Partial derivative of an arctan function

**daskywalker** · October 11th 2009, 07:05 AM

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

**Danny** · October 11th 2009, 07:13 AM

Originally Posted by daskywalker

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

Tour de-force

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

**Danny** · October 11th 2009, 07:19 AM

Originally Posted by daskywalker

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

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

**daskywalker** · October 11th 2009, 08:19 AM

Originally Posted by Danny

Tour de-force

$ \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)}

**Danny** · October 11th 2009, 08:38 AM

Originally Posted by daskywalker

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

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

**daskywalker** · October 11th 2009, 09:40 AM

Originally Posted by Danny

When you multiply the denominator through by $(1-xy)^2$ you get

$ (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!

Thread: Partial derivative of an arctan function