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Derivative of inverse trif function

  1. Oct 1, 2007 #1
    y= tan^(-1)[x^2-1]^(1/2) + csc^(-1)x

    i cannot get to the answer, can someone help me?

    well the answer should be zero. when i take the derivative of both parts (one of the tan inverse and one of the csc inverse) i dont get anywhere close to 0...
  2. jcsd
  3. Oct 1, 2007 #2
    could you show what you did?

    and what type of answer are you getting? w/o doing it I think something near 1? maybe 1 minus something.
  4. Oct 1, 2007 #3
    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
  5. Oct 1, 2007 #4

    well the derivative of arctan x = 1/(1+x^2), but in your case you'll have to do the chain rule.
    for csc^-1 I'm guessing that's inverse csc right? so then you'll want the derivative of 1/sin^-1 x which is arcsin to the power of negative 1 which will again need a use of the chain rule.
  6. Oct 7, 2007 #5
    The answer I am getting is not zero or one.
    You will have to use the chain rule multiple times. And keep in mind:
    [tex]\frac{d}{dx} arccsc(x) = \frac{-1}{x\sqrt{x^2 - 1}}[/tex]
    [tex]\frac{d}{dx} arctan(x) = \frac{1}{x^2 + 1}[/tex].
  7. Oct 7, 2007 #6


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    Well, it almost works out to zero, doesn't it? Maybe you made a slight error in your calculation; have you rechecked them?
  8. Oct 7, 2007 #7
    If the derivitive is graphed, it can be seen that after an approximate value of [tex]2[/tex], the derivitive gets very close to [tex]0[/tex].
    This is the solution i got for the problem:
    \frac{d}{dx}arccsc(x) + \sqrt{arctan(x^2 - 1)} = \frac{-1}{x\sqrt{x^2 - 1}} + \frac{x}{((x^2 - 1)^2 + 1)\sqrt{arctan(x^2 - 1)}}[/tex].
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