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Basic Derivative?

  1. Oct 26, 2005 #1


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    Alright, here is the equation:

    ln [(sin(x)^3)^3]

    And here is my answer with some steps, please tell me if i'm wrong..

    9 ln(sin'x)

    9 (cos'x)/(sin'x)

    Derivative= 9tan(x)^-1

    is that correct?
  2. jcsd
  3. Oct 26, 2005 #2
    I Get 9*x^2*cot(x^3)
    It Looks Like U Forgot The X^2 Term When You Got The Der Of X^3
  4. Oct 26, 2005 #3
    Usually we say that [tex]\frac{1}{\tan(x)}=\cot(x)[/tex]. Looks good to me.
  5. Oct 26, 2005 #4


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    soo.. which answer is correct? the whole sin'x is cubed, not just the x value within the sine..
  6. Oct 26, 2005 #5
    I see nothing wrong with your original answer, except for a better notation.
  7. Oct 26, 2005 #6


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    so 9cot(x) would be a better answer?
  8. Oct 26, 2005 #7
    Yes, it's more concise and very standard.
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