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

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to use the following derivative rules, such that, power rule, quotient rule and chain rule, in this order.

`f'(x) = (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)`

`f'(x) = 1/(2((cos2x)/(x^2+x+1))^(1/2))*(-2sin 2x(x^2 + x + 1) - (cos 2x)(2x + 1))/((x^2+x+1)^2)`

`f'(x) = (sqrt(x^2+x+1)(-2sin 2x(x^2 + x + 1) - (cos 2x)(2x + 1)))/((2cos 2x)(x^2+x+1)^2)`

Hence, evaluating the derivative of the given function, yields `f'(x) = (sqrt(x^2+x+1)(-2sin 2x(x^2 + x + 1) - (cos 2x)(2x + 1)))/((2cos 2x)(x^2+x+1)^2).`

giorgiana1976 | Student

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

We'll differentiate f(x) with respect to x:

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

We'll apply quotient rule for numerator:

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

We'll apply double angle identities:

f'(x) = {-4sinx*cosx*(x^2+x+1) - [2(cosx)^2-1](2x+1)}/2(x^2+x+1)^2*[sqrt[cos2x/(x^2+x+1)]