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Derivatives of trigonometric functions - Question

  1. Jul 12, 2011 #1
    1. The problem statement, all variables and given/known data



    Find the Derivative of:

    (ln(cos4x)) / 12x^2

    2. Relevant equations

    y' ln(x) = 1/x


    3. The attempt at a solution

    I have determined the correct answer, but I am still confused as to how I came to the solution. Starting with the numerator, the derivative of cos is -sin. This would produce ln(-4sin(4x)) in the numerator. Now, to take the derivative of the natural log, I would do ln(x)= 1/x, correct? So 1/-4sin(4x)?

    Why does the final numerator come out to -4tan(4x)? I know I am missing a basic rule or overlooking something... but why does it become tangent?

    Thank you
     
  2. jcsd
  3. Jul 12, 2011 #2

    Ray Vickson

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    Homework Helper

    You would NOT get ln(-4sin(4x)) in the numerator. You are forgetting the Chain Rule: [f(g(x))]' = f '(g(x) )* g '(x). Apply this to f(.) = ln (.) and g(x) = cos(4x).

    RGV
     
  4. Jul 12, 2011 #3

    Mentallic

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    Homework Helper

    First of all, this is a quotient so you need to use the quotient rule.

    [tex]y=\frac{u}{v}[/tex] where u and v are functions of x, then [tex]y'=\frac{v'u-u'v}{v^2}[/tex]

    Now, only tricky part there would be to determine u'. Since u=ln(cos(4x)) then we need to apply the chain rule, and Ray Vickson has already shown how this should be done.
     
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