Question

I suck at this so can some on give me the 1st and 2nd derivatives of:
3Asin2t + 3Bcos2t

Answer 1

t^' = 6Acos(2t)-6Bsin(2t)
t^'' = -12Asin(2t)-12cos(2t)

This is a chain rule problem
Derivative of the outside times the derivative of the inside.
d/dx sin(x) = cos(x)
d/dx cos(x) = -sin(x)

t' 3Asin(2t) = (3Asin(2t)) <- outside times (2t) <- inside = 3Acos(2t) x 2 = 6Acos(2t)
t' 3Bcos(2t) = (3Bcos(2t)) <- outside times (2t) <- inside = 3B(-sin(2t)) x 2 = -6Bsin(2t)

so t' = 6Acos(2t)-6Bsin(2t)

for the second derivative just do it all again.

t' 6Acos(2t) = 6Acos(2t) <- outside times (2t) <- inside = 6A(-sin(2t)) x 2 = -12Asin(2t)
t' -6Bsin(2t) = (-6Bsin(2t)) <- outside times (2t) <- inside = -6B(cos(2t)) x 2 = -12Bcos(2t)

so t'' = -12Asin(2t)-12Bcos(2t)

This link has some good reference materials to look at in helping you understand derivatives and later integrals. They use them at my college in the tutoring centers. http://tutorial.math.lamar.edu/cheat_table.aspx look at the calculus ones specifically.