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Simplifying a Derivative

  1. Jul 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Find derivative of sin^-1 (2x + 1)


    2. Relevant equations
    I do everything Wolfram does except I don't know how to simplify to get the final answer.

    http://www.wolframalpha.com/input/?i=derivative+sin^-1+(2x+++1)


    3. The attempt at a solution
    Everything Wolfram does basically.
     
  2. jcsd
  3. Jul 15, 2012 #2

    HallsofIvy

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    I don't see how you can expect anyone to tell you how to simplify what you got to Wolfram's formula, if you don't tell us what you got!
     
  4. Jul 15, 2012 #3
    Ah terribly sorry. 2 / sqrt(1 - (2x + 1)^2); the last thing in the show steps part for the derivative in Wolfram.
     
  5. Jul 15, 2012 #4

    HallsofIvy

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    Of course, you know that (2x+1)^2= 4x^2+ 4x+ 1. That means 1- (2x+1)^2= 1 -4x^2- 4x- 1= -4x^2- 4x= -4(x^2- x)= -4x(x-1).
     
  6. Jul 15, 2012 #5
    But if -4x(x-1) is under a radical (as well as being multiplied) how is it that I pull a 2 out of that?
     
  7. Jul 15, 2012 #6
    [itex]\sqrt{-4x(x-1)} = \sqrt{4 \cdot -x(x-1)} = \sqrt{4} \sqrt{-x(x-1)} = 2\sqrt{-x(x-1)}[/itex]

    It should be noted [itex]1 - (2x + 1)^2 = -4x(x+1)[/itex]. I'll leave it to you to show [itex]\sqrt{-4x(x+1)} = 2\sqrt{-x(x+1)}[/itex].
     
    Last edited: Jul 15, 2012
  8. Jul 15, 2012 #7

    HallsofIvy

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    What, exactly, do you think the square root of 4 is?
     
  9. Jul 15, 2012 #8
    Ah. I understand now. It took me a bit, but I understand now. Thanks.
     
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