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Math Help - help plz! tan^-1(s) = 1/1+s^2 how????

  1. #1
    Newbie
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    Oct 2007
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    help plz! tan^-1(s) = 1/1+s^2 how????

    I have a formula that I am trying to differentiate, but having lots of problems with it:
    r(s)= tan^-1 s, root of 2 /2 log(1+s^2), s - tan^-1 s)
    its derivates:
    r'(s) = ( ( 1+s^2)^-1, root of 2 s(1+s^2)^-1, 1- (1+s^2)-1

    = (1+s^2)^-1 (1, root of 2 s, s^2)

    I dont know how the derivative of tan^-1(s) = 1/1+s^2, can anyone explain?

    And cannot follow how all the derivatives work out, can anyone talk me through it? I have exam shortly and ally need to get to know these off by heart.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
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    Santiago, Chile
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    It's easy, first define y=\arctan x\implies\tan y=x.

    Now, differentiate implicity, and yields y'\sec^2y=1\implies y'=\frac1{1+\tan^2y}.

    And we're done.
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