Question

can someone help me with finding the derivative of:
f(x)=(e^-2t)cos(4t) ?

Answer 1

use the product rule, also the chain rule
if given y = uv then y ' = u'v + uv'
let u=e^(-2t) ----> u' = ... ?
v =cos4t ----> v' = ... ?

Answer 2

thank you :)
\[4\sin (4t)\] is this the derivative of cos(4t)?

Answer 3

\(\large \cos(4t) \qquad \rightarrow \qquad -4\sin(4t)\)

You were close! :) Just missed the negative when cosine goes to sine.

Answer 4

yay thanks. :)
I'm having trouble with the chain rule.
I got to the derivatives of both sides but im not sure what to do with
\[-2te ^{-2t-1}(e ^{-2t}\cos(4t))-4\sin(4t)\]

Answer 5

Woops! Little boo boo on the first part.
You can not apply the power rule to an exponential function.

Answer 6

well crud. what do i do with \[e ^{-2t}\]
I honestly dont remember.

Answer 7

Remember this derivative? :)

\(\large (e^t)'=e^t\)

We'll be doing the same thing, but since our exponent is more than just `t`, we need to apply the chain rule, multiply by the derivative of the exponent.

Example,\[\large (e^{3t})' \qquad = \qquad e^{3t}(3t)' \qquad = \qquad e^{3t}(3)\]

Answer 8

\[-2e ^{-2t}(e ^{-2t}\cos(4t))-4\sin(4t)\]
Hmmm. I hope this is what you're talking about. I feel like the intial equation in the middle should be comething different :(

Answer 9

\[\large f(t)=(e^{-2t})\cos4t\]
\[\large f'(t)=\color{royalblue}{(e^{-2t})'}\cos4t+(e^{-2t})\color{royalblue}{(\cos4t)'}\]Here is the setup for the product rule.
We need to differentiate the blue terms.\[\large f'(t)=\color{orangered}{(-2e^{-2t})}\cos4t+(e^{-2t})\color{royalblue}{(\cos4t)'}\]That's the derivative of the first blue term.\[\large f'(t)=\color{orangered}{(-2e^{-2t})}\cos4t+(e^{-2t})\color{orangered}{(-4\sin4t)}\]There is the other one.

Answer 10

I can't quite tell what is going on with yours. Not sure why there are 2 exponentials on the first term.. hmm

Answer 11

thank you so much :) i didnt quite understand the rules. I think ive solved it.
\[-2 e ^{-2t} (\cos(4 t)+2 \sin(4 t))\]

Answer 12

cool c: