We have to find the derivative of y = (cos 2x)^4. We need to use the chain rule here:

y' = [(cos 2x)^4]'

=> 4*(cos 2x)^(4 - 1)*(cos 2x)'

=> 4*(cos 2x)^3*(-sin 2x)*(2x)'

=> 4*(cos 2x)^3*(-sin 2x)*2

=> -8*(cos 2x)^3*(sin 2x)

**The required derivative is -8*(cos 2x)^3*(sin 2x)**

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We have to find the derivative of y = (cos 2x)^4. We need to use the chain rule here:

y' = [(cos 2x)^4]'

=> 4*(cos 2x)^(4 - 1)*(cos 2x)'

=> 4*(cos 2x)^3*(-sin 2x)*(2x)'

=> 4*(cos 2x)^3*(-sin 2x)*2

=> -8*(cos 2x)^3*(sin 2x)

**The required derivative is -8*(cos 2x)^3*(sin 2x)**

Given y= cos^4 (2x)

We need to find dy/dx

We will use the chain rule to find the derivative.

Let u= cos2x ==> u' = -2sin(2x)

==> y= u^4

==> y' = 4u^3 * u'

Now we will substitute with u= cos2x

==> y' = 4(cos^3 2x) * -2sin2x

**==> dy/dx = -8sin2x*cos^2 2x**