anonymous
  • anonymous
can someone help me with finding the derivative of: f(x)=(e^-2t)cos(4t) ?
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

RadEn
  • RadEn
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' = ... ?
anonymous
  • anonymous
thank you :) \[4\sin (4t)\] is this the derivative of cos(4t)?
zepdrix
  • zepdrix
\(\large \cos(4t) \qquad \rightarrow \qquad -4\sin(4t)\) You were close! :) Just missed the negative when cosine goes to sine.

Looking for something else?

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

More answers

anonymous
  • anonymous
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 \[-2te ^{-2t-1}(e ^{-2t}\cos(4t))-4\sin(4t)\]
zepdrix
  • zepdrix
Woops! Little boo boo on the first part. You can not apply the power rule to an exponential function.
anonymous
  • anonymous
well crud. what do i do with \[e ^{-2t}\] I honestly dont remember.
zepdrix
  • zepdrix
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,\[\large (e^{3t})' \qquad = \qquad e^{3t}(3t)' \qquad = \qquad e^{3t}(3)\]
anonymous
  • anonymous
\[-2e ^{-2t}(e ^{-2t}\cos(4t))-4\sin(4t)\] Hmmm. I hope this is what you're talking about. I feel like the intial equation in the middle should be comething different :(
zepdrix
  • zepdrix
\[\large f(t)=(e^{-2t})\cos4t\] \[\large f'(t)=\color{royalblue}{(e^{-2t})'}\cos4t+(e^{-2t})\color{royalblue}{(\cos4t)'}\]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.
zepdrix
  • zepdrix
I can't quite tell what is going on with yours. Not sure why there are 2 exponentials on the first term.. hmm
anonymous
  • anonymous
thank you so much :) i didnt quite understand the rules. I think ive solved it. \[-2 e ^{-2t} (\cos(4 t)+2 \sin(4 t))\]
zepdrix
  • zepdrix
cool c:

Looking for something else?

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