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Math Help - Partial derivative of an arctan function

  1. #1
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    Partial derivative of an arctan function

    arctan((x+y)/(1-xy))
    just do it in respect to x and I can do the rest.
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  2. #2
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    Quote Originally Posted by daskywalker View Post
    arctan((x+y)/(1-xy))
    just do it in respect to x and I can do the rest.
    Tour de-force

     <br />
\frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2}<br />
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  3. #3
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    Quote Originally Posted by daskywalker View Post
    arctan((x+y)/(1-xy))
    just do it in respect to x and I can do the rest.
    Or using this identity that makes it easier

     <br />
\tan ^{-1} \frac{x+y}{1-xy} = \tan ^{-1} x + \tan ^{-1} y.<br />
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    Quote Originally Posted by Danny View Post
    Tour de-force

     <br />
\frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2}<br />
    forgive my stupidity but I don't exactly see the simplification.I got to the point where

    (y^2+1)/{y^2(1+x^2)+(1+x^2)}
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  5. #5
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    Quote Originally Posted by daskywalker View Post
    forgive my stupidity but I don't exactly see the simplification.I got to the point where

    (y^2+1)/{y^2(1+x^2)+(1+x^2)}
    When you multiply the denominator through by (1-xy)^2 you get

     <br />
(1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2).<br />
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  6. #6
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    Quote Originally Posted by Danny View Post
    When you multiply the denominator through by (1-xy)^2 you get

     <br />
(1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2).<br />
    ok got it now. many thanks!
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