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

justaguide | College Teacher | (Level 2) Distinguished Educator

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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)

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

neela | High School Teacher | (Level 3) Valedictorian

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To find derivative of f(x)=[cos2x/(x^2+x+1)]^1/2.

f(x) = {cos2x/(x^2+x+1)}^(1/2).

We know that {f(g(x)} = (df/dg)g'(x).

Also {(u(x)/v(x)}' = {u'(x)v(x)-u(x)v'(x)}/(v(x))^2.

Therefore f'(x) ={ [cos2x/(x^2+x+1)]^1/2}'

= (1/2) )[cos2x/(x^2+x+1)]^(1/2-1) * {(cos2x)'*(x^2+x+1) - (cos2x)(x^2+x+1)}/(x^2+x+1)^2.

= (1/2) )[cos2x/(x^2+x+1)]^(1/2-1) * {(-2sin2x)*(x^2+x+1) - (cos2x)(2x+1)}/(x^2+x+1)^2.

=(x^2+x+1)^(1/2)  {(-2sin2x)*(x^2+x+1) - (cos2x)(2x+1)}/{2(x^2+x+1)^2*(cos2x)^(1/2)

= - (x^2+x+1)^(1/2) {(2sin2x)(x^2+x+1)+(2x+1) cos2x)}/(x^2+x+1)^2 * (c0sx)^(1/2).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll use the chain rule and the quotient rule to evaluate the first derivative of the composed function.

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

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

We'll take the numerator and we'll calculate it's derivative:

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

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

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

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