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Deriving cos(x)^ln(x)?

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data

    http://puu.sh/4M7BE.png [Broken]

    2. Relevant equations

    ln(ax) = x*ln(a)

    3. The attempt at a solution

    ln(y) = cos(x)*ln(ln(x))

    dy/dx * 1/y = -sinx*ln(ln(x)) + cosx/(x*lnx)

    No clue how to solve this, there's no ln(ln(x)) in the possible answers


    Attempt 2

    y = cos(x)^ln(x)
    ln(y) = ln(x)*ln[cos(x)]
    1/y * dy/dx = 1/x * ln(cos(x)) + ln(x) * 1/cos(x) * (-sin(x))
    dy/dx = y[ ln(cos(x))/x - ln(x) * sin(x) / cos(x)]
    dy/dx = [cos(x)ln(x)][ ln(cos(x))/x - (ln(x)sin(x))/cos(x)]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 9, 2013 #2

    Dick

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    You got off to a bad start. If y=(cos(x))^(ln(x)) then ln(y)=ln(x)*ln(cos(x)). NOT cos(x)*ln(ln(x)).
     
    Last edited by a moderator: May 6, 2017
  4. Oct 9, 2013 #3

    lurflurf

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    The power rule is

    $$(u^v)^\prime=v \, u^{v-1} \, u^\prime +u^{v} \log(u) \, v^\prime $$

    You can derive it by writing

    $$u^v=\exp(v \log(u))$$

    Then differentiate both sides.
     
  5. Oct 9, 2013 #4
    Can't tell what I did wrong in attempt 2 either :eek:

    Not sure how they got the inner cos outside of ln in the answer.

    http://puu.sh/4MaXv.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Oct 9, 2013 #5

    Dick

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    Your second attempt is correct. You are trying to compare it to a wrong answer.
     
    Last edited by a moderator: May 6, 2017
  7. Oct 9, 2013 #6
    I see, I guess the answer key is incorrect. Thanks for the help
     
  8. Oct 9, 2013 #7

    Dick

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    No, no, no. Compare your answer with the key answer c). Not with d).
     
  9. Oct 10, 2013 #8
    Ah, just went through the news feed. They noted that it should be C not D a couple of days ago, didn't catch that.
     
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