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D/dx of Inverse trig

  1. Jan 29, 2006 #1
    How do you derive a function where the trig is under a square root
    like this:

    i know that y=tan^-1u(x) => dy/dx=1/(1+u^2)*du/dx

    But how do I work it out? The book doesn't give the answers to pair numbers :(
  2. jcsd
  3. Jan 29, 2006 #2


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    chain rule?
  4. Jan 29, 2006 #3
    Yea that's what I was going to say.
  5. Jan 29, 2006 #4
    Yeah, chain rule.


    So using u-substitution you can write you function as this.


    I think you can make the neccessary substitution now.
    Last edited: Jan 29, 2006
  6. Jan 29, 2006 #5


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    The second line is wrong, Jameson.
    It should read:
    dx, not 1 :wink:.
    Have you worked out the problem, Chocolaty?
    Last edited: Jan 29, 2006
  7. Jan 29, 2006 #6
    Yes I have.
  8. Jan 29, 2006 #7


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    Ooops, sorry, I mean Chocolaty :biggrin:...
    Hope you don't mind. :rolleyes:
  9. Jan 29, 2006 #8
    lol That's alright.
  10. Jan 29, 2006 #9
    Tsk tsk. Sorry bout that. Fixed.
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