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RadEn
 one year ago
Best ResponseYou've already chosen the best response.1use the product rule, also the chain rule if given y = uv then y ' = u'v + uv' let u=e^(2t) > u' = ... ? v =cos4t > v' = ... ?

skay
 one year ago
Best ResponseYou've already chosen the best response.0thank you :) \[4\sin (4t)\] is this the derivative of cos(4t)?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \cos(4t) \qquad \rightarrow \qquad 4\sin(4t)\) You were close! :) Just missed the negative when cosine goes to sine.

skay
 one year ago
Best ResponseYou've already chosen the best response.0yay 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 \[2te ^{2t1}(e ^{2t}\cos(4t))4\sin(4t)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Woops! Little boo boo on the first part. You can not apply the power rule to an exponential function.

skay
 one year ago
Best ResponseYou've already chosen the best response.0well crud. what do i do with \[e ^{2t}\] I honestly dont remember.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Remember 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,\[\large (e^{3t})' \qquad = \qquad e^{3t}(3t)' \qquad = \qquad e^{3t}(3)\]

skay
 one year ago
Best ResponseYou've already chosen the best response.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 :(

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I can't quite tell what is going on with yours. Not sure why there are 2 exponentials on the first term.. hmm

skay
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much :) i didnt quite understand the rules. I think ive solved it. \[2 e ^{2t} (\cos(4 t)+2 \sin(4 t))\]
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