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lgbasalloteBest ResponseYou've already chosen the best response.1
i know the first term is \[e^t \cos (e^t)\]
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
What is a LGBADERIVATIVE?
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
Oh I would break it up And use log differentiation
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
\[a = e^{\sin t}\] \[\ln a = \sin t\] \[a' = \sin t \cos t?\]
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
well for the first one we don't need log differentiation but the second one we do
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Second will be: \[\huge cost(e^{sint})\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
oops yeah that's what i meant
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
\[g=e^{\sin(t)}\] \[\ln(g)=\ln(e^{\sin(t)})\] Or you can use the short way whatever lol
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
\[\huge e^t \cos t + \cos t (e^{\sin t})\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
No first one is not..
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[\huge e^t(cose^t)\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
oh argument is e^t
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
\[\huge e^t \cos (e^t) + \cos t(e^{\sin t})\]
 one year ago
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