Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

skay

  • one year ago

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

  • This Question is Closed
  1. RadEn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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' = ... ?

  2. skay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\large \cos(4t) \qquad \rightarrow \qquad -4\sin(4t)\) You were close! :) Just missed the negative when cosine goes to sine.

  4. skay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \[-2te ^{-2t-1}(e ^{-2t}\cos(4t))-4\sin(4t)\]

  5. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. skay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  8. skay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large f(t)=(e^{-2t})\cos4t\] \[\large f'(t)=\color{royalblue}{(e^{-2t})'}\cos4t+(e^{-2t})\color{royalblue}{(\cos4t)'}\]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.

  10. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. skay
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    cool c:

  13. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.