Question

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

Answer 1

`-2sin(2x)`

Explanation:

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)`

Answer 2

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

Explanation:

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`