Find the derivative of #cos^2x-sin^2x#?

I will split the derivative up into two:
#d/dx(cos^2(x)-sin^2(x))=d/dx(cos^2(x))-d/dx(sin^2(x))#

I will call the one involving #cos# for derivative 1 and the other derivative 2.

Derivative 1
I will let #u=cos(x)#, and then use the chain rule:
#d/dx(cos^2(x))=d/(du)(u^2)*d/dx(cos(x))=2u(-sin(x))#

If we resubstitute, we get:
#-2usin(x)=-2cos(x)sin(x)#

Derivative 2
I will do the same strategy and let #u=sin(x)#:
#d/dx(sin^2(x))=d/(du)(u^2)*d/dx(sin(x))=2ucos(x)#

Resubstituting, we get:
#2ucos(x)=2sin(x)cos(x)#

Completing the original derivative
We worked out the two parts of the derivative, so we can just plug them in:
#d/dx(cos^2(x)-sin^2(x))=-2cos(x)sin(x)-2sin(x)cos(x)#

#=-4sin(x)cos(x)#

And by the double angle identity for #sin#, this is:
#-2sin(2x)#

As #cos2x=cos^2x-sin^2x#

#d/(dx)(cos^2x-sin^2x)=d/(dx)cos2x=-2sin2x#

Let us also check it by differentiating #cos^2x-sin^2x# without converting it to #cos2x#

#d/(dx)(cos^2x-sin^2x)#

= #2cosx*(-sinx)-2sinx(cosx)#

= #-2sinxcosx-2sinxcosx#

= #-4sinxcosx#

and as #sin2x=2sinxcosx#,

#d/(dx)(cos^2x-sin^2x)=-4sinxcosx=-2sin2x#