Найдите d^2*z второй дифференциал функции z=arctg(xy)^1/2. Решите плизз❤️

Question 1

Question 2

Решите задачу:

$z=arctg\sqrt{xy}\\\\d^2z=\frac{\partial ^2z}{\partial x^2}\cdot dx^2+2\cdot \frac{\partial ^2z}{\partial x\partial y}\cdot dx\cdot dy+\frac{\partial^2z}{\partial ^2y}\cdot dy^2\\\\z'_{x}=\frac{1}{1+xy}\cdot \frac{y}{2\sqrt{xy}}=\frac{1}{2(1+xy)}\cdot \sqrt{\frac{y}{x}}\\\\z''_{xx}=\frac{1}{2}\cdot ( \frac{-y}{(1+xy)^2} \cdot \sqrt{\frac{y}{x}}+\frac{1}{1+xy}\cdot \sqrt{\frac{x}{y}}\cdot \frac{-y}{2x^2})=\\\\=-\frac{y}{2\sqrt{x}(1+xy)}\cdot ( \frac{\sqrt{y}}{1+xy}+\frac{1}{2x\sqrt{y}})$

$z''_{xy}=\frac{1}{2}\cdot {\frac{x}{(1+xy)^2}\cdot \sqrt{\frac{y}{x}}+\frac{1}{1+xy}\cdot \frac{1}{2}\cdot \sqrt{\frac{x}{y}}\cdot \frac{1}{x})=$

$=\frac{1}{2(1+xy)}\cdot (\frac{\sqrt{xy}}{1+xy}+\frac{1}{\sqrt{xy}})$

$z'_{y}=\frac{1}{1+xy}\cdot \frac{x}{2\sqrt{xy}}=\frac{1}{2(1+xy)}\cdot \sqrt{\frac{x}{y}}\\\\z''_{yy}=\frac{1}{2}\cdot (\frac{-x}{(1+xy)^2}\cdot \frac{\sqrt{x}}{\sqrt{y}}+\frac{1}{1+xy}\cdot \frac{\sqrt{y}}{2\sqrt{x}}\cdot \frac{-x}{y^2})=\\\\=\frac{-x}{2\sqrt{y}(1+xy)}\cdot (\frac{\sqrt{x}}{1+xy}+\frac{1}{2y\sqrt{x}})$

$d^2z=-\frac{y}{2\sqrt{x}(1+xy)}\cdot (\frac{\sqrt{y}}{1+xy}+\frac{1}{2x\sqrt{y}})dx^2+\\\\+\frac{1}{2(1+xy)}\cdot (\frac{\sqrt{xy}}{1+xy}+\frac{1}{\sqrt{xy}})dxdy-\frac{x}{2\sqrt{y}(1+xy)}\cdot (\frac{\sqrt{x}}{1+xy}+\frac{1}{2y\sqrt{x}})dy^2$