Verified Answer :

arthurpdc

We'll use that: 

\(g(x)=e^{f(x)}\\\\ g'(x)=f'(x)e^{f(x)}\)

So:

\(f(t)=e^{8t\sin(2t)}\\\\ f'(t)=[8t\sin(2t)]'e^{8t\sin(2t)}\\\\ f'(t)=[(8t)'\sin(2t)+8t(\sin(2t))']e^{8t\sin(2t)}\\\\ f'(t)=[8\sin(2t)+8t(2\cos(2t))]e^{8t\sin(2t)}\\\\ \boxed{f'(t)=8(\sin(2t)+2t\cos(2t))e^{8t\sin(2t)}}\)

Answer:

\(e^{8t} (2cos2t+8sin2t)\)

Step-by-step explanation:

Given is a funciton in t

\(F(t) = e^{8t} sin 2t\)

We have to find the derivative

Since this is product of two functions we use product rule as uv'+vu'

Here \(u = e^{8t} \\u'=8e^{8t}\)

\(v=sin2t\\v'=2cos 2t\)

Hence derivative =

\(F'(t) = e^{8t} (2cos2t+8sin2t)\)

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