Question

How do you find the derivative of `cos^2(2x)`?

Answer 1

`-2\sin(4x)`

Explanation:

Using chain rule of differentiation as follows

`\frac{d}{dx}\cos^2(2x)`

`=\frac{d}{dx}(\cos(2x))^2`

`=2\cos(2x)\frac{d}{dx}\cos (2x)`

`=2\cos(2x)(-2\sin (2x))`

`=-2(2\sin(2x)\cos (2x))`

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

`=-2\sin(4x)`

Answer 2

`(dy)/(dx)=-4sin(2x)cos(2x) or(dy)/(dx)=-2sin(4x)`

Explanation:

Here,

`y=cos^2(2x)`

Let , `y=u^2 , where , u=cos2x`

`=>(dy)/(du)=2u and (du)/(dx)=-sin2xd/(dx)(2x)=-2sin(2x)`

Using Chain Rule:

`color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)`

`=>(dy)/(dx)=2u*(-2sin(2x))`

Subst. back , `u=cos2x`

`(dy)/(dx)=2cos2x(-2sin2x)`

`=>color(brown)((dy)/(dx)=-4sin(2x)cos(2x)`

`=>(dy)/(dx)=-2{2sin(2x)cos(2x)}`

`=>color(brown)((dy)/(dx)=-2sin(4x)`