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Math Help - First three derivatives of arctan

  1. #1
    Member Jones's Avatar
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    First three derivatives of arctan

    Hi,

    I got stuck while trying to calculate the third derivative for arctan.

    The second derivative for arctan is \frac{-2x}{(1+x^2)^2}
    No problem. Then i try to rewrite it as:  -2x*(1+x^2)^{-2} and use the product rule. This is what i end up with:

     -2x*(1+x^2)^{-2}
     -2x*-2(1+x^2)^{-3} *2x - 2*(1+x^2)^{-2}

    which can be rearanged into: -4x^2 -\frac{2}{(1+x^2)^3} -\frac{2}{(1+x^2)^2}

    and it's wrong. What am i doing wrong?
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by Jones View Post
    Hi,

    I got stuck while trying to calculate the third derivative for arctan.

    The second derivative for arctan is \frac{-2x}{(1+x^2)^2}
    No problem. Then i try to rewrite it as:  -2x*(1+x^2)^{-2} and use the product rule. This is what i end up with:

     -2x*(1+x^2)^{-2}
     -2x*-2(1+x^2)^{-3} *2x - 2*(1+x^2)^{-2}

    which can be rearanged into: -4x^2 -\frac{2}{(1+x^2)^3} -\frac{2}{(1+x^2)^2}

    and it's wrong. What am i doing wrong?
    What you are doing wrong is not calculations, but brackets !!!!

    It's (-2x)*(-2)*(1+x^2)^{-3}*(2x)-2*(1+x^2)^{-2}

    And this gives (8x^2)*(1+x^2)^{-3}-2*(1+x^2)^{-2}=\frac{8x^2}{(1+x^2)^3}-\frac{2}{(1+x^2)^2}=\frac{8x^2-2(1+x^2)}{(1+x^2)^3}
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  3. #3
    MHF Contributor

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    Quote Originally Posted by Jones View Post
    Hi,

    I got stuck while trying to calculate the third derivative for arctan.

    The second derivative for arctan is \frac{-2x}{(1+x^2)^2}
    No problem. Then i try to rewrite it as:  -2x*(1+x^2)^{-2} and use the product rule. This is what i end up with:

     -2x*(1+x^2)^{-2}
     -2x*-2(1+x^2)^{-3} *2x - 2*(1+x^2)^{-2}

    which can be rearanged into: -4x^2 -\frac{2}{(1+x^2)^3} -\frac{2}{(1+x^2)^2}

    and it's wrong. What am i doing wrong?
    No, it can't be rearranged into that! It is
    \frac{2x^2}{(1+x^2)^3}- \frac{2}{(1+x^2)^2}
    which can be further written as \frac{2x^2}{(1+x^2)^3}- \frac{2(1+ x^2)}{(1+x^2)^3}= -\frac{2}{(1+x^2)^3}
    You seem to have misread -2x*-2(1+x^2)^{-3}*2x as
    (-2x- 2(1+x^2)^{-3})*2x, missing the first "*". It is, in fact, [/tex](-2x)(-2)(1+x^2)^{-3}(2x)[/tex]
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  4. #4
    Member Jones's Avatar
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    Ok, thanks guys. But are you sure thats correct?

    The key says \frac{-2+6x^2}{(1+x^2)^3}

    look:
    Arctan @ pici.se
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  5. #5
    Senior Member Twig's Avatar
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    Hi

    We are ok with the second derivative  f'' = \frac{-2x}{(1+x^{2})^{2}}

    Now, for the third derivative: Using quotient rule:

    \frac{(1+x^{2})^{2}\cdot (-2) + 2x\cdot (1+x^{2})\cdot 2x}{(1+x^{2})^{4}} Simplify by dividing out one  (1+x^{2}) term.

     \frac{-2\cdot (1+x^{2}) + 8x^{2}}{(1+x^{2})^{3}} = \frac{-2+6x^{2}}{(1+x^{2})^{3}}
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  6. #6
    Member Jones's Avatar
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    Of course, stupid me.
    It's the same as Moo showed, it just hadn't been simplified.
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