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.

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)
+ ...

To verify your calculations:

Starting with the equation y = 2[cos(1/2(f1 - f2)x)][cos(1/2(f1 + f2)x)], you correctly substituted f1 = 16 and f2 = 12:

y = 2[cos(1/2(16 - 12)x)][cos(1/2(16 + 12)x)]
= 2[cos(1/2(4)x)][cos(1/2(28)x)]
= 2[cos(2x)][cos(14x)]

So far, your calculations are correct.

To find the derivative of y = 2cos(2x)cos(14x), let's use the product rule:

Given y = 2cos(2x)cos(14x), we'll find y' (the derivative of y) step by step:

1. Differentiate the first term, 2cos(2x), with respect to x:
Let u = 2x.
Therefore, y' = 2 * d/dx[cos(u)] * cos(14x) [Applying the chain rule]
= 2 * (-sin(u)) * cos(14x) [Differentiating cos(u)]
= -2sin(2x) * cos(14x).

2. Differentiate the second term, cos(14x), with respect to x:
We simply use the chain rule here.
y' = 2cos(2x) * d/dx[cos(14x)]
= 2cos(2x) * (-sin(14x) * 14) [Differentiating cos(14x)]
= -28cos(2x) * sin(14x).

So, the derivative of y = 2cos(2x)cos(14x) is:
y' = -2sin(2x)cos(14x) - 28cos(2x)sin(14x)

Therefore, your final result for y' is correct:

y' = 2[-2sin(2x)][-14sin(14x)] = -2sin(2x)cos(14x) - 28cos(2x)sin(14x).

Regarding deriving the general equation, y = 2[cos(1/2(f1 - f2)x)][cos(1/2(f1 + f2)x)], it seems like you are looking for a general expression where f1 and f2 can take any values.

To derive this equation in general, you can follow the same steps as above, substituting f1 and f2 with their respective values. This procedure will give you the general result:

y' = -1/2(f1 - f2)sin(1/2(f1 - f2)x) cos(1/2(f1 + f2)x) - 1/2(f1 + f2)cos(1/2(f1 - f2)x)sin(1/2(f1 + f2)x)

Remember to replace f1 and f2 with the actual values when you want to evaluate the derivative for a specific case.

I hope this clarifies your query. Let me know if you have further questions!