Найти указанные частные производные сложной функции.

Решите задачу:

$z=arcsin \frac{u+v}{u^2+v^2} \\\\\frac{\partial z}{\partial u}= \frac{1}{\sqrt{1-(\frac{u+v}{u^2+v^2})^2}} \cdot \frac{(u^2+v^2)-(u+v)\cdot 2u}{(u^2+v^2)^2}= \frac{v^2-u^2-2uv}{(u^2+v^2)\sqrt{(u^2+v^2)^2-(u+v)^2}} \\\\\frac{\partial z}{\partial v}= \frac{1}{\sqrt{1-(\frac{u+v}{u^2+v^2})^2}} \cdot \frac{(u^2+v^2)-(u+v)\cdot 2v}{(u^2+v^2)^2} = \frac{u^2-v^2-2uv}{(u^2+v^2)\sqrt{(u^2+v^2)^2-(u+v)^2}}$

$\frac{\partial ^2z}{\partial u\partial v} = \frac{(2v-2u)(u^2+v^2)\sqrt{(u^2+v^2)^2-(u+v)^2}}{(u^2+v^2)^2((u^2+v^2)^2-(u+v)^2)}-\\\\-\frac{(v^2-u^2-2uv)\cdot [2v\sqrt{(u^2+v^2)^2-(u+v)^2}+ \frac{u^2+v^2}{2\sqrt{(u^2+v^2)^2-(u+v)^2}}\cdot (4v(u^2+v^2)-6(u+v))]}{(u^2+v^2)^2\cdot ((u^2+v^2)^2-(u+v)^2)}$