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**

We have to find the derivative of f(x)=[cos2x/(x^2+x+1)]^1/2.

Use the chain rule and the quotient rule.

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

= (1/2)[-2*sin x*(x^2+x+1)-(2x + 1)*cos(2 x)]/(x^2 + x +1)^(3/2)

= (-1/2)[2*sin x*(x^2+x+1)+(2x + 1)*cos(2 x)]/(x^2 + x +1)^(3/2)

**The required result is (-1/2)[2*sin x*(x^2+x+1)+(2x + 1)*cos(2 x)]/(x^2 + x +1)^(3/2)**

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