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Y = 2x arccos 3x Find the derivative

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    y = 2x arccos 3x Find the derivative. Can be left unsimplified.


    2. Relevant equations



    3. The attempt at a solution

    y = 2x arccos 3x
    y = 2x arccos (u) u = 3x
    dy/du = d/dx(2x)arccos (u) + 2x[d/dx(arccos (u)] du/dx = 3
    dy/du = 2arccos(u) + 2x[-1/sqrt of (1-u^2)]
    dy/dx = 2arccos(3x) + 2x[-1/sqrt of (1-(3x)^2)](3)
    dy/dx = 2arccos(3x) - [2x/sqrt of (1-(3x)^2)](3)
    dy/dx = 6arccos(3x) - [2x/sqrt of (1-(3x)^2)]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2012 #2
    Try using the definition of the inverse to switch to a normal trig function, then use implicit differentiation. You'll be left with a sin(something) in your solution, which you can evaluate by drawing a triangle & finding out its value and subbing that back into your equation, then just isolate dy/dx and simplify.
     
  4. Sep 15, 2012 #3

    SammyS

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    Some of your steps are strange, if not downright incorrect, [STRIKE]but you get to the correct result (with a corrected typo)[/STRIKE].

    Edit: See the "strike out".

    The first term should not be multiplied by 3. The second term should be.
     
    Last edited: Sep 15, 2012
  5. Sep 15, 2012 #4

    SammyS

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    e^(i Pi)+1=0,

    I would likely do this problem in a manner similar to what you suggest.

    Yes. there will be a sin(something). However, after simplifying the something you have sin(arccos(3x)), which is [itex]\displaystyle \sqrt{1-9x^2\ }\ .[/itex]
     
  6. Sep 16, 2012 #5
    Thanks for the help guys.
     
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