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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 I found on Wikipedia, which is different than the form found in Stewart's calculus [itex]\frac{d}{dx} \mathrm{arccsc} x = -\frac{1}{x \sqrt{x^2 - 1}}[/itex]. Where exactly does the |x| come from? There is proof of arcsin and arccos, but haven't yet found one which explains this detail for arccsc. Is there a proof that explains it?

    Thanks,
    - Farley
     
  2. jcsd
  3. Jan 11, 2009 #2

    jgens

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    Gold Member

    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]\frac{\text{d}}{\text{d}x}\text{arccsc}x=\frac{\text{d}}{\text{d}x}\arcsin\left(\frac{1}{x}\right)[/tex]
    Substitution:
    [tex]y=\arcsin u[/tex]
    Now I can use chain rule:
    [tex]\frac{\text{d}y}{\text{d}u}\cdot\frac{\text{d}u}{\text{d}x}=\frac{1}{\sqrt{1-u^2}}\cdot\left(-\frac{1}{x^2}\right)=\frac{1}{\sqrt{1-\frac{1}{x^2}}}\cdot\left(-\frac{1}{x^2}\right)=\frac{1}{\sqrt{\frac{x^2-1}{x^2}}}\cdot\left(-\frac{1}{x^2}\right)=\frac{|x|}{\sqrt{x^2-1}}\cdot\left(-\frac{1}{x^2}\right)[/tex]
    There is important to realise:
    [tex](x^2\geq0)\,\wedge\,(|x|\geq0)[/tex]
    so there quotient must be also positive.
    [tex]\frac{|x|}{x^2}=\frac{1}{|x|}[/tex]
    If I use this to my proof, I obtain:
    [tex]\frac{\text{d}}{\text{d}x}\text{arccsc}x=-\frac{1}{|x|\sqrt{x^2-1}}[/tex]
     
    Last edited: Jan 11, 2009
  5. Jan 11, 2009 #4

    jgens

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    Gold Member

    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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