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Homework Help: 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:
    y=sqr(tan^-1(x))
    y=(tan^-1(x))^(1/2)

    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

    StatusX

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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.

    [tex]u=\tan^{-1}(x)[/tex]
    [tex]du=\frac{dx}{x^2+1}[/tex]

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

    [tex]y=\sqrt{u}=u^{\frac{1}{2}}[/tex]
    [tex]y'=\frac{1}{2\sqrt{u}}*du[/tex]

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

    VietDao29

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    The second line is wrong, Jameson.
    It should read:
    [tex]du=\frac{dx}{x^2+1}[/tex].
    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

    VietDao29

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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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