The first term should differentiate into 3(2)(sin(2x-4))(cos(2x-4))(2) following the chain rule by first multiplying by the exponent of an argument and reducing the exponent by one and rewriting it. Then multiplying it by the derivative of sin(2x-4). Then multiplying it by the derivative of the argument in that one as well. 2. Following the same logic with the second term gives (2)(2 [the exponent of cos^{2}(3x+1)])(cos(3x+1) [rewriting the argument that was squared, but the exponent is n-1])(-sin(3x+1) [The derivative of the argument inside that was squared])(3 [Derivative of 3x+1])

So, a. y`= 3(2)(sin(2x-4))(cos(2x-4))(2) -(2)(2)(cos(3x+1))(-sin(3x+1))(3)

= 12sin(2x-4)cos(2x-4) + 12cos(3x+1)sin(3x+1)

= 12(sin(2x-4)cos(2x-4) + sin(3x+1)cos(3x+1))

b. y`=cos(x^{2}+π)(2x) = 2xcos(x^{2}+π)