# Derivative of the function.Determine the derivative of the function f(x)=[cos2x/(x^2+x+1)]^1/2.

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### 2 Answers

Since the function you need to differentiate is composed, hence, you need to use the following rule called chain rule, such that:

`(u(v(x)))' = u'(v(x))*v'(x)*x'`

Considering `u(v(x)) =` `((cos2x)/(x^2+x+1))^1/2` and `v(x) = (cos2x)/(x^2+x+1)` yields:

`f'(x) = (((cos2x)/(x^2+x+1))^1/2)'((cos2x)/(x^2+x+1))'`

You need to differentiate `(cos2x)/(x^2+x+1)` using quotient rule, such that:

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

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

**Hence, evaluating the derivatove of the given function yields **`f'(x) = (1/2((cos2x)/(x^2+x+1))^(-1/2))'(((-2sin 2x)(x^2+x+1) - (cos2x)(2x + 1))/((x^2+x+1)^2)).`

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**

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