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Math Help - derivative of inverse trig function

  1. #1
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    derivative of inverse trig function

    y= tan^(-1)[x^2-1]^(1/2) + csc^(-1)x

    i cannot get to the answer, please help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mstsoy View Post
    y= tan^(-1)[x^2-1]^(1/2) + csc^(-1)x

    i cannot get to the answer, please help.
    \frac d{dx}\tan^{-1}u = \frac {u'}{1 + u^2} .........by the chain rule

    \frac d{dx} \csc^{-1} x = - \frac {1}{x \sqrt {x^2 - 1}}
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  3. #3
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    no i have definitely tried working it out already.
    i took derivative of both sides of the plus sign
    the answer should come out to zero

    but im not getting newhere near that.

    im getting something like x/ [(x^2 -1)(x^2 -1)^2] - 1/[|x|(x^2-1)^2]

    before simplifying, and after simplifying its not zero..urgh i cannot get it
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