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lgbasallote
 4 years ago
LGBADERIVATIVE:
\[\huge f(t) = \sin (e^t) + e^{\sin t}\]
lgbasallote
 4 years ago
LGBADERIVATIVE: \[\huge f(t) = \sin (e^t) + e^{\sin t}\]

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1i know the first term is \[e^t \cos (e^t)\]

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1What is a LGBADERIVATIVE?

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1Oh I would break it up And use log differentiation

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1\[a = e^{\sin t}\] \[\ln a = \sin t\] \[a' = \sin t \cos t?\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Second will be: \[\huge cost(e^{sint})\]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1oops yeah that's what i meant

freckles
 4 years ago
Best 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

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1\[\huge e^t \cos t + \cos t (e^{\sin t})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No first one is not..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\huge e^t(cose^t)\]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1\[\huge e^t \cos (e^t) + \cos t(e^{\sin t})\]
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