What is derivative of f(x)= (sin2x+cos2x)(sin2x-cos2x)?

We notice that the given function is a special product that returns
a difference of two squares:

(a-b)(a+b) =`a^2 - b^2
`

Let a = sin x and b = cos x

(sin 2x + cos
2x)(sin 2x - cos 2x) = ` sin^2 (2x) - cos^2 (2x) `

We'll factor -1
and we'll get:

(sin 2x + cos 2x)(sin 2x - cos 2x) = -(` cos^2 (2x) -
sin^2 (2x)` )

This difference of two squares represents the formula
for the double angle:

-(`cos^2 (2x) - sin^2 (2x)` ) = - cos
(4x)

We'll apply the chain rule to differentiate with respect to
x:

f'(x) = -(-sin 4x)*(4x)'

f'(x) = 4sin
4x

The requested derivative of the given function is

f'(x) = 4sin
4x.