Very hard derivative w/ 'e' and trig function! Please help!?

Question

Find the derivate of:
q=[e^(-0.1t)](0.2sin120πt+0.8cos120πt)

I know these type of problems are hard to visualize with it all typed in one straight line, so to clarify:

(-0.1t) is the exponent for 'e', and everything else is multiplied by the result of e^(-0.1t)
Any help would be greatly appreciated.

It looks like part of the problem got cut off. The second set of parenthisis is:
0.2sin120πt+0.8cos120πt

Answer 1

q = exp(-0.1t) (0.2sin120πt+0.8cos120πt )
q' = -0.1 exp(-0.1t)(0.2 sin120 pit + 0.8 cos 120 pi t)
+ exp(-0.1t) 120 pi (0.2cos120pi t-0.8sin120pi t)
= exp(-0.1t)( (60pi-0.08)cos 120 pi t - (96pi +0.02) sin 12p pi t )

Answer 2

This is just an application of the product and chain rules so the derivative is
-0.1e^(-0.1t)(original bracket) + e^(-0.1t)(differential of sin and cos functions)

You don't want every detail done for you, do you?

Answer 3

This is the product of 2 functions namely f(t) = e^(-0.1t) ans g(t) = 0.2sin(120πt+0.8c) (it's not clear here if 0.8c is part of the argument of sin or not).
So d/dt[f(t)g(t)] = df(t)/dt x g(t) + dg(t)/dt x f(t)

df(t)/dt = -0.1e^(-0.1t) and dg(t)/dt = 120π x 0.2cos(120πt+0.8c)e^(-0.1t)

Thus the derivative sought for is -0.1e^(-0.1t) x 0.2sin(120πt+0.8c) +120π x 0.2cos(120πt+0.8c)e^(-0.1t) = e^(-0.1t) [-0.1x 0.2sin(120πt+0.8c) + 120π x 0.2cos(120πt+0.8c)]

which you can reduce to a simpler form if required by usin trig formulae

Answer 4

This question would be a lot easier to describe(visualize) using the "LaTex" mark-up language.

Please vote on this suggestion @ Yahoo! Answers Suggestion Board:
"LaTeX rendering in questions and answers"

'sorry for this non-answer :(

Answer 5

1) It's just a slightly complex product

2) Q(t) = U(t) * V(t)

3) U(t) = e^(-at); a = 0.1
dU/dt = U' = -aU

4) V(t) = sin(kt); k = 120 * PI
dV/dt = V' = k * cos(kt)

5) dQ/dt = Q'
Q' = U * V' + V * U'