find all second partial derivatives of $\arctan(\frac{x+y}{1-xy})$

Question

Given $f(x,y)=\arctan(\frac{x+y}{1-xy})$

find:

$f_{xx}$ $f_{yy}$ $f_{xy}$ $f_{yx}$

I checked other questions but they are not helpful I need the answer, I prefer working backwards

Answer 1

The first partial derivatives simplify quite a bit:

$$\begin{align}f_x(x,y)&=[1+(\frac{x+y}{1-xy})^2]^{-1}[\frac{(1-xy)-(x+y)(-y)}{(1-xy)^2}]\\&=[1+(\frac{(x+y)^2}{(1-xy)^2})]^{-1}[\frac{1+y^2}{(1-xy)^2}]\\&=[(1-xy)^2+(x+y)^2]^{-1}[1+y^2]\\&=[1-2xy+x^2y^2+x^2+2xy+y^2]^{-1}[1+y^2]\\&=[y^2(x^2+1)+(x^2+1)]^{-1}[y^2+1]\\&=[(y^2+1)(x^2+1)]^{-1}[y^2+1]\\&=(x^2+1)^{-1}\end{align}$$

$$\begin{align}f_y(x,y)&=[1+(\frac{x+y}{1-xy})^2]^{-1}[\frac{(1-xy)-(x+y)(-x)}{(1-xy)^2}]\\&=[1+(\frac{(x+y)^2}{(1-xy)^2})]^{-1}[\frac{1+x^2}{(1-xy)^2}]\\&=[(1-xy)^2+(x+y)^2]^{-1}[1+x^2]\\&=[1-2xy+x^2y^2+x^2+2xy+y^2]^{-1}[1+x^2]\\&=[y^2(x^2+1)+(x^2+1)]^{-1}[x^2+1]\\&=[(y^2+1)(x^2+1)]^{-1}[x^2+1]\\&=(y^2+1)^{-1}\end{align}$$

Now I'll leave it to you for the second partial derivatives.

Answer 2

Hint...note that $$f(x,y)=\arctan x+\arctan y$$

Answer 3

Divide and conquer.

Let $g(x,y)=\dfrac{x+y}{1-xy}$; then $$ \frac{\partial g}{\partial x}= \frac{(1-xy)-(x+y)(-y)}{(1-xy)^2}=\frac{1+y^2}{(1-xy)^2} $$ By symmetry, $$ \frac{\partial g}{\partial y}=\frac{1+x^2}{(1-xy)^2} $$ Since $$ \frac{1}{1+g(x,y)^2}=\frac{(1-xy)^2}{(1-xy)^2+(x+y)^2}= \frac{(1-xy)^2}{(1+x^2)(1+y^2)} $$ we have, by the chain rule, $$ \frac{\partial f}{\partial x}=\frac{1+y^2}{(1+x^2)(1+y^2)} \qquad \frac{\partial f}{\partial y}=\frac{1+x^2}{(1+x^2)(1+y^2)} $$