# Thread: Partial derivative of an arctan function

1. ## Partial derivative of an arctan function

arctan((x+y)/(1-xy))
just do it in respect to x and I can do the rest.

arctan((x+y)/(1-xy))
just do it in respect to x and I can do the rest.
Tour de-force

$\displaystyle \frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2}$

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

$\displaystyle \tan ^{-1} \frac{x+y}{1-xy} = \tan ^{-1} x + \tan ^{-1} y.$

4. Originally Posted by Danny
Tour de-force

$\displaystyle \frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2}$
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)}

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 $\displaystyle (1-xy)^2$ you get

$\displaystyle (1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2).$

6. Originally Posted by Danny
When you multiply the denominator through by $\displaystyle (1-xy)^2$ you get

$\displaystyle (1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2).$
ok got it now. many thanks!

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# first partials of F(x,y)=arctan[x y/1-xy]

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