Maths
posted by Lola .
y = 2[cos1/2(f1  f2)x][cos1/2(f1 + f2)x]
let f1 = 16 let f2 = 12
therefore
y = 2[cos1/2(16  12)x][cos1/2(16 + 12)x]
y = 2cos(2x)cos(14x)
derive y = 2cos(2x)cos(14x)
y' = 2[2sin(2x)][14sin(14x)]
is all of this correct?
Additionally, I am unsure how you would derive the general equation, y = 2[cos1/2(f1  f2)x][cos1/2(f1 + f2)x]
Thank you.

Maths 
Steve
y = 2cos(2x)cos(14x)
use the product rule:
y' = 2(sin(2x))(2)cos(14x) + 2cos(2x)(sin(14x))(14)
and you can factor out some numbers there.
The verb is differentiate, not derive.
It would work the same way in general:
y = 2[cos1/2(f1  f2)x][cos1/2(f1 + f2)x]
y' = 2(sin((f1f2)/2 x))(f1f2)/2 cos((f1+f2)/2 x)
+ ...
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