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

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nano75 | Student, Grade 9 | eNoter

Posted December 2, 2010 at 5:38 AM via web

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

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giorgiana1976 | College Teacher | Valedictorian

Posted December 2, 2010 at 3:51 PM (Answer #1)

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Since we have to determine the first derivative of a composed function, we'll apply 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 result of sifferentiating the given 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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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted December 2, 2010 at 5:39 AM (Answer #2)

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

We need to determine the first derivative ( f'(x)).

We notice that f(x) is a product of two functions. Then, we will use the product rule to solve.

Let f(x) = u * v  such that:

u = ( cos^2 x)   ==>   u' = -2*cosx*sinx = -sin2x

v= ln x^2  ==>     v' = 2x*1/x = 2

Then, we know that:

f'(x) = u'*v + u*v'

       = (cos^2 x)* 2 + (-sin2x( ln x^2)

         = 2cos^2 x - sin2x * ln x^2

==> f'(x) = 2cos^ 2x - 2lnx*sin2x

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