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

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Answer 1 · 2018-07-04T16:42:38

#-2\sin(4x)#

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 · 2018-07-04T16:52:08

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

Here,

#y=cos^2(2x)#

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

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

#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)#