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Derivative involving inverse trigonometric functions

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of:
    sqrt(x^2-4)-2tan^-1{.5*sqrt(x^2-4)}




    2. Relevant equations
    U'/1+U^2
    U'=x/2sqrt(x^2-4)
    1+U^2=x^2


    3. The attempt at a solution

    I combined the above components but my answer is incorrect. I feel that I might have the wrong answer for "1+U^2". I just cannot seem to catch my error at the moment.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 8, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    Did you mean the following function?


    [tex]\sqrt{(x^2-4)}-2\arctan\left(0.5 \sqrt{(x^2-4)}\right)[/tex]

    What do you mean on U'/1+U^2? Are no parentheses missing?

    Did you mean the derivative of arctan(U)?

    [tex]\frac{U'}{1+U^2}[/tex]

    with [tex]U=0.5\sqrt{(x^2-4)}[/tex]
    and
    [tex]U'=0.5 \frac{x}{\sqrt{(x^2-4)}}[/tex]?


    Recalculate U^2+1. It is not x^2.

    ehild
     
    Last edited: Mar 8, 2012
  4. Mar 8, 2012 #3
    My U'=[itex]\frac{x}{2\sqrt{x^2-4}}[/itex]

    When I square [itex]\frac{\sqrt{x^2-4}}{2}[/itex] and add one the only other answer I get is [itex]\frac{x^2}{4}[/itex]
    I've been working for a while and perhaps I'm missing something very simple.
     
    Last edited: Mar 8, 2012
  5. Mar 8, 2012 #4
    I've gotten the answer:

    [itex]\frac{\sqrt{x^2-4}}{x}[/itex]

    I made a simple mistake in calculating 1+U[itex]^{2}[/itex]

    Thanks for the help!
     
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