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