Verified answer

dy/dt = 2 sin(7 pi t + 9) cos ( 7 pi t + 9) (7 pi)

= 7 pi sin( 14 pi t + 18)

You can look at it like this:

y = sin^2(7πt + 9)

Let u = 7πt + 9

Then what you have is: y = y(u) = sin^2(u)

=> dy/du = 2sin(u)cos(u)......easy enough

Then you need du/dt

=> du/dt = 7π

Put it all together (chain rule)

=> dy/dt = dy/du du/dt = 2sin(u)cos(u) 7π

=> dy/dt = 2sin(7πt + 9)cos(7πt + 9) 7π

I find this way much easier, otherwise you have to remember the rules for handling the trig functions and this way is more direct.....in my opinion. :)

y = [ sin (7πt + 9) ] ²

dy/dt = 2 [ sin (7πt + 9) ] [ 7π ] [ cos (7πt + 9) ]

dy/dt = 7 π [ 2 ] [ sin (7πt + 9) ] cos (7πt + 9)

dy/dt = 7 π [ sin (14πt + 18) ]