# Partial derivative of an arctan function

• Oct 11th 2009, 07:05 AM
Partial derivative of an arctan function
arctan((x+y)/(1-xy))
just do it in respect to x and I can do the rest.
• Oct 11th 2009, 07:13 AM
Jester
Quote:

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

Tour de-force

$\displaystyle \frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2}$
• Oct 11th 2009, 07:19 AM
Jester
Quote:

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

$\displaystyle \tan ^{-1} \frac{x+y}{1-xy} = \tan ^{-1} x + \tan ^{-1} y.$
• Oct 11th 2009, 08:19 AM
Quote:

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)}
• Oct 11th 2009, 08:38 AM
Jester
Quote:

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

$\displaystyle (1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2).$
• Oct 11th 2009, 09:40 AM
When you multiply the denominator through by $\displaystyle (1-xy)^2$ you get
$\displaystyle (1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2).$