Find the derivative of f(x)=[cos2x/(x^2+x+1)]^1/2

f(x) =
[cos2x/(x^2+x+1)]^1//2

We need to find the first derivative
f'(x)

We will use the chain rule to find the
derivative.

Let f(x) = sqrt
u

==> f'(x) = (sqrtu)' = -1/2sqrtu *
u'

u= cos2x/(x^2+x+1)

Let us
calculate u'

u' = ( cos2x)'*(x^2+x+1) - (cos2x0*(x^2+x+1)'/
(x^2+x+1)^2

     = -2sinx*(x^2+x+1) - (cos2x)*(2x+1) ] /
(x^2 +x+1)^2

Now we will substitute into
f'(x).

==> f'(x) = -1/2sqrtu *
u'

                 = -1/2sqrt(cos2x/(x^2+x+1) *
[(-2sin2x(x^2+x+1)-(cos2x)(2x+1)] /
(x^2+x+1)^2

==>f'(x) =
(2sin2x*(x^2+x+1) +(cos2x*(2x+1)]/
(2sqrt(cos2x/(x^2+x+1)*(x^2+x+1)^2