I would like to show that the function:
$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$
is a $C^1$-function.
$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$
$ \frac{\partial f}{\partial y}(x,y)=-\frac{\partial f}{\partial y}(y,x)=... $
So I just have to show that:
$ \frac{\partial f}{\partial x}(x,y)\rightarrow_{(0,0)}0$
When $y\geq0$ :
$ -\frac{y^3}{6(x^2+y^2)}+\frac{x^2y}{x^2+y^2}-\frac{x^4y}{4!(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}$
When $y<0$ :
$ -\frac{y^3}{6(x^2+y^2)}+\frac{y^5}{5!(x^2+y^2)}+\frac{x^2y}{2(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}-\frac{yx^4}{4!(x^2+y^2)}$
So $ \frac{\sin(y)-y\cos(x)}{x^2+y^2}\rightarrow_{(0,0)}0 $
How can I directly find an upper bound of $ \left| \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2} \right|$ that tends to 0 ?