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

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RadEnBest ResponseYou've already chosen the best response.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' = ... ?
 one year ago

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

zepdrixBest 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.
 one year ago

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

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

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

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

skayBest 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 :(
 one year ago

zepdrixBest 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.
 one year ago

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

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