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Differentiation of tan^-1

  1. May 28, 2007 #1
    1. The problem statement, all variables and given/known data
    f(x) = tan^-1[(10000-200x)/(26x^2-2750x+77725)
    need to find f'(x)

    2. Relevant equations

    if f(x) = tan^-1 (x/a), then f'(x) = a/(a^2+ x^2)

    3. The attempt at a solution

    ok...the attempt im willing to do on my own, just needing help to get it in the form of x/a.
    preciate it. thx.
     
    Last edited: May 28, 2007
  2. jcsd
  3. May 28, 2007 #2

    matt grime

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    Why would you want to get it in that form? You know the chain rule, you know the derivative of tan^{-1}, and you can differentiate the expression inside your square brackets.
     
  4. May 28, 2007 #3

    Gib Z

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    Additional to what matt grime said, you need the quotient rule as well for the rational function that is the argument of the arctan.
     
  5. May 28, 2007 #4
    if f(x) = tan^-1[x/a], then f'(x) = [a/(a^2+x^2)]
    thats the only way i can think of going about it.
     
  6. May 29, 2007 #5

    danago

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    [tex]f(x)=tan^{-1}(x)[/tex]
    [tex]f'(x)=\frac{1}{1+x^2}[/tex]

    With that, you can easily do it with the chain rule.
     
  7. May 29, 2007 #6
    will do. thanks. = )
     
  8. May 30, 2007 #7

    NateTG

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    As an aside the following identity is occasionally handy:
    [tex]\frac{x}{1}=x[/tex]
    And brings your formula into line with danago's.
     
  9. Jun 3, 2007 #8

    HallsofIvy

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    What?? We have to memorize complicated identities like that??:frown:
     
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