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Differentiate the function: f(x) = (cos^ 2 x)* ln (x^2).

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lipiciuc | Student, Undergraduate | eNoter

Posted May 10, 2011 at 2:00 AM via web

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Differentiate the function:

 f(x) = (cos^ 2 x)* ln (x^2).

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giorgiana1976 | College Teacher | Valedictorian

Posted May 10, 2011 at 9:40 AM (Answer #2)

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Since the given function is the result of composition of functions, we'll determine the 1st derivative using the chain rule and also the product rule.

f'(x) = [(cos x)^2]'* ln (x^2) + (cos x)^2* [ln (x^2)]'

We'll apply the chain rule for the terms:

[(cos x)^2]' = 2 cos x*(cos x)'

[(cos x)^2]' = - 2cos x*sin x

[ln (x^2)]' = (x^2)'/x^2

[ln (x^2)]' = 2x/x^2

We'll simplify and we'll get:

[ln (x^2)]' = 2/x

The first derivative of the function is:

f'(x) =  - 2cos x*sin x*ln (x^2) + 2(cos x)^2/x

f'(x) = -sin 2x*ln (x^2) + 2(cos x)^2/x

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 11, 2011 at 5:06 AM (Answer #3)

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The derivative of f(x) = (cos^ 2 x)* ln (x^2) has to be found.

f(x) = (cos^ 2 x)* ln (x^2)

=> f(x) = (cos x)^2 * ln (x^2)

f'(x) = [(cos x)^2]' * ln (x^2) + (cos x)^2 * [ln (x^2)]'

f'(x) = [2*cos x*(-sin x)] * ln (x^2) + (cos x)^2 * [2x/x^2]

f'(x) = -sin (2x)* ln (x^2) + (cos x)^2 * [2/x]

The derivative of the given function is (cos x)^2 * [2/x] - sin (2x)* ln (x^2)

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