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Proof of d/dx (arccsc x)

  1. Jan 11, 2009 #1
    Hey all,

    I'm trying to understand how [itex]\frac{d}{dx} \mathrm{arccsc} x = -\frac{1}{|x| \sqrt{x^2 - 1}}[/itex], which http://en.wikipedia.org/wiki/Inverse_trigonometric_function" [Broken], but haven't yet found one which explains this detail for arccsc. Is there a proof that explains it?

    - Farley
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 11, 2009 #2


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    I think it may be largely notational, because if we allow x < 0 than the derivative becomes indentical to d(arcsec(x))/dx.

    Here's a proof for the derivative of arccsc(x):

    csc(y) = x
    d(csc(y))/dx = 1
    -csc(y)cot(y)y' = 1
    y' = -1/(csc(y)cot(y))
    Now, since 1 + cot(x)^2 = csc(x)^2, cot^2(x) = csc^2(x) - 1, therefore:
    y' = -1/(x(x^2-1)^1/2)
  4. Jan 11, 2009 #3
    [tex]y=\arcsin u[/tex]
    Now I can use chain rule:
    There is important to realise:
    so there quotient must be also positive.
    If I use this to my proof, I obtain:
    Last edited: Jan 11, 2009
  5. Jan 11, 2009 #4


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    Bah, an oversight on my part. I overlooked a restricted principle branch in the step following csc^2(x) - 1 = cot^2(x). Sorry!
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