how do i solve this?
do i have to use both chain and product rule- how do i do that?
find the deriviative of
y= sin^2(3x-2)
If you write it as
y = [sin(3x-2)]^2 it might be easier to see
dy/dx = 2[sin(3x-2)](cos(3x-2))(3)
= 6sin(3x-2)cos(3x-2)
now you could get fancy here and use the
sin 2A = 2sinAcosA identity to further reduce the above.
dy/dx = 3sin(6x-4) as your final simplified answer.