Question

derivative of the function. F(t) = e8t sin 2t

Answer 1

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 2

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 =

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