Respuesta :

arthurpdc
We'll use that: 

[tex]g(x)=e^{f(x)}\\\\ g'(x)=f'(x)e^{f(x)}[/tex]

So:

[tex]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)}}[/tex]

Answer:

[tex]e^{8t} (2cos2t+8sin2t)[/tex]

Step-by-step explanation:

Given is a funciton in t

[tex]F(t) = e^{8t} sin 2t[/tex]

We have to find the derivative

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

Here [tex]u = e^{8t} \\u'=8e^{8t}[/tex]

[tex]v=sin2t\\v'=2cos 2t[/tex]

Hence derivative =

[tex]F'(t) = e^{8t} (2cos2t+8sin2t)[/tex]

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