y = 2[cos1/2(f1 - f2)x][cos1/2(f1 + f2)x]

Question

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.

Answer 1

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((f1-f2)/2 x))(f1-f2)/2 cos((f1+f2)/2 x)
+ ...