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Math Help - proof of arccot(x)

  1. #1
    cpj
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    proof of arccot(x)

    Can someone show me how i can prove this?

    (d/dx)cot^-1 =-1/1+(x^2)

    thanks in advance
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  2. #2
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    I would like to know this as well, would doing the integral of his solution be a way to provide some sort of "proof" by using the identity tan{x^2}+1=sec{x^2}
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  3. #3
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    If you know  \frac{d}{dx}( \tan^{-1}(x) ) = \frac{1}{x^2 + 1} but not  \frac{d}{dx} ( \cot^{-1}(x) )

    Use this identity   \cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2}
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  4. #4
    cpj
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    thanks, but would it be possible for you to elaborate, maybe show some working, i'm not sure how to use that identity to help with the proof.
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  5. #5
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    Quote Originally Posted by simplependulum View Post
    If you know  \frac{d}{dx}( \tan^{-1}(x) ) = \frac{1}{x^2 + 1} but not  \frac{d}{dx} ( \cot^{-1}(x) )

    Use this identity  \cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2}

    sure

    We have  \cot(\theta) = \frac{1}{\tan(\theta)} = \tan( \frac{\pi}{2} - \theta )

     \theta = \cot^{-1}( \tan( \frac{\pi}{2} - \theta )) -(1)

    Let  x = \tan( \frac{\pi}{2} - \theta )

     \tan^{-1}(x) = \frac{\pi}{2} - \theta -(2)

     (2) - (1) , it is to eliminate  \theta

      \tan^{-1}(x) = \frac{\pi}{2} - \cot^{-1}( \tan( \frac{\pi}{2} - \theta ))

     \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}
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