y = 2[cos1/2(f1 - f2)x][cos1/2(f1 + f2)x]
let f1 = 16 let f2 = 12
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.

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asked by Lola
  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)
    + ...

    posted by Steve

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