730. Calculate the value of the total differential of the function $z=\operatorname{arcctg} \frac{x}{y}$ at $x=1, y=3, d x=0.01, d y=-0.05$.
Solution. We find the partial derivatives, then the partial differentials and the total differential of the given function:

$$
\frac{\partial z}{\partial x}=-\frac{y}{x^{2}+y^{2}} ; \quad \frac{\partial z}{\partial y}=\frac{x}{x^{2}+y^{2}} ; \quad d z=\frac{x d y-y d x}{x^{2}+y^{2}}
$$

Substituting the given values of the independent variables $x, y$, $d x$ and $d y$, which are functions of the total differential $d z$, we get

$$
d z=\frac{1 \cdot(-0.05)-3 \cdot 0.01}{1+9}=-0.008
$$