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Is the derivative of arctan

  1. Aug 19, 2008 #1
    1/(x^2+1)
     
  2. jcsd
  3. Aug 19, 2008 #2

    nicksauce

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  4. Aug 19, 2008 #3
    ok thanks
     
  5. Aug 20, 2008 #4
    You can look up a list of derivatives and integrals of common functions on wikipedia or on google. But suppose you had to prove this by hand:

    Let

    [tex] y = arctan(x) \rightarrow tan(y) = x [/tex]

    Now use implicit differentiation to get

    [tex] \frac{y'}{cos^{2}(y)} = 1 \rightarrow y' = cos^{2}(y) = cos^{2}(arctan(x)) [/tex]

    Now draw your triangle to figure out what that expression becomes.

    Note that you would do that by saying opp/adj = x so let opp = x and adj = 1 which means hyp = sqrt(1 + x^2), now since we need cosine = adj/hyp = 1/sqrt(1 + x^2) however since we have cosine^2, that gets rid of the sqrt( ) and you are left with 1/(1 + x^2). Hope this helped
     
  6. Aug 20, 2008 #5
    Or note that as NoMoreExams had [tex]y = arctan(x) \Rightarrow tan(y) = x[/tex]. Now the derivative is [tex]sec^2(y)y' = 1 \Rightarrow y' = \frac{1}{sec^{2}(y)} = \frac{1}{tan^2(y)+1} = \frac{1}{x^2 + 1}[/tex].
     
  7. Aug 22, 2008 #6

    HallsofIvy

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    Strictly speaking it is 1/(x2+ 1)+ C
     
  8. Aug 22, 2008 #7
    Why do you need the constant for differentiation? Wouldn't that introduce the variable again when integrating back?
     
  9. Aug 22, 2008 #8

    nicksauce

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    I've never heard of the constant of differentiation.
     
  10. Aug 22, 2008 #9
    where there are no borders to the integral we are solving we use +C

    but when we use a derivative i dont think we use +C
     
  11. Aug 22, 2008 #10

    rbj

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    Halls, you're kidding, aren't you?
     
  12. Aug 22, 2008 #11
    thanks it turned out tan^-1=arctan
     
  13. Aug 22, 2008 #12

    mathwonk

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    this equivalent to the fact that the derivative of tan is sec^2 = 1 + tan^2.

    i.e. tan' = 1 + tan^2 implies the deriv of tan^-1 is 1/1+x^2.
     
  14. Aug 23, 2008 #13

    HallsofIvy

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    Ooops! You are right. I was thinking integration. There goes me trying to be a smarthmouth again!
     
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