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

`e^(8t) (2cos2t+8sin2t)`

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)`