3t/2)[/sup:2kmw1qis][Acos(2t) + Bsin(2t)]
Where A and B are arbitrary constants.
Thank you!

Use product rule of differentiation:

\(\displaystyle \frac{d}{dt}[f(t)\cdot g(t)] \, = \, \frac{df(t)}{dt}\cdot g(t) \, + \, \frac{dg(t)}{dt}\cdot f(t)\)

In your case

f(t) = e[sup:2kmw1qis](-3t/2)[/sup:2kmw1qis]

and

g(t) = Acos(2t) + Bsin(2t)

find f'(t) and g'(t) and continue...

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.