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Find the derivative of function with respect to x

  1. Mar 9, 2013 #1
    Find the derivative of function with respect to x

    y = sin^-1 (x-1/x+1)

    Steps I took:
    = sin(x-1/x+1)^-1
    u = sin(x-1/x+1)
    u' = cos u * u'
    u = x-1/x+1
    u' = t'b - tb'/b^2

    t = x-1
    b = x+1
    t' = 1
    b' = 1
    b^2 = (x+1)(x+1) = x^2+2x+1
    u' = (1)(x+1) - (x-1)(1)/x^2+2x+1
    u' = x+1-x-1 / x^2+2x+1
    u' = 0 !!!

    how can it be zero? something has gone wrong here.

    The answer to the question of 1/(x^(1/2) (x+1))

    Thank you
     
  2. jcsd
  3. Mar 9, 2013 #2

    mfb

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    Staff: Mentor

    There are missing brackets: x-1/x+1 is ##x-\frac{1}{x}+1##. I think you mean (x-1)/(x+1), or ##\frac{x-1}{x+1}##.
    (I added brackets here)
    This step is wrong, check the signs in the numerator.
     
  4. Mar 9, 2013 #3

    SteamKing

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    Is the initial function sin or arcsin?

    If it is arcsin, I don't think you have the correct derivative.
     
  5. Mar 9, 2013 #4
    how would you tell the difference? (sin/arcsin)
     
  6. Mar 10, 2013 #5

    eumyang

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    Homework Helper

    sin-1 x is the same as arcsin x. I'm assuming that your original function is
    [itex]y = sin^{-1}\left( \frac{x-1}{x+1} \right)[/itex]
    so that looks like arcsin. Of course, the derivative of arcsin is not cosine.
     
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