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Homework Help: Derivative of inverse trig function

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Is there a derivative of cot^-1x

    2. Relevant equations

    3. The attempt at a solution
    so far i used the inverse property to detemermine that
    coty = x
    then i used the chain rule to get
    -csc^2y * dy/dx = 1
    and i isolated for dy/dx
    dy/dx = 1 / -csc^2y

    what now??
  2. jcsd
  3. Jan 21, 2008 #2


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  4. Jan 21, 2008 #3
    care to explain??
  5. Jan 21, 2008 #4

    [tex]x=\cot y[/tex]

    [tex]\frac{d}{dy}(x)=\frac{d}{dy}(\cot y)[/tex]

    [tex]1=-\csc^2 yy'[/tex]

    [tex]y'=-\frac{1}{\csc^2 y}[/tex]

    [tex]*1+\cot^2 y=\csc^2 y*[/tex]

    [tex]*x^2=\cot^2 y*[/tex]

    [tex]y'=-\frac{1}{x^2 +1}[/tex]
    Last edited: Jan 21, 2008
  6. Jan 21, 2008 #5
    yah hahaha i'm lookin at the things in the tex editor and i dono't know wtf is goin on but i'll learn sooon :P
  7. Jan 21, 2008 #6
    so will anyone show me/explain how it works
  8. Jan 21, 2008 #7
    This is a standard integral. By 'standard' i mean.. this is one of the integrals you are just supposed to know. However, it is proved using Integration by parts. Almost all books on calculus carry that proof...
  9. Jan 21, 2008 #8
  10. Jan 21, 2008 #9
    my calculus book says it's
    [tex] \pi / 2-\tan^-1x[/tex]

    But i have no idea how to prove it anyone know??
    Last edited: Jan 21, 2008
  11. Jan 21, 2008 #10
    or can anyone walk me through this .......
    i'm at [tex]\frac{d}{dx}[/tex] = [tex]\frac{1}{/-csc^2x}[/tex]
    Last edited: Jan 21, 2008
  12. Jan 21, 2008 #11
    did you not read my post?
  13. Jan 21, 2008 #12
    ahhh srry i only saw the first half where u said
    "still typing"

    thnx a lot
  14. Jan 21, 2008 #13

    Gib Z

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    What are u referring to? The integral, derivative, or the inverse cotangent itself can guess it, but you should make it clear.

    It would be good if in the next post you just re cap the things you need help on.
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