Using the limit definition of the derivative, find the derivative of f(x)=sin(3x) at x=2π

First, we need to find the derivative f'(x) for all x, and then we will substitute x=2π into the function we find. To do this, we need to set up our limit:
f'(x) = lim [sin(3(x+h))-sin(3x)]/h as h-> 0
We then use the identity sin(A+B) = sinAcosB + sinBcosA to simplify the limit
f'(x) = lim [sin(3x)cos(3h) + sin(3h)cos(3x) - sin(3x)]/h as h-> 0

Because h tends to 0, we can assume that h is as small as we want it to be. This means we're allowed to use the small angle approximations for sin3h and cos3h to simplify the expression. Remember that sin3h ~= 3h as h is small, and cos3h ~= 1- (3h)2/2 = 1- 9h2/2. Hence:
f'(x) = lim [ (1-9h²/2)sin(3x) + 3hcos(3x) - sin(3x)]/h as h->0
f'(x) = lim [sin(3x) -(9h²/2)sin(3x) + 3hcos(3x) - sin(3x)]/h as h->0 (multiply out (1-9h²/2)sin(3x) into two terms)
f'(x) = lim [3hcos(3x) -(9h²/2)sin(3x)]/h as h->0 (cancel out the sin(3x) terms)
f'(x) = lim 3cos(3x) - (9h/2)sin(3x) as h->0 (divide through by h)
f'(x) = 3cos(3x) (because (9h/2)sin(3x) ->0 as h-> 0)


This gives us our general function. We can then plug in x=2π into f'(x) to get out final answer.So
f'(2π) = 3cos(32π)
f'(2π) = 3cos(6π)
f'(2π) = 31 = 3

So our final answer is f'(2π)=3